Related
Example: We have the number 1122. I would like to check that if given number contains the digit 1 more than once. In this case, it should return true.
I need the code to be flexible, it has to work with any number, like 3340, 5660, 4177 etc.
You can easily "force" JS to coerce any numeric value to a string, either by calling the toString method, or concatenating:
var someNum = 1122;
var oneCount = (someNum + '').split('1').length;
by concatenating a number to an empty string, the variable is coerced to a string, so you can use all the string methods you like (.match, .substring, .indexOf, ...).
In this example, I've chosen to split the string on each '1' char, count and use the length of the resulting array. If the the length > 2, than you know what you need to know.
var multipleOnes = ((someNum + '').split('1').length > 2);//returns a bool, true in this case
In response to your comment, to make it flexible - writing a simple function will do:
function multipleDigit(number, digit, moreThan)
{
moreThan = (moreThan || 1) + 1;//default more than 1 time, +1 for the length at the end
digit = (digit !== undefined ? digit : 1).toString();
return ((someNum + '').split(digit).length > moreThan);
}
multipleDigit(1123, 1);//returns true
multipleDigit(1123, 1, 2);//returns false
multipleDigit(223344,3);//returns 3 -> more than 1 3 in number.
Use javascript's match() method. Essentially, what you'd need to do is first convert the number to a string. Numbers don't have the RegExp methods. After that, match for the number 1 globally and count the results (match returns an array with all matched results).
var number = 1100;
console.log(number.toString().match(/1/g).length);
function find(num, tofind) {
var b = parseInt(num, 10);
var c = parseInt(tofind, 10);
var a = c.split("");
var times = 0;
for (var i = 0; i < a.length; i++) {
if (a[i] == b) {
times++;
}
}
alert(times);
}
find('2', '1122');
Convert the number to a string and iterate over it. Return true once a second digit has been found, for efficiency.
function checkDigitRepeat(number, digit) {
var i, count = 0;
i = Math.abs(number);
if(isNaN(i)) {
throw(TypeError('expected Number for number, got: ' + number));
}
number = i.toString();
i = Math.abs(digit);
if(isNaN(i)) {
throw(TypeError('expected Number for digit, got: ' + digit));
}
digit = i.toString();
if(digit > 9) {
throw(SyntaxError('expected a digit for digit, got a sequence of digits: ' + digit));
}
for(i = 0; i < number.length; i += 1) {
if(number[i] === digit) {
count += 1;
if(count >= 2) { return true; }
}
}
return false;
}
In the event that you want to check for a sequence of digits, your solution may lie in using regular expressions.
var myNum = '0011';
var isMultipleTimes = function(num) {
return !!num.toString().match(/(\d)\1/g);
}
console.log(isMultipleTimes(myNum));
JavaScript Match
Using #Aspiring Aqib's answer, I made a function that actually works properly and in the way I want.
The way it works is:
Example execution: multDig('221','2')
Split the number (first argument) to an array where each element is one digit.Output: ['2','2','1']
Run a for loop, which checks each of the array elements if they match with the digit (second argument), and increment the times variable if there is a match.Output: 2
Check inside the for loop if the match was detected already to improve performance on longer numbers like 2211111111111111
Return true if the number was found more than once, otherwise, return false.
And finally the code itself:
function multDig(number, digit){
var finalSplit = number.toString().split(''), times = 0;
for (i = 0; i < finalSplit.length; i++){
if (finalSplit[i] == digit){
times++
}
if (times > 1){
return true;
}
}
return false;
}
I need to write a javascript program that will ask the user to enter a string of lower‐case
characters and then print its corresponding two‐digit code. For example, if
the input is “home”, the output should be 08151305.
So far I can get it to return the correct numbers, but I cannot get it to add the zero in front of the number if it is a single digit
This is what I have:
<html>
<head>
<script type="text/javascript">
function show_prompt()
{
var name = prompt("Please enter a string of lowercase characters");
document.write(name,'<br>');
document.write('Length of the input is ', name.length,'<br>');
document.write("<br>")
for (i=0; i < name.length; i++)
{
{
document.write(i, ", ",name.charCodeAt(i) - 96, '<br>');
}
}
}
</script>
</head>
<body>
<input type="button" onClick="show_prompt()"value="CSE 201 HW#4 Problem 3"/>
</body>
</html>
Well you can just check if it is a single digit and if so prepend "0":
function padToTwoDigits(c) {
// convert to string and check length
c = "" + c;
return c.length === 1 ? "0" + c : c;
// or work with it as a number:
return c >=0 && c <=9 ? "0" + c : c;
}
// then within your existing code
document.write(i, ", ",padToTwoDigits(name.charCodeAt(i) - 96), '<br>');
Of course those are just some ideas to get you started. You can pretty that up somewhat, e.g., you might create a more generic pad function with a parameter to say how many digits to pad to.
You can write your own pad function such as:
function pad(number) {
return (number < 10 ? '0' : '') + number
}
and use the pad function like:
document.write(i, ", ",pad(name.charCodeAt(i) - 96), '<br>');
Try this.
function show_prompt() {
var name = prompt("Please enter a string of lowercase characters");
document.write(name, '<br>');
document.write('Length of the input is ', name.length, '<br>');
document.write("<br>")
for (i = 0; i < name.length; i++) {
var n = name.charCodeAt(i) - 96;
document.write(i, ", ", n < 10 ? "0" + n : n, '<br>');
}
}
I wrote up a quick sample here:
http://jsfiddle.net/ZPvZ8/3/
I wrote up a prototype method to String that handles padding zeros.
String.prototype.pad = function(char, len){
var chars = this.split();
var initialLen = len - this.length;
for(var i = 0; i < initialLen; i++){
chars.unshift(char);
}
return chars.join('');
};
I convert it to an array and then add elements with the padding character. I chose to use an array since performing string manipulations is an expensive operation (expensive in memory and CPU usage).
