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Math operations from string using Javascript

I am trying to find a simple way to perform a set of javascript math operations without using eval() function. Example: 1+2x3x400+32/2+3 and it must follow the PEMDAS math principle. This is what I have, but it doesn't work exactly it should.
function mdas(equation) {
let operations = ["*", "/", "+", "-"];
for (let outerCount = 0; outerCount < operations.length; outerCount++) {
for (let innerCount = 0; innerCount < equation.length; ) {
if (equation[innerCount] == operations[outerCount]) {
let operationResult = runOperation(equation[innerCount - 1], operations[outerCount], equation[innerCount + 1]);
var leftSideOfEquation = equation.substr(0, equation.indexOf(innerCount - 1));
var rightSideOfEquation = equation.substr(equation.indexOf(innerCount), equation.length);
var rightSideOfEquation = rightSideOfEquation.replace(rightSideOfEquation[0],String(operationResult));
equation = leftSideOfEquation + rightSideOfEquation;
innerCount = 0;
}
else {
innerCount++;
}
}
}
return "Here is it: " + equation; //result of the equation
}

If you don't want to use a complete library like mathjs - and you don't want to tackle creating your own script which would involve: lexical analysis, tokenization, syntax analysis, recursive tree parsing, compiling and output...
the simplest banal suggestion: Function
const calc = s => Function(`return(${s})`)();
console.log( calc("1+2*3*400+32/2+3") ); // 2420
console.log( calc("-3*-2") ); // 6
console.log( calc("-3 * + 1") ); // -3
console.log( calc("-3 + -1") ); // -4
console.log( calc("2 * (3 + 1)") ); // 8
My take at a custom MDAS
Here I created a Regex to retrieve operands and operators, accounting for negative values: /(-?[\d.]+)([*\/+-])?/g.
Firstly we need to remove any whitespace from our string using str.replace(/ /g , "")
Using JavaScript's String.prototype.matchAll() we can get a 2D array with all the matches as [[fullMatch, operand, operator], [.. ] we can than further flatten it using Array.prototype.flat()
Having that flattened array, we can now filter it using Array.prototype.filter() to remove the fullMatch -es returned by the regular expression and remove the last undefined value.
Define a calc Object with the needed operation functions
Iterate over the MDAS groups */ and than +- as regular expressions /\/*/ and /+-/
Consume finally the array of matches until only one array key is left
let str = "-1+2 * 3*+400+-32 /2+3.1"; // 2386.1
str = str.replace(/ +/g, ""); // Remove all spaces!
// Get operands and operators as array.
// Remove full matches and undefined values.
const m = [...str.matchAll(/(-?[\d.]+)([*\/+-])?/g)].flat().filter((x, i) => x && i % 3);
const calc = {
"*": (a, b) => a * b,
"/": (a, b) => a / b,
"+": (a, b) => a + b,
"-": (a, b) => a - b,
};
// Iterate by MDAS groups order (first */ and than +-)
[/[*\/]/, /[+-]/].forEach(expr => {
for (let i = 0; i < m.length; i += 2) {
let [a, x, b] = [m[i], m[i + 1], m[i + 2]];
x = expr.exec(x);
if (!x) continue;
m[i] = calc[x.input](parseFloat(a), parseFloat(b)); // calculate and insert
m.splice(i + 1, 2); // remove operator and operand
i -= 2; // rewind loop
}
});
// Get the last standing result
console.log(m[0]); // 2386.1

It's a little hacky, but you can try something like this:
var eqs = [
'1+2*3*4+1+1+3',
'1+2*3*400+32/2+3',
'-5+2',
'3*-2',
];
for(var eq in eqs) { console.log(mdas(eqs[eq])); }
function mdas(equation) {
console.log(equation);
var failsafe = 100;
var num = '(((?<=[*+-])-|^-)?[0-9.]+)';
var reg = new RegExp(num + '([*/])' + num);
while(m = reg.exec(equation)) {
var n = (m[3] == "*") ? m[1]*m[4] : m[1]/m[4];
equation = equation.replace(m[0], n);
if(failsafe--<0) { return 'failsafe'; }
}
var reg = new RegExp(num + '([+-])' + num);
while(m = reg.exec(equation)) {
var n = (m[3] == "+") ? 1*m[1] + 1*m[4] : m[1]-m[4];
equation = equation.replace(m[0], n);
if(failsafe--<0) { return 'failsafe'; }
}
return equation;
}

Alternative way to padStart [duplicate]

I am in need of a JavaScript function which can take a value and pad it to a given length (I need spaces, but anything would do). I found this, but I have no idea what the heck it is doing and it doesn't seem to work for me.
String.prototype.pad = function(l, s, t) {
return s || (s = " "),
(l -= this.length) > 0 ?
(s = new Array(Math.ceil(l / s.length) + 1).join(s))
.substr(0, t = !t ? l : t == 1 ?
0 :
Math.ceil(l / 2)) + this + s.substr(0, l - t) :
this;
};
var s = "Jonas";
document.write(
'<h2>S = '.bold(), s, "</h2>",
'S.pad(20, "[]", 0) = '.bold(), s.pad(20, "[]", 0), "<br />",
'S.pad(20, "[====]", 1) = '.bold(), s.pad(20, "[====]", 1), "<br />",
'S.pad(20, "~", 2) = '.bold(), s.pad(20, "~", 2)
);

ECMAScript 2017 (ES8) added String.padStart (along with String.padEnd) for just this purpose:
"Jonas".padStart(10); // Default pad string is a space
"42".padStart(6, "0"); // Pad with "0"
"*".padStart(8, "-/|\\"); // produces '-/|\\-/|*'
If not present in the JavaScript host, String.padStart can be added as a polyfill.
Pre ES8
I found this solution here and this is for me much much simpler:
var n = 123
String("00000" + n).slice(-5); // returns 00123
("00000" + n).slice(-5); // returns 00123
(" " + n).slice(-5); // returns " 123" (with two spaces)
And here I made an extension to the string object:
String.prototype.paddingLeft = function (paddingValue) {
return String(paddingValue + this).slice(-paddingValue.length);
};
An example to use it:
function getFormattedTime(date) {
var hours = date.getHours();
var minutes = date.getMinutes();
hours = hours.toString().paddingLeft("00");
minutes = minutes.toString().paddingLeft("00");
return "{0}:{1}".format(hours, minutes);
};
String.prototype.format = function () {
var args = arguments;
return this.replace(/{(\d+)}/g, function (match, number) {
return typeof args[number] != 'undefined' ? args[number] : match;
});
};
This will return a time in the format "15:30".

A faster method
If you are doing this repeatedly, for example to pad values in an array, and performance is a factor, the following approach can give you nearly a 100x advantage in speed (jsPerf) over other solution that are currently discussed on the inter webs. The basic idea is that you are providing the pad function with a fully padded empty string to use as a buffer. The pad function just appends to string to be added to this pre-padded string (one string concat) and then slices or trims the result to the desired length.
function pad(pad, str, padLeft) {
if (typeof str === 'undefined')
return pad;
if (padLeft) {
return (pad + str).slice(-pad.length);
} else {
return (str + pad).substring(0, pad.length);
}
}
For example, to zero pad a number to a length of 10 digits,
pad('0000000000',123,true);
To pad a string with whitespace, so the entire string is 255 characters,
var padding = Array(256).join(' '), // make a string of 255 spaces
pad(padding,123,true);
Performance Test
See the jsPerf test here.
And this is faster than ES6 string.repeat by 2x as well, as shown by the revised JsPerf here
Please note that jsPerf is no longer online
Please note that the jsPerf site that we originally used to benchmark the various methods is no longer online. Unfortunately, this means we can't get to those test results. Sad but true.

String.prototype.padStart() and String.prototype.padEnd() are currently TC39 candidate proposals: see github.com/tc39/proposal-string-pad-start-end (only available in Firefox as of April 2016; a polyfill is available).

http://www.webtoolkit.info/javascript_pad.html
/**
*
* JavaScript string pad
* http://www.webtoolkit.info/
*
**/
var STR_PAD_LEFT = 1;
var STR_PAD_RIGHT = 2;
var STR_PAD_BOTH = 3;
function pad(str, len, pad, dir) {
if (typeof(len) == "undefined") { var len = 0; }
if (typeof(pad) == "undefined") { var pad = ' '; }
if (typeof(dir) == "undefined") { var dir = STR_PAD_RIGHT; }
if (len + 1 >= str.length) {
switch (dir){
case STR_PAD_LEFT:
str = Array(len + 1 - str.length).join(pad) + str;
break;
case STR_PAD_BOTH:
var padlen = len - str.length;
var right = Math.ceil( padlen / 2 );
var left = padlen - right;
str = Array(left+1).join(pad) + str + Array(right+1).join(pad);
break;
default:
str = str + Array(len + 1 - str.length).join(pad);
break;
} // switch
}
return str;
}
It's a lot more readable.

