Javascript either parseInt or + , appending instead of adding [duplicate] - javascript

This question already has answers here:
Addition is not working in JavaScript
(7 answers)
Closed 9 years ago.
i'm having some trouble with the parseInt function or + operand. I want to take in two numbers, multiply one by a third number and add them together. Instead of adding the numbers it appends one number to another.
<script language = 'JavaScript'>
function calculate_total()
{
var a = 0;
var b = 0;
var t = 0;
parseInt(a = prompt('Pay 1'), 10);
//Input of 10
if(isNaN(a))
{
alert('Error A');
}
//No Error
parseInt(b = prompt('Pay 2'), 10);
//input of 12
if(isNaN(b))
{
alert('Error B');
}
//No Error
parseInt(t = (a * 20 + b), 10);
if(isNaN(t))
{
alert('Error T');
}
else
{
alert('Total Pay: ' + t);
//expected answer 212
//Actual Answer 20012
}
//No Error
}
calculate_total();
</script>

parseInt(a = prompt('Pay 1'), 10);
...
parseInt(b = prompt('Pay 2'), 10);
...
parseInt(t = (a * 20 + b), 10);
Here a, b and t, all have got string data only and when it is converted to an int, its discarded immediately. So, fix them like this
a = parseInt(prompt('Pay 1'), 10);
...
b = parseInt(prompt('Pay 2'), 10);
...
t = a * 20 + b;
According to ECMA Script's specifications for Additive operation,
7 If Type(lprim) is String or Type(rprim) is String, then Return the
String that is the result of concatenating ToString(lprim) followed by
ToString(rprim)
So when you use a string and number with + operator, result will be concatenation of both the operands.
According to ECMA Script's specifications for Multiplicative operation,
Let left be the result of evaluating MultiplicativeExpression.
Let leftValue be GetValue(left).
Let right be the result of evaluating UnaryExpression.
Let rightValue be GetValue(right).
Let leftNum be ToNumber(leftValue).
Let rightNum be ToNumber(rightValue).
* operator basically converts both the operands to numbers. So, the result will be proper even if both the operands are numbers in strings.
You can confirm the above mentioned things with this
var a = "100", b = "12";
console.log(a * 20); // Prints 2000, both operands are converted to numbers
console.log(a * 20 + b); // Prints 200012, as b is string, both are concatenated

I'm not a Javascript expert by any means, but you seem to be ignoring the result of parseInt, instead storing just the result of prompt() in your a, b and t variables. I'd expect this:
parseInt(a = prompt('Pay 1'), 10);
to be:
a = parseInt(prompt('Pay 1'), 10);
(And the same for the other prompts.)
At that point, the variables' values will be the numbers rather than the strings, so + should add them appropriately.

You parsing result which is already wrong due to appending as string:
try updating following statements:
a = parseInt(prompt('Pay 1'), 10);
b = parseInt(prompt('Pay 2'), 10);
t = a * 20 + b; // no need to parse as a and b already integers

parseInt() returns an integer value. And you dony need parseInt(string, 10), for parseInt decimal system used by defaults.
a = parseInt(prompt('Pay 1'));
b = parseInt(prompt('Pay 2'));
t = a * 20 + b;

try
a = parseInt(prompt('pay 1'),10);
etc. for b and t.
Now you declare a as prompt('pay 1') and not as the int value returned by parseInt.