To use it in your program, you'd just call it like this:
var res = new Array();
for (var i = 0; i < name.length; i++) {
var strCode = (name.charCodeAt(i) - 96).toString();
res.push(strCode.pad('0', 2));
}
document.write(res.join(''));
var number = 1310;
should be left alone.
var number = 120;
should be changed to "0120";
var number = 10;
should be changed to "0010";
var number = 7;
should be changed to "0007";
In all modern browsers you can use
numberStr.padStart(4, "0");
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/padStart
function zeroPad(num) {
return num.toString().padStart(4, "0");
}
var numbers = [1310, 120, 10, 7];
numbers.forEach(
function(num) {
var paddedNum = zeroPad(num);
console.log(paddedNum);
}
);
function pad_with_zeroes(number, length) {
var my_string = '' + number;
while (my_string.length < length) {
my_string = '0' + my_string;
}
return my_string;
}
try these:
('0000' + number).slice(-4);
or
(number+'').padStart(4,'0');
Here's another way. Comes from something I did that needs to be done thousands of times on a page load. It's pretty CPU efficient to hard code a string of zeroes one time, and chop as many as you need for the pad as many times as needed. I do really like the power of 10 method -- that's pretty flexible.
Anyway, this is as efficient as I could come up with:
For the original question, CHOOSE ONE of the cases...
var number = 1310;
var number = 120;
var number = 10;
var number = 7;
then
// only needs to happen once
var zeroString = "00000";
// one assignment gets the padded number
var paddedNum = zeroString.substring((number + "").length, 4) + bareNum;
//output
alert("The padded number string is: " + paddedNum);
Of course you still need to validate the input. Because this ONLY works reliably under the following conditions:
Number of zeroes in the zeroString is desired_length + 1
Number of digits in your starting number is less than or equal to your desired length
Backstory:
I have a case that needs a fixed length (14 digit) zero-padded number. I wanted to see how basic I could make this. It's run tens of thousands of times on a page load, so efficiency matters. It's not quite re-usable as-is, and it's a bit inelegant. Except that it is very very simple.
For desired n digits padded string, this method requires a string of (at least) n+1 zeroes. Index 0 is the first character in the string, which won't ever be used, so really, it could be anything.
Note also that string.substring() is different from string.substr()!
var bareNum = 42 + '';
var zeroString = "000000000000000";
var paddedNum = zeroString.substring(bareNumber.length, 14) + bareNum
This pulls zeroes from zeroString starting at the position matching the length of the string, and continues to get zeroes to the necessary length of 14. As long as that "14" in the third line is a lower integer than the number of characters in zeroString, it will work.
function pad(n, len) {
return (new Array(len + 1).join('0') + n).slice(-len);
}
might not work in old IE versions.
//to: 0 - to left, 1 - to right
String.prototype.pad = function(_char, len, to) {
if (!this || !_char || this.length >= len) {
return this;
}
to = to || 0;
var ret = this;
var max = (len - this.length)/_char.length + 1;
while (--max) {
ret = (to) ? ret + _char : _char + ret;
}
return ret;
};
Usage:
someString.pad(neededChars, neededLength)
Example:
'332'.pad('0', 6); //'000332'
'332'.pad('0', 6, 1); //'332000'
An approach I like is to add 10^N to the number, where N is the number of zeros you want. Treat the resultant number as a string and slice off the zeroth digit. Of course, you'll want to be careful if your input number might be larger than your pad length, but it's still much faster than the loop method:
// You want to pad four places:
>>> var N = Math.pow(10, 4)
>>> var number = 1310
>>> number < N ? ("" + (N + number)).slice(1) : "" + number
"1310"
>>> var number = 120
>>> number < N ? ("" + (N + number)).slice(1) : "" + number
"0120"
>>> var number = 10
>>> number < N ? ("" + (N + number)).slice(1) : "" + number
"0010"
…
etc. You can make this into a function easily enough:
/**
* Pad a number with leading zeros to "pad" places:
*
* #param number: The number to pad
* #param pad: The maximum number of leading zeros
*/
function padNumber(number, pad) {
var N = Math.pow(10, pad);
return number < N ? ("" + (N + number)).slice(1) : "" + number
}
I wrote a general function for this. It takes an input control and pad length as input.
function padLeft(input, padLength) {
var num = $("#" + input).val();
$("#" + input).val(('0'.repeat(padLength) + num).slice(-padLength));
}
With RegExp/JavaScript:
var number = 7;
number = ('0000'+number).match(/\d{4}$/);
console.log(number);
With Function/RegExp/JavaScript:
var number = 7;
function padFix(n) {
return ('0000'+n).match(/\d{4}$/);
}
console.log(padFix(number));
No loop, no functions
let n = "" + 100;
let x = ("0000000000" + n).substring(n.length);//add your amount of zeros
alert(x + "-" + x.length);
Nate as the best way I found, it's just way too long to read. So I provide you with 3 simples solutions.
1. So here's my simplification of Nate's answer.
//number = 42
"0000".substring(number.toString().length, 4) + number;
2. Here's a solution that make it more reusable by using a function that takes the number and the desired length in parameters.
function pad_with_zeroes(number, len) {
var zeroes = "0".repeat(len);
return zeroes.substring(number.toString().length, len) + number;
}
// Usage: pad_with_zeroes(42,4);
// Returns "0042"
3. Here's a third solution, extending the Number prototype.
Number.prototype.toStringMinLen = function(len) {
var zeroes = "0".repeat(len);
return zeroes.substring(self.toString().length, len) + self;
}
//Usage: tmp=42; tmp.toStringMinLen(4)
Use String.JS librairy function padLeft:
S('123').padLeft(5, '0').s --> 00123
In Perl I can repeat a character multiple times using the syntax:
$a = "a" x 10; // results in "aaaaaaaaaa"
Is there a simple way to accomplish this in Javascript? I can obviously use a function, but I was wondering if there was any built in approach, or some other clever technique.