Here's a recursive approach to it.
function pad(width, string, padding) {
return (width <= string.length) ? string : pad(width, padding + string, padding)
}
An example...
pad(5, 'hi', '0')
=> "000hi"

ECMAScript 2017 adds a padStart method to the String prototype. This method will pad a string with spaces to a given length. This method also takes an optional string that will be used instead of spaces for padding.
'abc'.padStart(10); // " abc"
'abc'.padStart(10, "foo"); // "foofoofabc"
'abc'.padStart(6,"123465"); // "123abc"
'abc'.padStart(8, "0"); // "00000abc"
'abc'.padStart(1); // "abc"
A padEnd method was also added that works in the same manner.
For browser compatibility (and a useful polyfill) see this link.

Using the ECMAScript 6 method String#repeat, a pad function is as simple as:
String.prototype.padLeft = function(char, length) {
return char.repeat(Math.max(0, length - this.length)) + this;
}
String#repeat is currently supported in Firefox and Chrome only. for other implementation, one might consider the following simple polyfill:
String.prototype.repeat = String.prototype.repeat || function(n){
return n<=1 ? this : (this + this.repeat(n-1));
}

Using the ECMAScript 6 method String#repeat and Arrow functions, a pad function is as simple as:
var leftPad = (s, c, n) => c.repeat(n - s.length) + s;
leftPad("foo", "0", 5); //returns "00foo"
jsfiddle
edit:
suggestion from the comments:
const leftPad = (s, c, n) => n - s.length > 0 ? c.repeat(n - s.length) + s : s;
this way, it wont throw an error when s.lengthis greater than n
edit2:
suggestion from the comments:
const leftPad = (s, c, n) =>{ s = s.toString(); c = c.toString(); return s.length > n ? s : c.repeat(n - s.length) + s; }
this way, you can use the function for strings and non-strings alike.

The key trick in both those solutions is to create an array instance with a given size (one more than the desired length), and then to immediately call the join() method to make a string. The join() method is passed the padding string (spaces probably). Since the array is empty, the empty cells will be rendered as empty strings during the process of joining the array into one result string, and only the padding will remain. It's a really nice technique.

With ES8, there are two options for padding.
You can check them in the documentation.
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/padEnd
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/padStart

Taking up Samuel's ideas, upward here. And remember an old SQL script, I tried with this:
a=1234;
'0000'.slice(a.toString().length)+a;
It works in all the cases I could imagine:
a= 1 result 0001
a= 12 result 0012
a= 123 result 0123
a= 1234 result 1234
a= 12345 result 12345
a= '12' result 0012

Pad with default values
I noticed that I mostly need the padLeft for time conversion / number padding.
So I wrote this function:
function padL(a, b, c) { // string/number, length=2, char=0
return (new Array(b || 2).join(c || 0) + a).slice(-b)
}
This simple function supports Number or String as input.
The default pad is two characters.
The default char is 0.
So I can simply write:
padL(1);
// 01
If I add the second argument (pad width):
padL(1, 3);
// 001
The third parameter (pad character)
padL('zzz', 10, 'x');
// xxxxxxxzzz
#BananaAcid: If you pass a undefined value or a 0 length string, you get 0undefined, so:
As suggested
function padL(a, b, c) { // string/number, length=2, char=0
return (new Array((b || 1) + 1).join(c || 0) + (a || '')).slice(-(b || 2))
}
But this can also be achieved in a shorter way.
function padL(a, b, c) { // string/number, length=2, char=0
return (new Array(b || 2).join(c || 0) + (a || c || 0)).slice(-b)
}
It also works with:
padL(0)
padL(NaN)
padL('')
padL(undefined)
padL(false)
And if you want to be able to pad in both ways:
function pad(a, b, c, d) { // string/number, length=2, char=0, 0/false=Left-1/true=Right
return a = (a || c || 0), c = new Array(b || 2).join(c || 0), d ? (a + c).slice(0, b) : (c + a).slice(-b)
}
which can be written in a shorter way without using slice.
function pad(a, b, c, d) {
return a = (a || c || 0) + '', b = new Array((++b || 3) - a.length).join(c || 0), d ? a+b : b+a
}
/*
Usage:
pad(
input // (int or string) or undefined, NaN, false, empty string
// default:0 or PadCharacter
// Optional
,PadLength // (int) default:2
,PadCharacter // (string or int) default:'0'
,PadDirection // (bolean) default:0 (padLeft) - (true or 1) is padRight
)
*/
Now if you try to pad 'averylongword' with 2... that’s not my problem.
I said that I would give you a tip.
Most of the time, if you pad, you do it for the same value N times.
Using any type of function inside a loop slows down the loop!!!
So if you just want to pad left some numbers inside a long list, don't use functions to do this simple thing.
Use something like this:
var arrayOfNumbers = [1, 2, 3, 4, 5, 6, 7],
paddedArray = [],
len = arrayOfNumbers.length;
while(len--) {
paddedArray[len] = ('0000' + arrayOfNumbers[len]).slice(-4);
}
If you don't know how the maximum padding size based on the numbers inside the array.
var arrayOfNumbers = [1, 2, 3, 4, 5, 6, 7, 49095],
paddedArray = [],
len = arrayOfNumbers.length;
// Search the highest number
var arrayMax = Function.prototype.apply.bind(Math.max, null),
// Get that string length
padSize = (arrayMax(arrayOfNumbers) + '').length,
// Create a Padding string
padStr = new Array(padSize).join(0);
// And after you have all this static values cached start the loop.
while(len--) {
paddedArray[len] = (padStr + arrayOfNumbers[len]).slice(-padSize); // substr(-padSize)
}
console.log(paddedArray);
/*
0: "00001"
1: "00002"
2: "00003"
3: "00004"
4: "00005"
5: "00006"
6: "00007"
7: "49095"
*/

padding string has been inplemented in new javascript version.
str.padStart(targetLength [, padString])
https://developer.mozilla.org/es/docs/Web/JavaScript/Referencia/Objetos_globales/String/padStart
If you want your own function check this example:
const myString = 'Welcome to my house';
String.prototype.padLeft = function(times = 0, str = ' ') {
return (Array(times).join(str) + this);
}
console.log(myString.padLeft(12, ':'));
//:::::::::::Welcome to my house

Here is a build in method you can use -
str1.padStart(2, '0')

Here's a simple function that I use.
var pad=function(num,field){
var n = '' + num;
var w = n.length;
var l = field.length;
var pad = w < l ? l-w : 0;
return field.substr(0,pad) + n;
};
For example:
pad (20,' '); // 20
pad (321,' '); // 321
pad (12345,' '); //12345
pad ( 15,'00000'); //00015
pad ( 999,'*****'); //**999
pad ('cat','_____'); //__cat

A short way:
(x=>(new Array(int-x.length+1)).join(char)+x)(String)
Example:
(x=>(new Array(6-x.length+1)).join("0")+x)("1234")
return: "001234"

Here is a simple answer in basically one line of code.
var value = 35 // the numerical value
var x = 5 // the minimum length of the string
var padded = ("00000" + value).substr(-x);
Make sure the number of characters in you padding, zeros here, is at least as many as your intended minimum length. So really, to put it into one line, to get a result of "00035" in this case is:
var padded = ("00000" + 35).substr(-5);