I've fixed the code for you:
<script language = 'JavaScript'>
function calculate_total()
{
var a = 0;
var b = 0;
var t = 0;
a = parseInt(prompt('Pay 1'), 10);
//Input of 10
if(isNaN(a))
{
alert('Error A');
}
//No Error
b = parseInt(prompt('Pay 2'), 10);
//input of 12
if(isNaN(b))
{
alert('Error B');
}
//No Error
t = parseInt(a * 20 + b, 10);
if(isNaN(t))
{
alert('Error T');
}
else
{
alert('Total Pay: ' + t);
//expected answer 212
//Actual Answer 20012
}
//No Error
}
calculate_total();
</script>

Related

Saving money as integer

We have the following code in Mongoose schema trying to convert money to integer to save in MongoDB:
amount: { type: Number, get: getAmount, set: setAmount, required: true}
function setAmount(num) {
return num * 100;
}
function getAmount(num) {
return (num / 100).toFixed(2);
}
However saving 64.49 we still end up with this in MongoDB documents:
"amount": 6448.999999999999
How do we fix this?
To avoid these kind of precision issue on "money" variables, I always use "cents" as unit.
Don't store 15.24$, store 1524 cents in your code.
Then do all operation using integers.
Then use cents => dollars conversion only for display, by adding "." separator before the last two characters.
Consider using BigNumber: http://mikemcl.github.io/bignumber.js/
Small example (in the console):
> 0.2 * 0.4
< 0.08000000000000002
And with BigNumber:
> var a = new BigNumber(0.2);
> var b = new BigNumber(0.4);
> var c = a.times(b);
> c.valueOf();
< "0.08"
Does this satisfy your needs?
function convert(number) {
const [i, d] = ('' + number).split(/\./);
const a = +i * 100;
const b = +(d || '0').slice(0, 2);
const c = +(d || '0').slice(2)? 1 : 0;
return a + b + c;
}
Here is the old version that didn't round up the trailing .0099999:
function convert(number) {
const [a, b] = ('' + number).split(/\./);
return +a * 100 + +(b || '0').slice(0, 2);
}
It receives 1 number and returns the same number multiplied by 100.

which are alternative of tofixed() in javascript [duplicate]