These days, the repeat string method is implemented almost everywhere. (It is not in Internet Explorer.) So unless you need to support older browsers, you can simply write:
"a".repeat(10)
Before repeat, we used this hack:
Array(11).join("a") // create string with 10 a's: "aaaaaaaaaa"
(Note that an array of length 11 gets you only 10 "a"s, since Array.join puts the argument between the array elements.)
Simon also points out that according to this benchmark, it appears that it's faster in Safari and Chrome (but not Firefox) to repeat a character multiple times by simply appending using a for loop (although a bit less concise).
In a new ES6 harmony, you will have native way for doing this with repeat. Also ES6 right now only experimental, this feature is already available in Edge, FF, Chrome and Safari
"abc".repeat(3) // "abcabcabc"
And surely if repeat function is not available you can use old-good Array(n + 1).join("abc")
Convenient if you repeat yourself a lot:
String.prototype.repeat = String.prototype.repeat || function(n){
n= n || 1;
return Array(n+1).join(this);
}
alert( 'Are we there yet?\nNo.\n'.repeat(10) )
Array(10).fill('a').join('')
Although the most voted answer is a bit more compact, with this approach you don't have to add an extra array item.
An alternative is:
for(var word = ''; word.length < 10; word += 'a'){}
If you need to repeat multiple chars, multiply your conditional:
for(var word = ''; word.length < 10 * 3; word += 'foo'){}
NOTE: You do not have to overshoot by 1 as with word = Array(11).join('a')
The most performance-wice way is https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/repeat
Short version is below.
String.prototype.repeat = function(count) {
if (count < 1) return '';
var result = '', pattern = this.valueOf();
while (count > 1) {
if (count & 1) result += pattern;
count >>>= 1, pattern += pattern;
}
return result + pattern;
};
var a = "a";
console.debug(a.repeat(10));
Polyfill from Mozilla:
if (!String.prototype.repeat) {
String.prototype.repeat = function(count) {
'use strict';
if (this == null) {
throw new TypeError('can\'t convert ' + this + ' to object');
}
var str = '' + this;
count = +count;
if (count != count) {
count = 0;
}
if (count < 0) {
throw new RangeError('repeat count must be non-negative');
}
if (count == Infinity) {
throw new RangeError('repeat count must be less than infinity');
}
count = Math.floor(count);
if (str.length == 0 || count == 0) {
return '';
}
// Ensuring count is a 31-bit integer allows us to heavily optimize the
// main part. But anyway, most current (August 2014) browsers can't handle
// strings 1 << 28 chars or longer, so:
if (str.length * count >= 1 << 28) {
throw new RangeError('repeat count must not overflow maximum string size');
}
var rpt = '';
for (;;) {
if ((count & 1) == 1) {
rpt += str;
}
count >>>= 1;
if (count == 0) {
break;
}
str += str;
}
// Could we try:
// return Array(count + 1).join(this);
return rpt;
}
}
If you're not opposed to including a library in your project, lodash has a repeat function.
_.repeat('*', 3);
// → '***
https://lodash.com/docs#repeat
For all browsers
The following function will perform a lot faster than the option suggested in the accepted answer:
var repeat = function(str, count) {
var array = [];
for(var i = 0; i < count;)
array[i++] = str;
return array.join('');
}
You'd use it like this :
var repeatedString = repeat("a", 10);
To compare the performance of this function with that of the option proposed in the accepted answer, see this Fiddle and this Fiddle for benchmarks.
For moderns browsers only
In modern browsers, you can now do this using String.prototype.repeat method:
var repeatedString = "a".repeat(10);
Read more about this method on MDN.
This option is even faster. Unfortunately, it doesn't work in any version of Internet explorer. The numbers in the table specify the first browser version that fully supports the method:
In ES2015/ES6 you can use "*".repeat(n)
So just add this to your projects, and your are good to go.
String.prototype.repeat = String.prototype.repeat ||
function(n) {
if (n < 0) throw new RangeError("invalid count value");
if (n == 0) return "";
return new Array(n + 1).join(this.toString())
};
String.repeat() is supported by 96.39% of browsers as of now.
function pad(text, maxLength){
return text + "0".repeat(maxLength - text.length);
}
console.log(pad('text', 7)); //text000
/**
* Repeat a string `n`-times (recursive)
* #param {String} s - The string you want to repeat.
* #param {Number} n - The times to repeat the string.
* #param {String} d - A delimiter between each string.
*/
var repeat = function (s, n, d) {
return --n ? s + (d || "") + repeat(s, n, d) : "" + s;
};
var foo = "foo";
console.log(
"%s\n%s\n%s\n%s",
repeat(foo), // "foo"
repeat(foo, 2), // "foofoo"
repeat(foo, "2"), // "foofoo"
repeat(foo, 2, "-") // "foo-foo"
);
Just for the fun of it, here is another way by using the toFixed(), used to format floating point numbers.