ES7 is just drafts and proposals right now, but if you wanted to track compatibility with the specification, your pad functions need:
Multi-character pad support.
Don't truncate the input string
Pad defaults to space
From my polyfill library, but apply your own due diligence for prototype extensions.
// Tests
'hello'.lpad(4) === 'hello'
'hello'.rpad(4) === 'hello'
'hello'.lpad(10) === ' hello'
'hello'.rpad(10) === 'hello '
'hello'.lpad(10, '1234') === '41234hello'
'hello'.rpad(10, '1234') === 'hello12341'
String.prototype.lpad || (String.prototype.lpad = function(length, pad)
{
if(length < this.length)
return this;
pad = pad || ' ';
let str = this;
while(str.length < length)
{
str = pad + str;
}
return str.substr( -length );
});
String.prototype.rpad || (String.prototype.rpad = function(length, pad)
{
if(length < this.length)
return this;
pad = pad || ' ';
let str = this;
while(str.length < length)
{
str += pad;
}
return str.substr(0, length);
});

Array manipulations are really slow compared to simple string concat. Of course, benchmark for your use case.
function(string, length, pad_char, append) {
string = string.toString();
length = parseInt(length) || 1;
pad_char = pad_char || ' ';
while (string.length < length) {
string = append ? string+pad_char : pad_char+string;
}
return string;
};

A variant of #Daniel LaFavers' answer.
var mask = function (background, foreground) {
bg = (new String(background));
fg = (new String(foreground));
bgl = bg.length;
fgl = fg.length;
bgs = bg.substring(0, Math.max(0, bgl - fgl));
fgs = fg.substring(Math.max(0, fgl - bgl));
return bgs + fgs;
};
For example:
mask('00000', 11 ); // '00011'
mask('00011','00' ); // '00000'
mask( 2 , 3 ); // '3'
mask('0' ,'111'); // '1'
mask('fork' ,'***'); // 'f***'
mask('_____','dog'); // '__dog'

If you don't mind including a utility library, lodash library has _.pad, _.padLeft and _.padRight functions.

I think its better to avoid recursion because its costly.
function padLeft(str,size,padwith) {
if(size <= str.length) {
// not padding is required.
return str;
} else {
// 1- take array of size equal to number of padding char + 1. suppose if string is 55 and we want 00055 it means we have 3 padding char so array size should be 3 + 1 (+1 will explain below)
// 2- now join this array with provided padding char (padwith) or default one ('0'). so it will produce '000'
// 3- now append '000' with orginal string (str = 55), will produce 00055
// why +1 in size of array?
// it is a trick, that we are joining an array of empty element with '0' (in our case)
// if we want to join items with '0' then we should have at least 2 items in the array to get joined (array with single item doesn't need to get joined).
// <item>0<item>0<item>0<item> to get 3 zero we need 4 (3+1) items in array
return Array(size-str.length+1).join(padwith||'0')+str
}
}
alert(padLeft("59",5) + "\n" +
padLeft("659",5) + "\n" +
padLeft("5919",5) + "\n" +
padLeft("59879",5) + "\n" +
padLeft("5437899",5));

It's 2014, and I suggest a JavaScript string-padding function. Ha!
Bare-bones: right-pad with spaces
function pad (str, length) {
var padding = (new Array(Math.max(length - str.length + 1, 0))).join(" ");
return str + padding;
}
Fancy: pad with options
/**
* #param {*} str Input string, or any other type (will be converted to string)
* #param {number} length Desired length to pad the string to
* #param {Object} [opts]
* #param {string} [opts.padWith=" "] Character to use for padding
* #param {boolean} [opts.padLeft=false] Whether to pad on the left
* #param {boolean} [opts.collapseEmpty=false] Whether to return an empty string if the input was empty
* #returns {string}
*/
function pad(str, length, opts) {
var padding = (new Array(Math.max(length - (str + "").length + 1, 0))).join(opts && opts.padWith || " "),
collapse = opts && opts.collapseEmpty && !(str + "").length;
return collapse ? "" : opts && opts.padLeft ? padding + str : str + padding;
}
Usage (fancy):
pad("123", 5);
// Returns "123 "
pad(123, 5);
// Returns "123 " - non-string input
pad("123", 5, { padWith: "0", padLeft: true });
// Returns "00123"
pad("", 5);
// Returns " "
pad("", 5, { collapseEmpty: true });
// Returns ""
pad("1234567", 5);
// Returns "1234567"

/**************************************************************************************************
Pad a string to pad_length fillig it with pad_char.
By default the function performs a left pad, unless pad_right is set to true.
If the value of pad_length is negative, less than, or equal to the length of the input string, no padding takes place.
**************************************************************************************************/
if(!String.prototype.pad)
String.prototype.pad = function(pad_char, pad_length, pad_right)
{
var result = this;
if( (typeof pad_char === 'string') && (pad_char.length === 1) && (pad_length > this.length) )
{
var padding = new Array(pad_length - this.length + 1).join(pad_char); //thanks to http://stackoverflow.com/questions/202605/repeat-string-javascript/2433358#2433358
result = (pad_right ? result + padding : padding + result);
}
return result;
}
And then you can do:
alert( "3".pad("0", 3) ); //shows "003"
alert( "hi".pad(" ", 3) ); //shows " hi"
alert( "hi".pad(" ", 3, true) ); //shows "hi "

If you just want a very simple hacky one-liner to pad, just make a string of the desired padding character of the desired max padding length and then substring it to the length of what you want to pad.
Example: padding the string store in e with spaces to 25 characters long.
var e = "hello"; e = e + " ".substring(e.length)
Result: "hello "
If you want to do the same with a number as input just call .toString() on it before.

A friend asked about using a JavaScript function to pad left. It turned into a little bit of an endeavor between some of us in chat to code golf it. This was the result:
function l(p,t,v){
v+="";return v.length>=t?v:l(p,t,p+v);
}
It ensures that the value to be padded is a string, and then if it isn't the length of the total desired length it will pad it once and then recurse. Here is what it looks like with more logical naming and structure
function padLeft(pad, totalLength, value){
value = value.toString();
if( value.length >= totalLength ){
return value;
}else{
return padLeft(pad, totalLength, pad + value);
}
}
The example we were using was to ensure that numbers were padded with 0 to the left to make a max length of 6. Here is an example set:
function l(p,t,v){v+="";return v.length>=t?v:l(p,t,p+v);}
var vals = [6451,123,466750];
var pad = l(0,6,vals[0]);// pad with 0's, max length 6
var pads = vals.map(function(i){ return l(0,6,i) });
document.write(pads.join("<br />"));