Suppose I have a value of 15.7784514, I want to display it 15.77 with no rounding.
var num = parseFloat(15.7784514);
document.write(num.toFixed(1)+"<br />");
document.write(num.toFixed(2)+"<br />");
document.write(num.toFixed(3)+"<br />");
document.write(num.toFixed(10));
Results in -
15.8
15.78
15.778
15.7784514000
How do I display 15.77?
Convert the number into a string, match the number up to the second decimal place:
function calc(theform) {
var num = theform.original.value, rounded = theform.rounded
var with2Decimals = num.toString().match(/^-?\d+(?:\.\d{0,2})?/)[0]
rounded.value = with2Decimals
}
<form onsubmit="return calc(this)">
Original number: <input name="original" type="text" onkeyup="calc(form)" onchange="calc(form)" />
<br />"Rounded" number: <input name="rounded" type="text" placeholder="readonly" readonly>
</form>
The toFixed method fails in some cases unlike toString, so be very careful with it.
Update 5 Nov 2016
New answer, always accurate
function toFixed(num, fixed) {
var re = new RegExp('^-?\\d+(?:\.\\d{0,' + (fixed || -1) + '})?');
return num.toString().match(re)[0];
}
As floating point math in javascript will always have edge cases, the previous solution will be accurate most of the time which is not good enough.
There are some solutions to this like num.toPrecision, BigDecimal.js, and accounting.js.
Yet, I believe that merely parsing the string will be the simplest and always accurate.
Basing the update on the well written regex from the accepted answer by #Gumbo, this new toFixed function will always work as expected.
Old answer, not always accurate.
Roll your own toFixed function:
function toFixed(num, fixed) {
fixed = fixed || 0;
fixed = Math.pow(10, fixed);
return Math.floor(num * fixed) / fixed;
}
Another single-line solution :
number = Math.trunc(number*100)/100
I used 100 because you want to truncate to the second digit, but a more flexible solution would be :
number = Math.trunc(number*Math.pow(10, digits))/Math.pow(10, digits)
where digits is the amount of decimal digits to keep.
See Math.trunc specs for details and browser compatibility.
I opted to write this instead to manually remove the remainder with strings so I don't have to deal with the math issues that come with numbers:
num = num.toString(); //If it's not already a String
num = num.slice(0, (num.indexOf("."))+3); //With 3 exposing the hundredths place
Number(num); //If you need it back as a Number
This will give you "15.77" with num = 15.7784514;
Update (Jan 2021)
Depending on its range, a number in javascript may be shown in scientific notation. For example, if you type 0.0000001 in the console, you may see it as 1e-7, whereas 0.000001 appears unchanged (0.000001).
If your application works on a range of numbers for which scientific notation is not involved, you can just ignore this update and use the original answer below.
This update is about adding a function that checks if the number is in scientific format and, if so, converts it into decimal format. Here I'm proposing this one, but you can use any other function that achieves the same goal, according to your application's needs:
function toFixed(x) {
if (Math.abs(x) < 1.0) {
let e = parseInt(x.toString().split('e-')[1]);
if (e) {
x *= Math.pow(10,e-1);
x = '0.' + (new Array(e)).join('0') + x.toString().substring(2);
}
} else {
let e = parseInt(x.toString().split('+')[1]);
if (e > 20) {
e -= 20;
x /= Math.pow(10,e);
x += (new Array(e+1)).join('0');
}
}
return x;
}
Now just apply that function to the parameter (that's the only change with respect to the original answer):
function toFixedTrunc(x, n) {
x = toFixed(x)
// From here on the code is the same than the original answer
const v = (typeof x === 'string' ? x : x.toString()).split('.');
if (n <= 0) return v[0];
let f = v[1] || '';
if (f.length > n) return `${v[0]}.${f.substr(0,n)}`;
while (f.length < n) f += '0';
return `${v[0]}.${f}`
}
This updated version addresses also a case mentioned in a comment:
toFixedTrunc(0.000000199, 2) => "0.00"
Again, choose what fits your application needs at best.
Original answer (October 2017)