By doing
(0).toFixed(2)
(0).toFixed(3)
(0).toFixed(4)
we get
0.00
0.000
0.0000
If the first two characters 0. are deleted, we can use this repeating pattern to generate any repetition.
function repeat(str, nTimes) {
return (0).toFixed(nTimes).substr(2).replaceAll('0', str);
}
console.info(repeat('3', 5));
console.info(repeat('hello ', 4));
Another interesting way to quickly repeat n character is to use idea from quick exponentiation algorithm:
var repeatString = function(string, n) {
var result = '', i;
for (i = 1; i <= n; i *= 2) {
if ((n & i) === i) {
result += string;
}
string = string + string;
}
return result;
};
For repeat a value in my projects i use repeat
For example:
var n = 6;
for (i = 0; i < n; i++) {
console.log("#".repeat(i+1))
}
but be careful because this method has been added to the ECMAScript 6 specification.
function repeatString(n, string) {
var repeat = [];
repeat.length = n + 1;
return repeat.join(string);
}
repeatString(3,'x'); // => xxx
repeatString(10,'🌹'); // => "🌹🌹🌹🌹🌹🌹🌹🌹🌹🌹"
This is how you can call a function and get the result by the helps of Array() and join()
using Typescript and arrow fun
const repeatString = (str: string, num: number) => num > 0 ?
Array(num+1).join(str) : "";
console.log(repeatString("🌷",10))
//outputs: 🌷🌷🌷🌷🌷🌷🌷🌷🌷🌷
function repeatString(str, num) {
// Array(num+1) is the string you want to repeat and the times to repeat the string
return num > 0 ? Array(num+1).join(str) : "";
}
console.log(repeatString("a",10))
// outputs: aaaaaaaaaa
console.log(repeatString("🌷",10))
//outputs: 🌷🌷🌷🌷🌷🌷🌷🌷🌷🌷
Here is what I use:
function repeat(str, num) {
var holder = [];
for(var i=0; i<num; i++) {
holder.push(str);
}
return holder.join('');
}
I realize that it's not a popular task, what if you need to repeat your string not an integer number of times?
It's possible with repeat() and slice(), here's how:
String.prototype.fracRepeat = function(n){
if(n < 0) n = 0;
var n_int = ~~n; // amount of whole times to repeat
var n_frac = n - n_int; // amount of fraction times (e.g., 0.5)
var frac_length = ~~(n_frac * this.length); // length in characters of fraction part, floored
return this.repeat(n) + this.slice(0, frac_length);
}
And below a shortened version:
String.prototype.fracRepeat = function(n){
if(n < 0) n = 0;
return this.repeat(n) + this.slice(0, ~~((n - ~~n) * this.length));
}
var s = "abcd";
console.log(s.fracRepeat(2.5))
I'm going to expand on #bonbon's answer. His method is an easy way to "append N chars to an existing string", just in case anyone needs to do that. For example since "a google" is a 1 followed by 100 zeros.
for(var google = '1'; google.length < 1 + 100; google += '0'){}
document.getElementById('el').innerText = google;
<div>This is "a google":</div>
<div id="el"></div>
NOTE: You do have to add the length of the original string to the conditional.
Lodash offers a similar functionality as the Javascript repeat() function which is not available in all browers. It is called _.repeat and available since version 3.0.0:
_.repeat('a', 10);
var stringRepeat = function(string, val) {
var newString = [];
for(var i = 0; i < val; i++) {
newString.push(string);
}
return newString.join('');
}
var repeatedString = stringRepeat("a", 1);
Can be used as a one-liner too:
function repeat(str, len) {
while (str.length < len) str += str.substr(0, len-str.length);
return str;
}
In CoffeeScript:
( 'a' for dot in [0..10]).join('')
String.prototype.repeat = function (n) { n = Math.abs(n) || 1; return Array(n + 1).join(this || ''); };
// console.log("0".repeat(3) , "0".repeat(-3))
// return: "000" "000"
How do you convert decimal values to their hexadecimal equivalent in JavaScript?
Convert a number to a hexadecimal string with:
hexString = yourNumber.toString(16);
And reverse the process with:
yourNumber = parseInt(hexString, 16);
If you need to handle things like bit fields or 32-bit colors, then you need to deal with signed numbers. The JavaScript function toString(16) will return a negative hexadecimal number which is usually not what you want. This function does some crazy addition to make it a positive number.
function decimalToHexString(number)
{
if (number < 0)
{
number = 0xFFFFFFFF + number + 1;
}
return number.toString(16).toUpperCase();
}
console.log(decimalToHexString(27));
console.log(decimalToHexString(48.6));
The code below will convert the decimal value d to hexadecimal. It also allows you to add padding to the hexadecimal result. So 0 will become 00 by default.
function decimalToHex(d, padding) {
var hex = Number(d).toString(16);
padding = typeof (padding) === "undefined" || padding === null ? padding = 2 : padding;
while (hex.length < padding) {
hex = "0" + hex;
}
return hex;
}
function toHex(d) {
return ("0"+(Number(d).toString(16))).slice(-2).toUpperCase()
}
For completeness, if you want the two's-complement hexadecimal representation of a negative number, you can use the zero-fill-right shift >>> operator. For instance:
> (-1).toString(16)
"-1"
> ((-2)>>>0).toString(16)
"fffffffe"
There is however one limitation: JavaScript bitwise operators treat their operands as a sequence of 32 bits, that is, you get the 32-bits two's complement.
With padding:
function dec2hex(i) {
return (i+0x10000).toString(16).substr(-4).toUpperCase();
}
The accepted answer did not take into account single digit returned hexadecimal codes. This is easily adjusted by:
function numHex(s)
{
var a = s.toString(16);
if ((a.length % 2) > 0) {
a = "0" + a;
}
return a;
}
and
function strHex(s)
{
var a = "";
for (var i=0; i<s.length; i++) {
a = a + numHex(s.charCodeAt(i));
}
return a;
}
I believe the above answers have been posted numerous times by others in one form or another. I wrap these in a toHex() function like so:
function toHex(s)
{
var re = new RegExp(/^\s*(\+|-)?((\d+(\.\d+)?)|(\.\d+))\s*$/);
if (re.test(s)) {
return '#' + strHex( s.toString());
}
else {
return 'A' + strHex(s);
}
}
Note that the numeric regular expression came from 10+ Useful JavaScript Regular Expression Functions to improve your web applications efficiency.