A little late, but thought I might share anyway. I found it useful to add a prototype extension to Object. That way I can pad numbers and strings, left or right. I have a module with similar utilities I include in my scripts.
// include the module in your script, there is no need to export
var jsAddOns = require('<path to module>/jsAddOns');
~~~~~~~~~~~~ jsAddOns.js ~~~~~~~~~~~~
/*
* method prototype for any Object to pad it's toString()
* representation with additional characters to the specified length
*
* #param padToLength required int
* entire length of padded string (original + padding)
* #param padChar optional char
* character to use for padding, default is white space
* #param padLeft optional boolean
* if true padding added to left
* if omitted or false, padding added to right
*
* #return padded string or
* original string if length is >= padToLength
*/
Object.prototype.pad = function(padToLength, padChar, padLeft) {
// get the string value
s = this.toString()
// default padToLength to 0
// if omitted, original string is returned
padToLength = padToLength || 0;
// default padChar to empty space
padChar = padChar || ' ';
// ignore padding if string too long
if (s.length >= padToLength) {
return s;
}
// create the pad of appropriate length
var pad = Array(padToLength - s.length).join(padChar);
// add pad to right or left side
if (padLeft) {
return pad + s;
} else {
return s + pad;
}
};

Never insert data somewhere (especially not at beginning, like str = pad + str;), since the data will be reallocated everytime. Append always at end!
Don't pad your string in the loop. Leave it alone and build your pad string first. In the end concatenate it with your main string.
Don't assign padding string each time (like str += pad;). It is much faster to append the padding string to itself and extract first x-chars (the parser can do this efficiently if you extract from first char). This is exponential growth, which means that it wastes some memory temporarily (you should not do this with extremely huge texts).
if (!String.prototype.lpad) {
String.prototype.lpad = function(pad, len) {
while (pad.length < len) {
pad += pad;
}
return pad.substr(0, len-this.length) + this;
}
}
if (!String.prototype.rpad) {
String.prototype.rpad = function(pad, len) {
while (pad.length < len) {
pad += pad;
}
return this + pad.substr(0, len-this.length);
}
}

Here is a JavaScript function that adds a specified number of paddings with a custom symbol. The function takes three parameters.
padMe --> string or number to left pad
pads --> number of pads
padSymble --> custom symbol, default is "0"
function leftPad(padMe, pads, padSymble) {
if(typeof padMe === "undefined") {
padMe = "";
}
if (typeof pads === "undefined") {
pads = 0;
}
if (typeof padSymble === "undefined") {
padSymble = "0";
}
var symble = "";
var result = [];
for(var i=0; i < pads; i++) {
symble += padSymble;
}
var length = symble.length - padMe.toString().length;
result = symble.substring(0, length);
return result.concat(padMe.toString());
}
Here are some results:
> leftPad(1)
"1"
> leftPad(1, 4)
"0001"
> leftPad(1, 4, "0")
"0001"
> leftPad(1, 4, "#")
"###1"

Yet another take at with combination of a couple of solutions:
/**
* pad string on left
* #param {number} number of digits to pad, default is 2
* #param {string} string to use for padding, default is '0' *
* #returns {string} padded string
*/
String.prototype.paddingLeft = function (b, c) {
if (this.length > (b||2))
return this + '';
return (this || c || 0) + '', b = new Array((++b || 3) - this.length).join(c || 0), b + this
};
/**
* pad string on right
* #param {number} number of digits to pad, default is 2
* #param {string} string to use for padding, default is '0' *
* #returns {string} padded string
*/
String.prototype.paddingRight = function (b, c) {
if (this.length > (b||2))
return this + '';
return (this||c||0) + '', b = new Array((++b || 3) - this.length).join(c || 0), this + b
};

print integer in a string javascript [duplicate]

I'm looking for a good JavaScript equivalent of the C/PHP printf() or for C#/Java programmers, String.Format() (IFormatProvider for .NET).
My basic requirement is a thousand separator format for numbers for now, but something that handles lots of combinations (including dates) would be good.
I realize Microsoft's Ajax library provides a version of String.Format(), but we don't want the entire overhead of that framework.

Current JavaScript
From ES6 on you could use template strings:
let soMany = 10;
console.log(`This is ${soMany} times easier!`);
// "This is 10 times easier!"
See Kim's answer below for details.
Older answer
Try sprintf() for JavaScript.
If you really want to do a simple format method on your own, don’t do the replacements successively but do them simultaneously.
Because most of the other proposals that are mentioned fail when a replace string of previous replacement does also contain a format sequence like this:
"{0}{1}".format("{1}", "{0}")
Normally you would expect the output to be {1}{0} but the actual output is {1}{1}. So do a simultaneous replacement instead like in fearphage’s suggestion.

Building on the previously suggested solutions:
// First, checks if it isn't implemented yet.
if (!String.prototype.format) {
String.prototype.format = function() {
var args = arguments;
return this.replace(/{(\d+)}/g, function(match, number) {
return typeof args[number] != 'undefined'
? args[number]
: match
;
});
};
}
"{0} is dead, but {1} is alive! {0} {2}".format("ASP", "ASP.NET")
outputs
ASP is dead, but ASP.NET is alive! ASP {2}
If you prefer not to modify String's prototype:
if (!String.format) {
String.format = function(format) {
var args = Array.prototype.slice.call(arguments, 1);
return format.replace(/{(\d+)}/g, function(match, number) {
return typeof args[number] != 'undefined'
? args[number]
: match
;
});
};
}
Gives you the much more familiar:
String.format('{0} is dead, but {1} is alive! {0} {2}', 'ASP', 'ASP.NET');
with the same result:
ASP is dead, but ASP.NET is alive! ASP {2}

It's funny because Stack Overflow actually has their own formatting function for the String prototype called formatUnicorn. Try it! Go into the console and type something like:
"Hello, {name}, are you feeling {adjective}?".formatUnicorn({name:"Gabriel", adjective: "OK"});
You get this output:
Hello, Gabriel, are you feeling OK?
You can use objects, arrays, and strings as arguments! I got its code and reworked it to produce a new version of String.prototype.format:
String.prototype.formatUnicorn = String.prototype.formatUnicorn ||
function () {
"use strict";
var str = this.toString();
if (arguments.length) {
var t = typeof arguments[0];
var key;
var args = ("string" === t || "number" === t) ?
Array.prototype.slice.call(arguments)
: arguments[0];
for (key in args) {
str = str.replace(new RegExp("\\{" + key + "\\}", "gi"), args[key]);
}
}
return str;
};
Note the clever Array.prototype.slice.call(arguments) call -- that means if you throw in arguments that are strings or numbers, not a single JSON-style object, you get C#'s String.Format behavior almost exactly.
"a{0}bcd{1}ef".formatUnicorn("FOO", "BAR"); // yields "aFOObcdBARef"
That's because Array's slice will force whatever's in arguments into an Array, whether it was originally or not, and the key will be the index (0, 1, 2...) of each array element coerced into a string (eg, "0", so "\\{0\\}" for your first regexp pattern).
Neat.

Number Formatting in JavaScript
I got to this question page hoping to find how to format numbers in JavaScript, without introducing yet another library. Here's what I've found:
Rounding floating-point numbers
The equivalent of sprintf("%.2f", num) in JavaScript seems to be num.toFixed(2), which formats num to 2 decimal places, with rounding (but see #ars265's comment about Math.round below).
(12.345).toFixed(2); // returns "12.35" (rounding!)
(12.3).toFixed(2); // returns "12.30" (zero padding)
Exponential form
The equivalent of sprintf("%.2e", num) is num.toExponential(2).
(33333).toExponential(2); // "3.33e+4"
Hexadecimal and other bases
To print numbers in base B, try num.toString(B). JavaScript supports automatic conversion to and from bases 2 through 36 (in addition, some browsers have limited support for base64 encoding).
(3735928559).toString(16); // to base 16: "deadbeef"
parseInt("deadbeef", 16); // from base 16: 3735928559
Reference Pages
Quick tutorial on JS number formatting