General solution to truncate (no rounding) a number to the n-th decimal digit and convert it to a string with exactly n decimal digits, for any n≥0.
function toFixedTrunc(x, n) {
const v = (typeof x === 'string' ? x : x.toString()).split('.');
if (n <= 0) return v[0];
let f = v[1] || '';
if (f.length > n) return `${v[0]}.${f.substr(0,n)}`;
while (f.length < n) f += '0';
return `${v[0]}.${f}`
}
where x can be either a number (which gets converted into a string) or a string.
Here are some tests for n=2 (including the one requested by OP):
0 => 0.00
0.01 => 0.01
0.5839 => 0.58
0.999 => 0.99
1.01 => 1.01
2 => 2.00
2.551 => 2.55
2.99999 => 2.99
4.27 => 4.27
15.7784514 => 15.77
123.5999 => 123.59
And for some other values of n:
15.001097 => 15.0010 (n=4)
0.000003298 => 0.0000032 (n=7)
0.000003298257899 => 0.000003298257 (n=12)
parseInt is faster then Math.floor
function floorFigure(figure, decimals){
if (!decimals) decimals = 2;
var d = Math.pow(10,decimals);
return (parseInt(figure*d)/d).toFixed(decimals);
};
floorFigure(123.5999) => "123.59"
floorFigure(123.5999, 3) => "123.599"
num = 19.66752
f = num.toFixed(3).slice(0,-1)
alert(f)
This will return 19.66
Simple do this
number = parseInt(number * 100)/100;
Just truncate the digits:
function truncDigits(inputNumber, digits) {
const fact = 10 ** digits;
return Math.floor(inputNumber * fact) / fact;
}
This is not a safe alternative, as many others commented examples with numbers that turn into exponential notation, that scenery is not covered by this function
// typescript
// function formatLimitDecimals(value: number, decimals: number): number {
function formatLimitDecimals(value, decimals) {
const stringValue = value.toString();
if(stringValue.includes('e')) {
// TODO: remove exponential notation
throw 'invald number';
} else {
const [integerPart, decimalPart] = stringValue.split('.');
if(decimalPart) {
return +[integerPart, decimalPart.slice(0, decimals)].join('.')
} else {
return integerPart;
}
}
}
console.log(formatLimitDecimals(4.156, 2)); // 4.15
console.log(formatLimitDecimals(4.156, 8)); // 4.156
console.log(formatLimitDecimals(4.156, 0)); // 4
console.log(formatLimitDecimals(0, 4)); // 0
// not covered
console.log(formatLimitDecimals(0.000000199, 2)); // 0.00
These solutions do work, but to me seem unnecessarily complicated. I personally like to use the modulus operator to obtain the remainder of a division operation, and remove that. Assuming that num = 15.7784514:
num-=num%.01;
This is equivalent to saying num = num - (num % .01).
I fixed using following simple way-
var num = 15.7784514;
Math.floor(num*100)/100;
Results will be 15.77
My version for positive numbers:
function toFixed_norounding(n,p)
{
var result = n.toFixed(p);
return result <= n ? result: (result - Math.pow(0.1,p)).toFixed(p);
}
Fast, pretty, obvious. (version for positive numbers)
The answers here didn't help me, it kept rounding up or giving me the wrong decimal.
my solution converts your decimal to a string, extracts the characters and then returns the whole thing as a number.
function Dec2(num) {
num = String(num);
if(num.indexOf('.') !== -1) {
var numarr = num.split(".");
if (numarr.length == 1) {
return Number(num);
}
else {
return Number(numarr[0]+"."+numarr[1].charAt(0)+numarr[1].charAt(1));
}
}
else {
return Number(num);
}
}
Dec2(99); // 99
Dec2(99.9999999); // 99.99
Dec2(99.35154); // 99.35
Dec2(99.8); // 99.8
Dec2(10265.985475); // 10265.98
The following code works very good for me:
num.toString().match(/.\*\\..{0,2}|.\*/)[0];
This worked well for me. I hope it will fix your issues too.
function toFixedNumber(number) {
const spitedValues = String(number.toLocaleString()).split('.');
let decimalValue = spitedValues.length > 1 ? spitedValues[1] : '';
decimalValue = decimalValue.concat('00').substr(0,2);
return '$'+spitedValues[0] + '.' + decimalValue;
}
// 5.56789 ----> $5.56
// 0.342 ----> $0.34
// -10.3484534 ----> $-10.34
// 600 ----> $600.00
function convertNumber(){
var result = toFixedNumber(document.getElementById("valueText").value);
document.getElementById("resultText").value = result;
}
function toFixedNumber(number) {