Update: After testing this thing several times I found an error (double quotes in the RegExp), so I fixed that. HOWEVER! After quite a bit of testing and having read the post by almaz - I realized I could not get negative numbers to work.
Further - I did some reading up on this and since all JavaScript numbers are stored as 64 bit words no matter what - I tried modifying the numHex code to get the 64 bit word. But it turns out you can not do that. If you put "3.14159265" AS A NUMBER into a variable - all you will be able to get is the "3", because the fractional portion is only accessible by multiplying the number by ten(IE:10.0) repeatedly. Or to put that another way - the hexadecimal value of 0xF causes the floating point value to be translated into an integer before it is ANDed which removes everything behind the period. Rather than taking the value as a whole (i.e.: 3.14159265) and ANDing the floating point value against the 0xF value.
So the best thing to do, in this case, is to convert the 3.14159265 into a string and then just convert the string. Because of the above, it also makes it easy to convert negative numbers because the minus sign just becomes 0x26 on the front of the value.
So what I did was on determining that the variable contains a number - just convert it to a string and convert the string. This means to everyone that on the server side you will need to unhex the incoming string and then to determine the incoming information is numeric. You can do that easily by just adding a "#" to the front of numbers and "A" to the front of a character string coming back. See the toHex() function.
Have fun!
After another year and a lot of thinking, I decided that the "toHex" function (and I also have a "fromHex" function) really needed to be revamped. The whole question was "How can I do this more efficiently?" I decided that a to/from hexadecimal function should not care if something is a fractional part but at the same time it should ensure that fractional parts are included in the string.
So then the question became, "How do you know you are working with a hexadecimal string?". The answer is simple. Use the standard pre-string information that is already recognized around the world.
In other words - use "0x". So now my toHex function looks to see if that is already there and if it is - it just returns the string that was sent to it. Otherwise, it converts the string, number, whatever. Here is the revised toHex function:
/////////////////////////////////////////////////////////////////////////////
// toHex(). Convert an ASCII string to hexadecimal.
/////////////////////////////////////////////////////////////////////////////
toHex(s)
{
if (s.substr(0,2).toLowerCase() == "0x") {
return s;
}
var l = "0123456789ABCDEF";
var o = "";
if (typeof s != "string") {
s = s.toString();
}
for (var i=0; i<s.length; i++) {
var c = s.charCodeAt(i);
o = o + l.substr((c>>4),1) + l.substr((c & 0x0f),1);
}
return "0x" + o;
}
This is a very fast function that takes into account single digits, floating point numbers, and even checks to see if the person is sending a hex value over to be hexed again. It only uses four function calls and only two of those are in the loop. To un-hex the values you use:
/////////////////////////////////////////////////////////////////////////////
// fromHex(). Convert a hex string to ASCII text.
/////////////////////////////////////////////////////////////////////////////
fromHex(s)
{
var start = 0;
var o = "";
if (s.substr(0,2).toLowerCase() == "0x") {
start = 2;
}
if (typeof s != "string") {
s = s.toString();
}
for (var i=start; i<s.length; i+=2) {
var c = s.substr(i, 2);
o = o + String.fromCharCode(parseInt(c, 16));
}
return o;
}
Like the toHex() function, the fromHex() function first looks for the "0x" and then it translates the incoming information into a string if it isn't already a string. I don't know how it wouldn't be a string - but just in case - I check. The function then goes through, grabbing two characters and translating those in to ASCII characters. If you want it to translate Unicode, you will need to change the loop to going by four(4) characters at a time. But then you also need to ensure that the string is NOT divisible by four. If it is - then it is a standard hexadecimal string. (Remember the string has "0x" on the front of it.)
A simple test script to show that -3.14159265, when converted to a string, is still -3.14159265.
<?php
echo <<<EOD
<html>
<head><title>Test</title>
<script>
var a = -3.14159265;
alert( "A = " + a );
var b = a.toString();
alert( "B = " + b );
</script>
</head>
<body>
</body>
</html>
EOD;
?>
Because of how JavaScript works in respect to the toString() function, all of those problems can be eliminated which before were causing problems. Now all strings and numbers can be converted easily. Further, such things as objects will cause an error to be generated by JavaScript itself. I believe this is about as good as it gets. The only improvement left is for W3C to just include a toHex() and fromHex() function in JavaScript.
Without the loop:
function decimalToHex(d) {
var hex = Number(d).toString(16);
hex = "000000".substr(0, 6 - hex.length) + hex;
return hex;
}
// Or "#000000".substr(0, 7 - hex.length) + hex;
// Or whatever
// *Thanks to MSDN
Also isn't it better not to use loop tests that have to be evaluated?
For example, instead of:
for (var i = 0; i < hex.length; i++){}
have
for (var i = 0, var j = hex.length; i < j; i++){}
Combining some of these good ideas for an RGB-value-to-hexadecimal function (add the # elsewhere for HTML/CSS):
function rgb2hex(r,g,b) {
if (g !== undefined)
return Number(0x1000000 + r*0x10000 + g*0x100 + b).toString(16).substring(1);
else
return Number(0x1000000 + r[0]*0x10000 + r[1]*0x100 + r[2]).toString(16).substring(1);
}
Constrained/padded to a set number of characters:
function decimalToHex(decimal, chars) {
return (decimal + Math.pow(16, chars)).toString(16).slice(-chars).toUpperCase();
}
For anyone interested, here's a JSFiddle comparing most of the answers given to this question.
And here's the method I ended up going with:
function decToHex(dec) {
return (dec + Math.pow(16, 6)).toString(16).substr(-6)
}
Also, bear in mind that if you're looking to convert from decimal to hex for use in CSS as a color data type, you might instead prefer to extract the RGB values from the decimal and use rgb().