Mozilla reference page for toFixed() (with links to toPrecision(), toExponential(), toLocaleString(), ...)

From ES6 on you could use template strings:
let soMany = 10;
console.log(`This is ${soMany} times easier!`);
// "This is 10 times easier!"
Be aware that template strings are surrounded by backticks ` instead of (single) quotes.
For further information:
https://developers.google.com/web/updates/2015/01/ES6-Template-Strings
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/template_strings
Note:
Check the mozilla-site to find a list of supported browsers.

jsxt, Zippo
This option fits better.
String.prototype.format = function() {
var formatted = this;
for (var i = 0; i < arguments.length; i++) {
var regexp = new RegExp('\\{'+i+'\\}', 'gi');
formatted = formatted.replace(regexp, arguments[i]);
}
return formatted;
};
With this option I can replace strings like these:
'The {0} is dead. Don\'t code {0}. Code {1} that is open source!'.format('ASP', 'PHP');
With your code the second {0} wouldn't be replaced. ;)

I use this simple function:
String.prototype.format = function() {
var formatted = this;
for( var arg in arguments ) {
formatted = formatted.replace("{" + arg + "}", arguments[arg]);
}
return formatted;
};
That's very similar to string.format:
"{0} is dead, but {1} is alive!".format("ASP", "ASP.NET")

For Node.js users there is util.format which has printf-like functionality:
util.format("%s world", "Hello")

I'm surprised no one used reduce, this is a native concise and powerful JavaScript function.
ES6 (EcmaScript2015)
String.prototype.format = function() {
return [...arguments].reduce((p,c) => p.replace(/%s/,c), this);
};
console.log('Is that a %s or a %s?... No, it\'s %s!'.format('plane', 'bird', 'SOman'));
< ES6
function interpolate(theString, argumentArray) {
var regex = /%s/;
var _r=function(p,c){return p.replace(regex,c);}
return argumentArray.reduce(_r, theString);
}
interpolate("%s, %s and %s", ["Me", "myself", "I"]); // "Me, myself and I"
How it works:
reduce applies a function against an accumulator and each element in the array (from left to right) to reduce it to a single value.
var _r= function(p,c){return p.replace(/%s/,c)};
console.log(
["a", "b", "c"].reduce(_r, "[%s], [%s] and [%s]") + '\n',
[1, 2, 3].reduce(_r, "%s+%s=%s") + '\n',
["cool", 1337, "stuff"].reduce(_r, "%s %s %s")
);

Here's a minimal implementation of sprintf in JavaScript: it only does "%s" and "%d", but I have left space for it to be extended. It is useless to the OP, but other people who stumble across this thread coming from Google might benefit from it.
function sprintf() {
var args = arguments,
string = args[0],
i = 1;
return string.replace(/%((%)|s|d)/g, function (m) {
// m is the matched format, e.g. %s, %d
var val = null;
if (m[2]) {
val = m[2];
} else {
val = args[i];
// A switch statement so that the formatter can be extended. Default is %s
switch (m) {
case '%d':
val = parseFloat(val);
if (isNaN(val)) {
val = 0;
}
break;
}
i++;
}
return val;
});
}
Example:
alert(sprintf('Latitude: %s, Longitude: %s, Count: %d', 41.847, -87.661, 'two'));
// Expected output: Latitude: 41.847, Longitude: -87.661, Count: 0
In contrast with similar solutions in previous replies, this one does all substitutions in one go, so it will not replace parts of previously replaced values.

3 different ways to format javascript string
There are 3 different ways to format a string by replacing placeholders with the variable value.
Using template literal (backticks ``)
let name = 'John';
let age = 30;
// using backticks
console.log(`${name} is ${age} years old.`);
// John is 30 years old.
Using concatenation
let name = 'John';
let age = 30;
// using concatenation
console.log(name + ' is ' + age + ' years old.');
// John is 30 years old.
Creating own format function
String.prototype.format = function () {
var args = arguments;
return this.replace(/{([0-9]+)}/g, function (match, index) {
// check if the argument is there
return typeof args[index] == 'undefined' ? match : args[index];
});
};
console.log('{0} is {1} years old.'.format('John', 30));

JavaScript programmers can use String.prototype.sprintf at https://github.com/ildar-shaimordanov/jsxt/blob/master/js/String.js. Below is example:
var d = new Date();
var dateStr = '%02d:%02d:%02d'.sprintf(
d.getHours(),
d.getMinutes(),
d.getSeconds());

I want to share my solution for the 'problem'. I haven't re-invented the wheel but tries to find a solution based on what JavaScript already does. The advantage is, that you get all implicit conversions for free. Setting the prototype property $ of String gives a very nice and compact syntax (see examples below). It is maybe not the most efficient way, but in most cases dealing with output it does not have to be super optimized.
String.form = function(str, arr) {
var i = -1;
function callback(exp, p0, p1, p2, p3, p4) {
if (exp=='%%') return '%';
if (arr[++i]===undefined) return undefined;
exp = p2 ? parseInt(p2.substr(1)) : undefined;
var base = p3 ? parseInt(p3.substr(1)) : undefined;
var val;
switch (p4) {
case 's': val = arr[i]; break;
case 'c': val = arr[i][0]; break;
case 'f': val = parseFloat(arr[i]).toFixed(exp); break;
case 'p': val = parseFloat(arr[i]).toPrecision(exp); break;
case 'e': val = parseFloat(arr[i]).toExponential(exp); break;
case 'x': val = parseInt(arr[i]).toString(base?base:16); break;
case 'd': val = parseFloat(parseInt(arr[i], base?base:10).toPrecision(exp)).toFixed(0); break;
}
val = typeof(val)=='object' ? JSON.stringify(val) : val.toString(base);
var sz = parseInt(p1); /* padding size */
var ch = p1 && p1[0]=='0' ? '0' : ' '; /* isnull? */
while (val.length<sz) val = p0 !== undefined ? val+ch : ch+val; /* isminus? */
return val;
}
var regex = /%(-)?(0?[0-9]+)?([.][0-9]+)?([#][0-9]+)?([scfpexd%])/g;
return str.replace(regex, callback);
}
String.prototype.$ = function() {
return String.form(this, Array.prototype.slice.call(arguments));
}
Here are a few examples:
String.format("%s %s", [ "This is a string", 11 ])
console.log("%s %s".$("This is a string", 11))
var arr = [ "12.3", 13.6 ]; console.log("Array: %s".$(arr));
var obj = { test:"test", id:12 }; console.log("Object: %s".$(obj));
console.log("%c", "Test");
console.log("%5d".$(12)); // ' 12'
console.log("%05d".$(12)); // '00012'
console.log("%-5d".$(12)); // '12 '
console.log("%5.2d".$(123)); // ' 120'
console.log("%5.2f".$(1.1)); // ' 1.10'
console.log("%10.2e".$(1.1)); // ' 1.10e+0'
console.log("%5.3p".$(1.12345)); // ' 1.12'
console.log("%5x".$(45054)); // ' affe'
console.log("%20#2x".$("45054")); // ' 1010111111111110'
console.log("%6#2d".$("111")); // ' 7'
console.log("%6#16d".$("affe")); // ' 45054'

Adding to zippoxer's answer, I use this function:
String.prototype.format = function () {
var a = this, b;
for (b in arguments) {
a = a.replace(/%[a-z]/, arguments[b]);
}
return a; // Make chainable
};
var s = 'Hello %s The magic number is %d.';
s.format('world!', 12); // Hello World! The magic number is 12.
I also have a non-prototype version which I use more often for its Java-like syntax:
function format() {
var a, b, c;
a = arguments[0];
b = [];
for(c = 1; c < arguments.length; c++){
b.push(arguments[c]);
}
for (c in b) {
a = a.replace(/%[a-z]/, b[c]);
}
return a;
}
format('%d ducks, 55 %s', 12, 'cats'); // 12 ducks, 55 cats
ES 2015 update
All the cool new stuff in ES 2015 makes this a lot easier:
function format(fmt, ...args){
return fmt
.split("%%")
.reduce((aggregate, chunk, i) =>
aggregate + chunk + (args[i] || ""), "");
}
format("Hello %%! I ate %% apples today.", "World", 44);
// "Hello World, I ate 44 apples today."
I figured that since this, like the older ones, doesn't actually parse the letters, it might as well just use a single token %%. This has the benefit of being obvious and not making it difficult to use a single %. However, if you need %% for some reason, you would need to replace it with itself:
format("I love percentage signs! %%", "%%");
// "I love percentage signs! %%"

+1 Zippo with the exception that the function body needs to be as below or otherwise it appends the current string on every iteration:
String.prototype.format = function() {
var formatted = this;
for (var arg in arguments) {
formatted = formatted.replace("{" + arg + "}", arguments[arg]);
}
return formatted;
};

I use a small library called String.format for JavaScript which supports most of the format string capabilities (including format of numbers and dates), and uses the .NET syntax. The script itself is smaller than 4 kB, so it doesn't create much of overhead.