const spitedValues = String(number.toLocaleString()).split('.');
let decimalValue = spitedValues.length > 1 ? spitedValues[1] : '';
decimalValue = decimalValue.concat('00').substr(0,2);
return '$'+spitedValues[0] + '.' + decimalValue;
}
<div>
<input type="text" id="valueText" placeholder="Input value here..">
<br>
<button onclick="convertNumber()" >Convert</button>
<br><hr>
<input type="text" id="resultText" placeholder="result" readonly="true">
</div>
An Easy way to do it is the next but is necessary ensure that the amount parameter is given as a string.
function truncate(amountAsString, decimals = 2){
var dotIndex = amountAsString.indexOf('.');
var toTruncate = dotIndex !== -1 && ( amountAsString.length > dotIndex + decimals + 1);
var approach = Math.pow(10, decimals);
var amountToTruncate = toTruncate ? amountAsString.slice(0, dotIndex + decimals +1) : amountAsString;
return toTruncate
? Math.floor(parseFloat(amountToTruncate) * approach ) / approach
: parseFloat(amountAsString);
}
console.log(truncate("7.99999")); //OUTPUT ==> 7.99
console.log(truncate("7.99999", 3)); //OUTPUT ==> 7.999
console.log(truncate("12.799999999999999")); //OUTPUT ==> 7.99
Here you are. An answer that shows yet another way to solve the problem:
// For the sake of simplicity, here is a complete function:
function truncate(numToBeTruncated, numOfDecimals) {
var theNumber = numToBeTruncated.toString();
var pointIndex = theNumber.indexOf('.');
return +(theNumber.slice(0, pointIndex > -1 ? ++numOfDecimals + pointIndex : undefined));
}
Note the use of + before the final expression. That is to convert our truncated, sliced string back to number type.
Hope it helps!
truncate without zeroes
function toTrunc(value,n){
return Math.floor(value*Math.pow(10,n))/(Math.pow(10,n));
}
or
function toTrunc(value,n){
x=(value.toString()+".0").split(".");
return parseFloat(x[0]+"."+x[1].substr(0,n));
}
test:
toTrunc(17.4532,2) //17.45
toTrunc(177.4532,1) //177.4
toTrunc(1.4532,1) //1.4
toTrunc(.4,2) //0.4
truncate with zeroes
function toTruncFixed(value,n){
return toTrunc(value,n).toFixed(n);
}
test:
toTrunc(17.4532,2) //17.45
toTrunc(177.4532,1) //177.4
toTrunc(1.4532,1) //1.4
toTrunc(.4,2) //0.40
If you exactly wanted to truncate to 2 digits of precision, you can go with a simple logic:
function myFunction(number) {
var roundedNumber = number.toFixed(2);
if (roundedNumber > number)
{
roundedNumber = roundedNumber - 0.01;
}
return roundedNumber;
}
I used (num-0.05).toFixed(1) to get the second decimal floored.
It's more reliable to get two floating points without rounding.
Reference Answer
var number = 10.5859;
var fixed2FloatPoints = parseInt(number * 100) / 100;
console.log(fixed2FloatPoints);
Thank You !
My solution in typescript (can easily be ported to JS):
/**
* Returns the price with correct precision as a string
*
* #param price The price in decimal to be formatted.
* #param decimalPlaces The number of decimal places to use
* #return string The price in Decimal formatting.
*/
type toDecimal = (price: number, decimalPlaces?: number) => string;
const toDecimalOdds: toDecimal = (
price: number,
decimalPlaces: number = 2,
): string => {
const priceString: string = price.toString();
const pointIndex: number = priceString.indexOf('.');
// Return the integer part if decimalPlaces is 0
if (decimalPlaces === 0) {
return priceString.substr(0, pointIndex);
}
// Return value with 0s appended after decimal if the price is an integer
if (pointIndex === -1) {
const padZeroString: string = '0'.repeat(decimalPlaces);
return `${priceString}.${padZeroString}`;
}
// If numbers after decimal are less than decimalPlaces, append with 0s
const padZeroLen: number = priceString.length - pointIndex - 1;
if (padZeroLen > 0 && padZeroLen < decimalPlaces) {
const padZeroString: string = '0'.repeat(padZeroLen);
return `${priceString}${padZeroString}`;
}
return priceString.substr(0, pointIndex + decimalPlaces + 1);
};
Test cases:
expect(filters.toDecimalOdds(3.14159)).toBe('3.14');
expect(filters.toDecimalOdds(3.14159, 2)).toBe('3.14');
expect(filters.toDecimalOdds(3.14159, 0)).toBe('3');