For example (JSFiddle):
let c = 4210330 // your color in decimal format
let rgb = [(c & 0xff0000) >> 16, (c & 0x00ff00) >> 8, (c & 0x0000ff)]
// Vanilla JS:
document.getElementById('some-element').style.color = 'rgb(' + rgb + ')'
// jQuery:
$('#some-element').css('color', 'rgb(' + rgb + ')')
This sets #some-element's CSS color property to rgb(64, 62, 154).
var number = 3200;
var hexString = number.toString(16);
The 16 is the radix and there are 16 values in a hexadecimal number :-)
function dec2hex(i)
{
var result = "0000";
if (i >= 0 && i <= 15) { result = "000" + i.toString(16); }
else if (i >= 16 && i <= 255) { result = "00" + i.toString(16); }
else if (i >= 256 && i <= 4095) { result = "0" + i.toString(16); }
else if (i >= 4096 && i <= 65535) { result = i.toString(16); }
return result
}
If you want to convert a number to a hexadecimal representation of an RGBA color value, I've found this to be the most useful combination of several tips from here:
function toHexString(n) {
if(n < 0) {
n = 0xFFFFFFFF + n + 1;
}
return "0x" + ("00000000" + n.toString(16).toUpperCase()).substr(-8);
}
AFAIK comment 57807 is wrong and should be something like:
var hex = Number(d).toString(16);
instead of
var hex = parseInt(d, 16);
function decimalToHex(d, padding) {
var hex = Number(d).toString(16);
padding = typeof (padding) === "undefined" || padding === null ? padding = 2 : padding;
while (hex.length < padding) {
hex = "0" + hex;
}
return hex;
}
And if the number is negative?
Here is my version.
function hexdec (hex_string) {
hex_string=((hex_string.charAt(1)!='X' && hex_string.charAt(1)!='x')?hex_string='0X'+hex_string : hex_string);
hex_string=(hex_string.charAt(2)<8 ? hex_string =hex_string-0x00000000 : hex_string=hex_string-0xFFFFFFFF-1);
return parseInt(hex_string, 10);
}
As the accepted answer states, the easiest way to convert from decimal to hexadecimal is var hex = dec.toString(16). However, you may prefer to add a string conversion, as it ensures that string representations like "12".toString(16) work correctly.
// Avoids a hard-to-track-down bug by returning `c` instead of `12`
(+"12").toString(16);
To reverse the process you may also use the solution below, as it is even shorter.
var dec = +("0x" + hex);
It seems to be slower in Google Chrome and Firefox, but is significantly faster in Opera. See http://jsperf.com/hex-to-dec.
I'm doing conversion to hex string in a pretty large loop, so I tried several techniques in order to find the fastest one. My requirements were to have a fixed-length string as a result, and encode negative values properly (-1 => ff..f).
Simple .toString(16) didn't work for me since I needed negative values to be properly encoded. The following code is the quickest I've tested so far on 1-2 byte values (note that symbols defines the number of output symbols you want to get, that is for 4-byte integer it should be equal to 8):
var hex = ['0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'a', 'b', 'c', 'd', 'e', 'f'];
function getHexRepresentation(num, symbols) {
var result = '';
while (symbols--) {
result = hex[num & 0xF] + result;
num >>= 4;
}
return result;
}
It performs faster than .toString(16) on 1-2 byte numbers and slower on larger numbers (when symbols >= 6), but still should outperform methods that encode negative values properly.
Converting hex color numbers to hex color strings:
A simple solution with toString and ES6 padStart for converting hex color numbers to hex color strings.
const string = `#${color.toString(16).padStart(6, '0')}`;
For example:
0x000000 will become #000000
0xFFFFFF will become #FFFFFF
Check this example in a fiddle here
How to convert decimal to hexadecimal in JavaScript
I wasn't able to find a brutally clean/simple decimal to hexadecimal conversion that didn't involve a mess of functions and arrays ... so I had to make this for myself.
function DecToHex(decimal) { // Data (decimal)
length = -1; // Base string length
string = ''; // Source 'string'
characters = [ '0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'A', 'B', 'C', 'D', 'E', 'F' ]; // character array
do { // Grab each nibble in reverse order because JavaScript has no unsigned left shift
string += characters[decimal & 0xF]; // Mask byte, get that character
++length; // Increment to length of string
} while (decimal >>>= 4); // For next character shift right 4 bits, or break on 0
decimal += 'x'; // Convert that 0 into a hex prefix string -> '0x'
do
decimal += string[length];
while (length--); // Flip string forwards, with the prefixed '0x'
return (decimal); // return (hexadecimal);
}
/* Original: */
D = 3678; // Data (decimal)
C = 0xF; // Check
A = D; // Accumulate
B = -1; // Base string length
S = ''; // Source 'string'
H = '0x'; // Destination 'string'
do {
++B;
A& = C;
switch(A) {
case 0xA: A='A'
break;
case 0xB: A='B'
break;
case 0xC: A='C'
break;
case 0xD: A='D'
break;
case 0xE: A='E'
break;
case 0xF: A='F'
break;
A = (A);
}
S += A;
D >>>= 0x04;
A = D;
} while(D)
do
H += S[B];
while (B--)
S = B = A = C = D; // Zero out variables
alert(H); // H: holds hexadecimal equivalent
You can do something like this in ECMAScript 6:
const toHex = num => (num).toString(16).toUpperCase();
If you are looking for converting Large integers i.e. Numbers greater than Number.MAX_SAFE_INTEGER -- 9007199254740991, then you can use the following code
const hugeNumber = "9007199254740991873839" // Make sure its in String
const hexOfHugeNumber = BigInt(hugeNumber).toString(16);
console.log(hexOfHugeNumber)
To sum it all up;
function toHex(i, pad) {
if (typeof(pad) === 'undefined' || pad === null) {
pad = 2;
}
var strToParse = i.toString(16);
while (strToParse.length < pad) {
strToParse = "0" + strToParse;
}
var finalVal = parseInt(strToParse, 16);
if ( finalVal < 0 ) {
finalVal = 0xFFFFFFFF + finalVal + 1;
}
return finalVal;
}
However, if you don't need to convert it back to an integer at the end (i.e. for colors), then just making sure the values aren't negative should suffice.