I'll add my own discoveries which I've found since I asked:
number_format (for thousand separator/currency formatting)
sprintf (same author as above)
Sadly it seems sprintf doesn't handle thousand separator formatting like .NET's string format.

If you are looking to handle the thousands separator, you should really use toLocaleString() from the JavaScript Number class since it will format the string for the user's region.
The JavaScript Date class can format localized dates and times.

Very elegant:
String.prototype.format = function (){
var args = arguments;
return this.replace(/\{\{|\}\}|\{(\d+)\}/g, function (curlyBrack, index) {
return ((curlyBrack == "{{") ? "{" : ((curlyBrack == "}}") ? "}" : args[index]));
});
};
// Usage:
"{0}{1}".format("{1}", "{0}")
Credit goes to (broken link) https://gist.github.com/0i0/1519811

There is "sprintf" for JavaScript which you can find at http://www.webtoolkit.info/javascript-sprintf.html.

The PHPJS project has written JavaScript implementations for many of PHP's functions. Since PHP's sprintf() function is basically the same as C's printf(), their JavaScript implementation of it should satisfy your needs.

I use this one:
String.prototype.format = function() {
var newStr = this, i = 0;
while (/%s/.test(newStr))
newStr = newStr.replace("%s", arguments[i++])
return newStr;
}
Then I call it:
"<h1>%s</h1><p>%s</p>".format("Header", "Just a test!");

I have a solution very close to Peter's, but it deals with number and object case.
if (!String.prototype.format) {
String.prototype.format = function() {
var args;
args = arguments;
if (args.length === 1 && args[0] !== null && typeof args[0] === 'object') {
args = args[0];
}
return this.replace(/{([^}]*)}/g, function(match, key) {
return (typeof args[key] !== "undefined" ? args[key] : match);
});
};
}
Maybe it could be even better to deal with the all deeps cases, but for my needs this is just fine.
"This is an example from {name}".format({name:"Blaine"});
"This is an example from {0}".format("Blaine");
PS: This function is very cool if you are using translations in templates frameworks like AngularJS:
<h1> {{('hello-message'|translate).format(user)}} <h1>
<h1> {{('hello-by-name'|translate).format( user ? user.name : 'You' )}} <h1>
Where the en.json is something like
{
"hello-message": "Hello {name}, welcome.",
"hello-by-name": "Hello {0}, welcome."
}

One very slightly different version, the one I prefer (this one uses {xxx} tokens rather than {0} numbered arguments, this is much more self-documenting and suits localization much better):
String.prototype.format = function(tokens) {
var formatted = this;
for (var token in tokens)
if (tokens.hasOwnProperty(token))
formatted = formatted.replace(RegExp("{" + token + "}", "g"), tokens[token]);
return formatted;
};
A variation would be:
var formatted = l(this);
that calls an l() localization function first.

For basic formatting:
var template = jQuery.validator.format("{0} is not a valid value");
var result = template("abc");

We can use a simple lightweight String.Format string operation library for Typescript.
String.Format():
var id = image.GetId()
String.Format("image_{0}.jpg", id)
output: "image_2db5da20-1c5d-4f1a-8fd4-b41e34c8c5b5.jpg";
String Format for specifiers:
var value = String.Format("{0:L}", "APPLE"); //output "apple"
value = String.Format("{0:U}", "apple"); // output "APPLE"
value = String.Format("{0:d}", "2017-01-23 00:00"); //output "23.01.2017"
value = String.Format("{0:s}", "21.03.2017 22:15:01") //output "2017-03-21T22:15:01"
value = String.Format("{0:n}", 1000000);
//output "1.000.000"
value = String.Format("{0:00}", 1);
//output "01"
String Format for Objects including specifiers:
var fruit = new Fruit();
fruit.type = "apple";
fruit.color = "RED";
fruit.shippingDate = new Date(2018, 1, 1);
fruit.amount = 10000;
String.Format("the {type:U} is {color:L} shipped on {shippingDate:s} with an amount of {amount:n}", fruit);
// output: the APPLE is red shipped on 2018-01-01 with an amount of 10.000

Just in case someone needs a function to prevent polluting global scope, here is the function that does the same:
function _format (str, arr) {
return str.replace(/{(\d+)}/g, function (match, number) {
return typeof arr[number] != 'undefined' ? arr[number] : match;
});
};

For those who like Node.JS and its util.format feature, I've just extracted it out into its vanilla JavaScript form (with only functions that util.format uses):
exports = {};
function isString(arg) {
return typeof arg === 'string';
}
function isNull(arg) {
return arg === null;
}
function isObject(arg) {
return typeof arg === 'object' && arg !== null;
}
function isBoolean(arg) {
return typeof arg === 'boolean';
}
function isUndefined(arg) {
return arg === void 0;
}
function stylizeNoColor(str, styleType) {
return str;
}
function stylizeWithColor(str, styleType) {
var style = inspect.styles[styleType];
if (style) {
return '\u001b[' + inspect.colors[style][0] + 'm' + str +
'\u001b[' + inspect.colors[style][3] + 'm';
} else {
return str;
}
}
function isFunction(arg) {
return typeof arg === 'function';
}
function isNumber(arg) {
return typeof arg === 'number';
}
function isSymbol(arg) {
return typeof arg === 'symbol';
}
function formatPrimitive(ctx, value) {
if (isUndefined(value))
return ctx.stylize('undefined', 'undefined');
if (isString(value)) {
var simple = '\'' + JSON.stringify(value).replace(/^"|"$/g, '')
.replace(/'/g, "\\'")
.replace(/\\"/g, '"') + '\'';
return ctx.stylize(simple, 'string');
}
if (isNumber(value)) {
// Format -0 as '-0'. Strict equality won't distinguish 0 from -0,
// so instead we use the fact that 1 / -0 < 0 whereas 1 / 0 > 0 .
if (value === 0 && 1 / value < 0)
return ctx.stylize('-0', 'number');
return ctx.stylize('' + value, 'number');
}
if (isBoolean(value))
return ctx.stylize('' + value, 'boolean');
// For some reason typeof null is "object", so special case here.
if (isNull(value))
return ctx.stylize('null', 'null');
// es6 symbol primitive
if (isSymbol(value))
return ctx.stylize(value.toString(), 'symbol');
}
function arrayToHash(array) {
var hash = {};
array.forEach(function (val, idx) {
hash[val] = true;
});
return hash;
}
function objectToString(o) {
return Object.prototype.toString.call(o);
}
function isDate(d) {
return isObject(d) && objectToString(d) === '[object Date]';
}
function isError(e) {
return isObject(e) &&
(objectToString(e) === '[object Error]' || e instanceof Error);
}
function isRegExp(re) {
return isObject(re) && objectToString(re) === '[object RegExp]';
}
function formatError(value) {
return '[' + Error.prototype.toString.call(value) + ']';
}
function formatPrimitiveNoColor(ctx, value) {
var stylize = ctx.stylize;
ctx.stylize = stylizeNoColor;
var str = formatPrimitive(ctx, value);
ctx.stylize = stylize;
return str;
}
function isArray(ar) {
return Array.isArray(ar);
}
function hasOwnProperty(obj, prop) {
return Object.prototype.hasOwnProperty.call(obj, prop);
}
function formatProperty(ctx, value, recurseTimes, visibleKeys, key, array) {
var name, str, desc;
desc = Object.getOwnPropertyDescriptor(value, key) || {value: value[key]};
if (desc.get) {
if (desc.set) {
str = ctx.stylize('[Getter/Setter]', 'special');
} else {
str = ctx.stylize('[Getter]', 'special');
}
} else {
if (desc.set) {
str = ctx.stylize('[Setter]', 'special');
}
}
if (!hasOwnProperty(visibleKeys, key)) {
name = '[' + key + ']';
}
if (!str) {
if (ctx.seen.indexOf(desc.value) < 0) {
if (isNull(recurseTimes)) {
str = formatValue(ctx, desc.value, null);
} else {
str = formatValue(ctx, desc.value, recurseTimes - 1);
}
if (str.indexOf('\n') > -1) {
if (array) {
str = str.split('\n').map(function (line) {
return ' ' + line;
}).join('\n').substr(2);
} else {
str = '\n' + str.split('\n').map(function (line) {
return ' ' + line;
}).join('\n');
}
}
} else {
str = ctx.stylize('[Circular]', 'special');
}
}
if (isUndefined(name)) {
if (array && key.match(/^\d+$/)) {
return str;
}
name = JSON.stringify('' + key);
if (name.match(/^"([a-zA-Z_][a-zA-Z_0-9]*)"$/)) {
name = name.substr(1, name.length - 2);
name = ctx.stylize(name, 'name');
} else {
name = name.replace(/'/g, "\\'")
.replace(/\\"/g, '"')
.replace(/(^"|"$)/g, "'")
.replace(/\\\\/g, '\\');
name = ctx.stylize(name, 'string');
}
}
return name + ': ' + str;
}
function formatArray(ctx, value, recurseTimes, visibleKeys, keys) {
var output = [];