I haven't found a clear answer, without checks if it is negative or positive, that uses two's complement (negative numbers included). For that, I show my solution to one byte:
((0xFF + number +1) & 0x0FF).toString(16);
You can use this instruction to any number bytes, only you add FF in respective places. For example, to two bytes:
((0xFFFF + number +1) & 0x0FFFF).toString(16);
If you want cast an array integer to string hexadecimal:
s = "";
for(var i = 0; i < arrayNumber.length; ++i) {
s += ((0xFF + arrayNumber[i] +1) & 0x0FF).toString(16);
}
In case you're looking to convert to a 'full' JavaScript or CSS representation, you can use something like:
numToHex = function(num) {
var r=((0xff0000&num)>>16).toString(16),
g=((0x00ff00&num)>>8).toString(16),
b=(0x0000ff&num).toString(16);
if (r.length==1) { r = '0'+r; }
if (g.length==1) { g = '0'+g; }
if (b.length==1) { b = '0'+b; }
return '0x'+r+g+b; // ('#' instead of'0x' for CSS)
};
var dec = 5974678;
console.log( numToHex(dec) ); // 0x5b2a96
This is based on Prestaul and Tod's solutions. However, this is a generalisation that accounts for varying size of a variable (e.g. Parsing signed value from a microcontroller serial log).
function decimalToPaddedHexString(number, bitsize)
{
let byteCount = Math.ceil(bitsize/8);
let maxBinValue = Math.pow(2, bitsize)-1;
/* In node.js this function fails for bitsize above 32bits */
if (bitsize > 32)
throw "number above maximum value";
/* Conversion to unsigned form based on */
if (number < 0)
number = maxBinValue + number + 1;
return "0x"+(number >>> 0).toString(16).toUpperCase().padStart(byteCount*2, '0');
}
Test script:
for (let n = 0 ; n < 64 ; n++ ) {
let s=decimalToPaddedHexString(-1, n);
console.log(`decimalToPaddedHexString(-1,${(n+"").padStart(2)}) = ${s.padStart(10)} = ${("0b"+parseInt(s).toString(2)).padStart(34)}`);
}
Test results:
decimalToPaddedHexString(-1, 0) = 0x0 = 0b0
decimalToPaddedHexString(-1, 1) = 0x01 = 0b1
decimalToPaddedHexString(-1, 2) = 0x03 = 0b11
decimalToPaddedHexString(-1, 3) = 0x07 = 0b111
decimalToPaddedHexString(-1, 4) = 0x0F = 0b1111
decimalToPaddedHexString(-1, 5) = 0x1F = 0b11111
decimalToPaddedHexString(-1, 6) = 0x3F = 0b111111
decimalToPaddedHexString(-1, 7) = 0x7F = 0b1111111
decimalToPaddedHexString(-1, 8) = 0xFF = 0b11111111
decimalToPaddedHexString(-1, 9) = 0x01FF = 0b111111111
decimalToPaddedHexString(-1,10) = 0x03FF = 0b1111111111
decimalToPaddedHexString(-1,11) = 0x07FF = 0b11111111111
decimalToPaddedHexString(-1,12) = 0x0FFF = 0b111111111111
decimalToPaddedHexString(-1,13) = 0x1FFF = 0b1111111111111
decimalToPaddedHexString(-1,14) = 0x3FFF = 0b11111111111111
decimalToPaddedHexString(-1,15) = 0x7FFF = 0b111111111111111
decimalToPaddedHexString(-1,16) = 0xFFFF = 0b1111111111111111
decimalToPaddedHexString(-1,17) = 0x01FFFF = 0b11111111111111111
decimalToPaddedHexString(-1,18) = 0x03FFFF = 0b111111111111111111
decimalToPaddedHexString(-1,19) = 0x07FFFF = 0b1111111111111111111
decimalToPaddedHexString(-1,20) = 0x0FFFFF = 0b11111111111111111111
decimalToPaddedHexString(-1,21) = 0x1FFFFF = 0b111111111111111111111
decimalToPaddedHexString(-1,22) = 0x3FFFFF = 0b1111111111111111111111
decimalToPaddedHexString(-1,23) = 0x7FFFFF = 0b11111111111111111111111
decimalToPaddedHexString(-1,24) = 0xFFFFFF = 0b111111111111111111111111
decimalToPaddedHexString(-1,25) = 0x01FFFFFF = 0b1111111111111111111111111
decimalToPaddedHexString(-1,26) = 0x03FFFFFF = 0b11111111111111111111111111
decimalToPaddedHexString(-1,27) = 0x07FFFFFF = 0b111111111111111111111111111
decimalToPaddedHexString(-1,28) = 0x0FFFFFFF = 0b1111111111111111111111111111
decimalToPaddedHexString(-1,29) = 0x1FFFFFFF = 0b11111111111111111111111111111
decimalToPaddedHexString(-1,30) = 0x3FFFFFFF = 0b111111111111111111111111111111
decimalToPaddedHexString(-1,31) = 0x7FFFFFFF = 0b1111111111111111111111111111111
decimalToPaddedHexString(-1,32) = 0xFFFFFFFF = 0b11111111111111111111111111111111
Thrown: 'number above maximum value'
Note: Not too sure why it fails above 32 bitsize
rgb(255, 255, 255) // returns FFFFFF
rgb(255, 255, 300) // returns FFFFFF
rgb(0,0,0) // returns 000000
rgb(148, 0, 211) // returns 9400D3
function rgb(...values){
return values.reduce((acc, cur) => {
let val = cur >= 255 ? 'ff' : cur <= 0 ? '00' : Number(cur).toString(16);
return acc + (val.length === 1 ? '0'+val : val);
}, '').toUpperCase();
}
Arbitrary precision
This solution take on input decimal string, and return hex string. A decimal fractions are supported. Algorithm
split number to sign (s), integer part (i) and fractional part (f) e.g for -123.75 we have s=true, i=123, f=75
integer part to hex:
if i='0' stop
get modulo: m=i%16 (in arbitrary precision)
convert m to hex digit and put to result string
for next step calc integer part i=i/16 (in arbitrary precision)
fractional part
count fractional digits n
multiply k=f*16 (in arbitrary precision)
split k to right part with n digits and put them to f, and left part with rest of digits and put them to d
convert d to hex and add to result.
finish when number of result fractional digits is enough
// #param decStr - string with non-negative integer
// #param divisor - positive integer
function dec2HexArbitrary(decStr, fracDigits=0) {
// Helper: divide arbitrary precision number by js number
// #param decStr - string with non-negative integer
// #param divisor - positive integer
function arbDivision(decStr, divisor)
{
// algorithm https://www.geeksforgeeks.org/divide-large-number-represented-string/
let ans='';
let idx = 0;
let temp = +decStr[idx];
while (temp < divisor) temp = temp * 10 + +decStr[++idx];
while (decStr.length > idx) {
ans += (temp / divisor)|0 ;
temp = (temp % divisor) * 10 + +decStr[++idx];
}
if (ans.length == 0) return "0";
return ans;
}
// Helper: calc module of arbitrary precision number
// #param decStr - string with non-negative integer
// #param mod - positive integer
function arbMod(decStr, mod) {
// algorithm https://www.geeksforgeeks.org/how-to-compute-mod-of-a-big-number/
let res = 0;
for (let i = 0; i < decStr.length; i++)
res = (res * 10 + +decStr[i]) % mod;
return res;
}
// Helper: multiply arbitrary precision integer by js number
// #param decStr - string with non-negative integer
// #param mult - positive integer
function arbMultiply(decStr, mult) {
let r='';
let m=0;
for (let i = decStr.length-1; i >=0 ; i--) {
let n = m+mult*(+decStr[i]);
r= (i ? n%10 : n) + r
m= n/10|0;
}
return r;
}
// dec2hex algorithm starts here
let h= '0123456789abcdef'; // hex 'alphabet'
let m= decStr.match(/-?(.*?)\.(.*)?/) || decStr.match(/-?(.*)/); // separate sign,integer,ractional
let i= m[1].replace(/^0+/,'').replace(/^$/,'0'); // integer part (without sign and leading zeros)
let f= (m[2]||'0').replace(/0+$/,'').replace(/^$/,'0'); // fractional part (without last zeros)
let s= decStr[0]=='-'; // sign
let r=''; // result
if(i=='0') r='0';
while(i!='0') { // integer part
r=h[arbMod(i,16)]+r;
i=arbDivision(i,16);
}
if(fracDigits) r+=".";
let n = f.length;
for(let j=0; j<fracDigits; j++) { // frac part
let k= arbMultiply(f,16);
f = k.slice(-n);
let d= k.slice(0,k.length-n);
r+= d.length ? h[+d] : '0';
}
return (s?'-':'')+r;
}
// -----------
// TESTS
// -----------
let tests = [
["0",2],
["000",2],
["123",0],
["-123",0],
["00.000",2],
["255.75",5],
["-255.75",5],
["127.999",32],
];
console.log('Input Standard Abitrary');
tests.forEach(t=> {
let nonArb = (+t[0]).toString(16).padEnd(17,' ');
let arb = dec2HexArbitrary(t[0],t[1]);
console.log(t[0].padEnd(10,' '), nonArb, arb);
});
// Long Example (40 digits after dot)
let example = "123456789012345678901234567890.09876543210987654321"
console.log(`\nLong Example:`);
console.log('dec:',example);
console.log('hex: ',dec2HexArbitrary(example,40));
The problem basically how many padding zeros to expect.
If you expect string 01 and 11 from Number 1 and 17. it's better to use Buffer as a bridge, with which number is turn into bytes, and then the hex is just an output format of it. And the bytes organization is well controlled by Buffer functions, like writeUInt32BE, writeInt16LE, etc.
import { Buffer } from 'buffer';
function toHex(n) { // 4byte
const buff = Buffer.alloc(4);
buff.writeInt32BE(n);
return buff.toString('hex');
}
> toHex(1)
'00000001'
> toHex(17)
'00000011'
> toHex(-1)
'ffffffff'
> toHex(-1212)
'fffffb44'
> toHex(1212)
'000004bc'
Here's my solution:
hex = function(number) {
return '0x' + Math.abs(number).toString(16);
}
The question says: "How to convert decimal to hexadecimal in JavaScript". While, the question does not specify that the hexadecimal string should begin with a 0x prefix, anybody who writes code should know that 0x is added to hexadecimal codes to distinguish hexadecimal codes from programmatic identifiers and other numbers (1234 could be hexadecimal, decimal, or even octal).
Therefore, to correctly answer this question, for the purpose of script-writing, you must add the 0x prefix.
The Math.abs(N) function converts negatives to positives, and as a bonus, it doesn't look like somebody ran it through a wood-chipper.
The answer I wanted, would have had a field-width specifier, so we could for example show 8/16/32/64-bit values the way you would see them listed in a hexadecimal editing application. That, is the actual, correct answer.