for (var i = 0, l = value.length; i < l; ++i) {
if (hasOwnProperty(value, String(i))) {
output.push(formatProperty(ctx, value, recurseTimes, visibleKeys,
String(i), true));
} else {
output.push('');
}
}
keys.forEach(function (key) {
if (!key.match(/^\d+$/)) {
output.push(formatProperty(ctx, value, recurseTimes, visibleKeys,
key, true));
}
});
return output;
}
function reduceToSingleString(output, base, braces) {
var length = output.reduce(function (prev, cur) {
return prev + cur.replace(/\u001b\[\d\d?m/g, '').length + 1;
}, 0);
if (length > 60) {
return braces[0] +
(base === '' ? '' : base + '\n ') +
' ' +
output.join(',\n ') +
' ' +
braces[1];
}
return braces[0] + base + ' ' + output.join(', ') + ' ' + braces[1];
}
function formatValue(ctx, value, recurseTimes) {
// Provide a hook for user-specified inspect functions.
// Check that value is an object with an inspect function on it
if (ctx.customInspect &&
value &&
isFunction(value.inspect) &&
// Filter out the util module, it's inspect function is special
value.inspect !== exports.inspect &&
// Also filter out any prototype objects using the circular check.
!(value.constructor && value.constructor.prototype === value)) {
var ret = value.inspect(recurseTimes, ctx);
if (!isString(ret)) {
ret = formatValue(ctx, ret, recurseTimes);
}
return ret;
}
// Primitive types cannot have properties
var primitive = formatPrimitive(ctx, value);
if (primitive) {
return primitive;
}
// Look up the keys of the object.
var keys = Object.keys(value);
var visibleKeys = arrayToHash(keys);
if (ctx.showHidden) {
keys = Object.getOwnPropertyNames(value);
}
// This could be a boxed primitive (new String(), etc.), check valueOf()
// NOTE: Avoid calling `valueOf` on `Date` instance because it will return
// a number which, when object has some additional user-stored `keys`,
// will be printed out.
var formatted;
var raw = value;
try {
// the .valueOf() call can fail for a multitude of reasons
if (!isDate(value))
raw = value.valueOf();
} catch (e) {
// ignore...
}
if (isString(raw)) {
// for boxed Strings, we have to remove the 0-n indexed entries,
// since they just noisey up the output and are redundant
keys = keys.filter(function (key) {
return !(key >= 0 && key < raw.length);
});
}
// Some type of object without properties can be shortcutted.
if (keys.length === 0) {
if (isFunction(value)) {
var name = value.name ? ': ' + value.name : '';
return ctx.stylize('[Function' + name + ']', 'special');
}
if (isRegExp(value)) {
return ctx.stylize(RegExp.prototype.toString.call(value), 'regexp');
}
if (isDate(value)) {
return ctx.stylize(Date.prototype.toString.call(value), 'date');
}
if (isError(value)) {
return formatError(value);
}
// now check the `raw` value to handle boxed primitives
if (isString(raw)) {
formatted = formatPrimitiveNoColor(ctx, raw);
return ctx.stylize('[String: ' + formatted + ']', 'string');
}
if (isNumber(raw)) {
formatted = formatPrimitiveNoColor(ctx, raw);
return ctx.stylize('[Number: ' + formatted + ']', 'number');
}
if (isBoolean(raw)) {
formatted = formatPrimitiveNoColor(ctx, raw);
return ctx.stylize('[Boolean: ' + formatted + ']', 'boolean');
}
}
var base = '', array = false, braces = ['{', '}'];
// Make Array say that they are Array
if (isArray(value)) {
array = true;
braces = ['[', ']'];
}
// Make functions say that they are functions
if (isFunction(value)) {
var n = value.name ? ': ' + value.name : '';
base = ' [Function' + n + ']';
}
// Make RegExps say that they are RegExps
if (isRegExp(value)) {
base = ' ' + RegExp.prototype.toString.call(value);
}
// Make dates with properties first say the date
if (isDate(value)) {
base = ' ' + Date.prototype.toUTCString.call(value);
}
// Make error with message first say the error
if (isError(value)) {
base = ' ' + formatError(value);
}
// Make boxed primitive Strings look like such
if (isString(raw)) {
formatted = formatPrimitiveNoColor(ctx, raw);
base = ' ' + '[String: ' + formatted + ']';
}
// Make boxed primitive Numbers look like such
if (isNumber(raw)) {
formatted = formatPrimitiveNoColor(ctx, raw);
base = ' ' + '[Number: ' + formatted + ']';
}
// Make boxed primitive Booleans look like such
if (isBoolean(raw)) {
formatted = formatPrimitiveNoColor(ctx, raw);
base = ' ' + '[Boolean: ' + formatted + ']';
}
if (keys.length === 0 && (!array || value.length === 0)) {
return braces[0] + base + braces[1];
}
if (recurseTimes < 0) {
if (isRegExp(value)) {
return ctx.stylize(RegExp.prototype.toString.call(value), 'regexp');
} else {
return ctx.stylize('[Object]', 'special');
}
}
ctx.seen.push(value);
var output;
if (array) {
output = formatArray(ctx, value, recurseTimes, visibleKeys, keys);
} else {
output = keys.map(function (key) {
return formatProperty(ctx, value, recurseTimes, visibleKeys, key, array);
});
}
ctx.seen.pop();
return reduceToSingleString(output, base, braces);
}
function inspect(obj, opts) {
// default options
var ctx = {
seen: [],
stylize: stylizeNoColor
};
// legacy...
if (arguments.length >= 3) ctx.depth = arguments[2];
if (arguments.length >= 4) ctx.colors = arguments[3];
if (isBoolean(opts)) {
// legacy...
ctx.showHidden = opts;
} else if (opts) {
// got an "options" object
exports._extend(ctx, opts);
}
// set default options
if (isUndefined(ctx.showHidden)) ctx.showHidden = false;
if (isUndefined(ctx.depth)) ctx.depth = 2;
if (isUndefined(ctx.colors)) ctx.colors = false;
if (isUndefined(ctx.customInspect)) ctx.customInspect = true;
if (ctx.colors) ctx.stylize = stylizeWithColor;
return formatValue(ctx, obj, ctx.depth);
}
exports.inspect = inspect;
// http://en.wikipedia.org/wiki/ANSI_escape_code#graphics
inspect.colors = {
'bold': [1, 22],
'italic': [3, 23],
'underline': [4, 24],
'inverse': [7, 27],
'white': [37, 39],
'grey': [90, 39],
'black': [30, 39],
'blue': [34, 39],
'cyan': [36, 39],
'green': [32, 39],
'magenta': [35, 39],
'red': [31, 39],
'yellow': [33, 39]
};
// Don't use 'blue' not visible on cmd.exe
inspect.styles = {
'special': 'cyan',
'number': 'yellow',
'boolean': 'yellow',
'undefined': 'grey',
'null': 'bold',
'string': 'green',
'symbol': 'green',
'date': 'magenta',
// "name": intentionally not styling
'regexp': 'red'
};
var formatRegExp = /%[sdj%]/g;
exports.format = function (f) {
if (!isString(f)) {
var objects = [];
for (var j = 0; j < arguments.length; j++) {
objects.push(inspect(arguments[j]));
}
return objects.join(' ');
}
var i = 1;
var args = arguments;
var len = args.length;
var str = String(f).replace(formatRegExp, function (x) {
if (x === '%%') return '%';
if (i >= len) return x;
switch (x) {
case '%s':
return String(args[i++]);
case '%d':
return Number(args[i++]);
case '%j':
try {
return JSON.stringify(args[i++]);
} catch (_) {
return '[Circular]';
}
default:
return x;
}
});
for (var x = args[i]; i < len; x = args[++i]) {
if (isNull(x) || !isObject(x)) {
str += ' ' + x;
} else {
str += ' ' + inspect(x);
}
}
return str;
};
Harvested from: https://github.com/joyent/node/blob/master/lib/util.js

I have a slightly longer formatter for JavaScript here...
You can do formatting several ways:
String.format(input, args0, arg1, ...)
String.format(input, obj)
"literal".format(arg0, arg1, ...)
"literal".format(obj)
Also, if you have say a ObjectBase.prototype.format (such as with DateJS) it will use that.
Examples...
var input = "numbered args ({0}-{1}-{2}-{3})";
console.log(String.format(input, "first", 2, new Date()));
//Outputs "numbered args (first-2-Thu May 31 2012...Time)-{3})"
console.log(input.format("first", 2, new Date()));
//Outputs "numbered args(first-2-Thu May 31 2012...Time)-{3})"
console.log(input.format(
"object properties ({first}-{second}-{third:yyyy-MM-dd}-{fourth})"
,{
'first':'first'
,'second':2
,'third':new Date() //assumes Date.prototype.format method
}
));
//Outputs "object properties (first-2-2012-05-31-{3})"
I've also aliased with .asFormat and have some detection in place in case there's already a string.format (such as with MS Ajax Toolkit (I hate that library).

Using Lodash you can get template functionality:
Use the ES template literal delimiter as an "interpolate" delimiter.
Disable support by replacing the "interpolate" delimiter.
var compiled = _.template('hello ${ user }!');
compiled({ 'user': 'pebbles' });
// => 'hello pebbles!

How do you write an arithmetic expression parser in JavaScript, without using eval or a constructor function?

Given a string:
var str1 = "25*5+5*7";
Without using eval or the constructor function in JavaScript, how would I be able to write a function called "output" that takes in the string and outputs the arithmetic value of the string, which in this case is 160?

Here's a full precedence expression evaluator following the recursive parsing
idea I linked-to in a comment on the OP's question.
To do this, first I wrote a simple BNF grammar for the expressions I wanted to process:
sum = product | sum "+" product | sum "-" product ;
product = term | product "*" term | product "/" term ;
term = "-" term | "(" sum ")" | number ;
This by itself requires a bit of experience to do simply and straightforwardly. If you have no experience with BNF you will find
it incredibly useful for describing complex streams of items like expressions, messages, programming langauges, ...
Using that grammar, I followed the procedure outlined in the other message
to produce the following code. It should be obvious that it is driven by grammar in a dumb mechanical way, and therefore pretty easy to write if you have that grammar.
(Untested. I'm not a JavaScript coder. This will surely contain a few syntax/semantic hiccups. Took me at about 15 minutes to code.)
var SE="Syntax Error";
function parse(str) { // returns integer expression result or SE
var text=str;
var scan=1;
return parse_sum();
function parse_sum() {
var number, number2;
if (number=parse_product()==SE) return SE;
while (true) {
skip_blanks();
if (match("+") {
number2=parse_product();
if (number2==SE) return SE;
number+=number2;
}
else if (match('-')) {
{ number2=parse_product();
if (number2==SE) return SE;
number-=number2;
}
else return number;
}
}
function parse_product() {
var number, number2;
if (number=parse_number()==SE) return SE;
while (true) {
if (match("*") {
number2=parse_term();
if (number2==SE) return SE;
number*=number2;
}
else if (match('/')) {
number2=parse_term();
if (number2==SE) return SE;
number/=number2;
}
else return number;
}
}
function parse_term() {
var number;
skip_blanks();
if (match("(")) {
number=parse_sum();
if (number=SE) return SE;
skip_blanks();
if (!match(")") return SE;
}
else if match("-") {
number= - parse_term();
}
else if (number=parse_number()==SE) return SE;
return number;
}
function skip_blanks() {
while (match(" ")) { };
return;
}
function parse_number() {
number=0;
if (is_digit()) {
while (is_digit()) {}
return number;
}
else return SE;
}
var number;
function is_digit() { // following 2 lines are likely wrong in detail but not intent
if (text[scan]>="0" && text[scan]<="9") {
number=number*10+text[scan].toInt();
return true;
}
else return false;
}
function match(c) {
if (text[scan]==c)
{ scan++; return true }
else return false;
}
}
It is straightforward to code such parsers/evaluators. See my SO answer on how to build a parser (which links to how to how to build an evaluator).

This is a simple parser with * over + precedence. I've tried to make it as educational as possible. I'll leave it up to you to add division and subtraction. Or brackets, if you're particularly ambitious.
function parse(str) {
var signs = ["*", "+"]; // signs in the order in which they should be evaluated
var funcs = [multiply, add]; // the functions associated with the signs
var tokens = str.split(/\b/); // split the string into "tokens" (numbers or signs)
for (var round = 0; round < signs.length; round++) { // do this for every sign
document.write("tokens at this point: " + tokens.join(" ") + "<BR>");
for (var place = 0; place < tokens.length; place++) { // do this for every token
if (tokens[place] == signs[round]) { // a sign is found
var a = parseInt(tokens[place - 1]); // convert previous token to number
var b = parseInt(tokens[place + 1]); // convert next token to number
var result = funcs[round](a, b); // call the appropriate function
document.write("calculating: " + a + signs[round] + b + "=" + result + "<BR>");
tokens[place - 1] = result.toString(); // store the result as a string
tokens.splice(place--, 2); // delete obsolete tokens and back up one place
}
}
}
return tokens[0]; // at the end tokens[] has only one item: the result
function multiply(x, y) { // the functions which actually do the math
return x * y;
}
function add(x, y) { // the functions which actually do the math
return x + y;
}
}
var str = "25*5+5*7";
document.write("result: " + str + " = " + parse(str));

You can use the expression parser of math.js:
var str1= "25*5+5*7"
document.write(str1 + ' = ' + math.eval(str1));
// output: "25*5+5*7 = 160"
<script src="http://cdnjs.cloudflare.com/ajax/libs/mathjs/2.1.1/math.min.js"></script>

You can create a new script:
function parse(str) {
var s = document.createElement('script');
s.text = "window.result = " + str;
document.body.appendChild(s); // Run script
document.body.removeChild(s); // Clean up
return result; // Return the result
}
document.body.innerHTML = parse("5*5+5*5");
Or use event handler content attributes:
function parse(str) {
var el = document.createElement('div');
el.setAttribute('onclick', "this.result = " + str);
el.onclick(); // Run script
return el.result; // Return the result
}
document.body.innerHTML = parse("5*5+5*5");
Note these approaches are unsafe and as evil as eval but uglier. So I don't recommend them.

How to trim specific characters from the end of the JavaScript string?

In JavaScript, how do I trim from the right(string end)?
I have the following example:
var s1 = "this is a test~";
var s = s1.rtrim('~');

Use a RegExp. Don't forget to escape special characters.
s1 = s1.replace(/~+$/, ''); //$ marks the end of a string
// ~+$ means: all ~ characters at the end of a string

You can modify the String prototype if you like. Modifying the String prototype is generally frowned upon, but I personally prefer this method, as it makes the code cleaner IMHO.
String.prototype.rtrim = function(s) {
return this.replace(new RegExp(s + "*$"),'');
};
Then call...
var s1 = "this is a test~";
var s = s1.rtrim('~');
alert(s);

IMO this is the best way to do a right/left trim and therefore, having a full functionality for trimming (since javascript supports string.trim natively)
String.prototype.rtrim = function (s) {
if (s == undefined)
s = '\\s';
return this.replace(new RegExp("[" + s + "]*$"), '');
};
String.prototype.ltrim = function (s) {
if (s == undefined)
s = '\\s';
return this.replace(new RegExp("^[" + s + "]*"), '');
};
Usage example:
var str1 = ' jav '
var r1 = mystring.trim(); // result = 'jav'
var r2 = mystring.rtrim(); // result = ' jav'
var r3 = mystring.rtrim(' v'); // result = ' ja'
var r4 = mystring.ltrim(); // result = 'jav '

There are no trim, ltrim, or rtrim functions in Javascript. Many libraries provide them, but generally they will look something like:
str.replace(/~*$/, '');
For right trims, the following is generally faster than a regex because of how regex deals with end characters in most browsers:
function rtrim(str, ch)
{
let i = str.length;
while (i-- && str.charAt(i) === ch);
return str.substring(0, i + 1);
}
console.log(rtrim("moo", "x"));
console.log(rtrim("moo", "o"));
console.log(rtrim("oo", "o"));

A solution using a regular expression:
"hi there~".replace(/~*$/, "")

str.trimEnd();
str.trimRight();
These are currently stage 4 proposals expected to be part of ES2019. They work in NodeJS and several browsers.
See below for more info:
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/trimEnd

This removes a specified string or character from the right side of a string
function rightTrim(sourceString,searchString)
{
for(;;)
{
var pos = sourceString.lastIndexOf(searchString);
if(pos === sourceString.length -1)
{
var result = sourceString.slice(0,pos);
sourceString = result;
}
else
{
break;
}
}
return sourceString;
}
Please use like so:
rightTrim('sourcecodes.....','.'); //outputs 'sourcecodes'
rightTrim('aaabakadabraaa','a'); //outputs 'aaabakadabr'

My 2 cents:
function rtrim(str: string, ch: string): string
{
var i:number = str.length - 1;
while (ch === str.charAt(i) && i >= 0) i--
return str.substring(0, i + 1);
}
const tests = ["/toto/", "/toto///l/", "/toto////", "/////", "/"]
tests.forEach(test => {
console.log(`${test} = ${rtrim(test, "/")}`)
})
Gives
"/toto/ = /toto"
"/toto///l/ = /toto///l"
"/toto//// = /toto"
"///// = "
"/ = "

You can add lodash library. It is a JavaScript utility library, There are some trim functions. You can use:
_.trimEnd('this is a test~', '~')

This is old, I know. But I don't see what's wrong with substr...?
function rtrim(str, length) {
return str.substr(0, str.length - length);
}