expect(filters.toDecimalOdds(3.14159, 10)).toBe('3.1415900000');
expect(filters.toDecimalOdds(8.2)).toBe('8.20');
Any improvements?
Another solution, that truncates and round:
function round (number, decimals, truncate) {
if (truncate) {
number = number.toFixed(decimals + 1);
return parseFloat(number.slice(0, -1));
}
var n = Math.pow(10.0, decimals);
return Math.round(number * n) / n;
};
function limitDecimalsWithoutRounding(val, decimals){
let parts = val.toString().split(".");
return parseFloat(parts[0] + "." + parts[1].substring(0, decimals));
}
var num = parseFloat(15.7784514);
var new_num = limitDecimalsWithoutRounding(num, 2);
Roll your own toFixed function: for positive values Math.floor works fine.
function toFixed(num, fixed) {
fixed = fixed || 0;
fixed = Math.pow(10, fixed);
return Math.floor(num * fixed) / fixed;
}
For negative values Math.floor is round of the values. So you can use Math.ceil instead.
Example,
Math.ceil(-15.778665 * 10000) / 10000 = -15.7786
Math.floor(-15.778665 * 10000) / 10000 = -15.7787 // wrong.
Gumbo's second solution, with the regular expression, does work but is slow because of the regular expression. Gumbo's first solution fails in certain situations due to imprecision in floating points numbers. See the JSFiddle for a demonstration and a benchmark. The second solution takes about 1636 nanoseconds per call on my current system, Intel Core i5-2500 CPU at 3.30 GHz.
The solution I've written involves adding a small compensation to take care of floating point imprecision. It is basically instantaneous, i.e. on the order of nanoseconds. I clocked 2 nanoseconds per call but the JavaScript timers are not very precise or granular. Here is the JS Fiddle and the code.
function toFixedWithoutRounding (value, precision)
{
var factorError = Math.pow(10, 14);
var factorTruncate = Math.pow(10, 14 - precision);
var factorDecimal = Math.pow(10, precision);
return Math.floor(Math.floor(value * factorError + 1) / factorTruncate) / factorDecimal;
}
var values = [1.1299999999, 1.13, 1.139999999, 1.14, 1.14000000001, 1.13 * 100];
for (var i = 0; i < values.length; i++)
{
var value = values[i];
console.log(value + " --> " + toFixedWithoutRounding(value, 2));
}
for (var i = 0; i < values.length; i++)
{
var value = values[i];
console.log(value + " --> " + toFixedWithoutRounding(value, 4));
}
console.log("type of result is " + typeof toFixedWithoutRounding(1.13 * 100 / 100, 2));
// Benchmark
var value = 1.13 * 100;
var startTime = new Date();
var numRun = 1000000;
var nanosecondsPerMilliseconds = 1000000;
for (var run = 0; run < numRun; run++)
toFixedWithoutRounding(value, 2);
var endTime = new Date();
var timeDiffNs = nanosecondsPerMilliseconds * (endTime - startTime);
var timePerCallNs = timeDiffNs / numRun;
console.log("Time per call (nanoseconds): " + timePerCallNs);
Building on David D's answer:
function NumberFormat(num,n) {
var num = (arguments[0] != null) ? arguments[0] : 0;
var n = (arguments[1] != null) ? arguments[1] : 2;
if(num > 0){
num = String(num);
if(num.indexOf('.') !== -1) {
var numarr = num.split(".");
if (numarr.length > 1) {
if(n > 0){
var temp = numarr[0] + ".";
for(var i = 0; i < n; i++){
if(i < numarr[1].length){
temp += numarr[1].charAt(i);
}
}
num = Number(temp);
}
}
}
}
return Number(num);
}
console.log('NumberFormat(123.85,2)',NumberFormat(123.85,2));
console.log('NumberFormat(123.851,2)',NumberFormat(123.851,2));
console.log('NumberFormat(0.85,2)',NumberFormat(0.85,2));
console.log('NumberFormat(0.851,2)',NumberFormat(0.851,2));
console.log('NumberFormat(0.85156,2)',NumberFormat(0.85156,2));
console.log('NumberFormat(0.85156,4)',NumberFormat(0.85156,4));
console.log('NumberFormat(0.85156,8)',NumberFormat(0.85156,8));
console.log('NumberFormat(".85156",2)',NumberFormat(".85156",2));
console.log('NumberFormat("0.85156",2)',NumberFormat("0.85156",2));
console.log('NumberFormat("1005.85156",2)',NumberFormat("1005.85156",2));
console.log('NumberFormat("0",2)',NumberFormat("0",2));
console.log('NumberFormat("",2)',NumberFormat("",2));
console.log('NumberFormat(85156,8)',NumberFormat(85156,8));
console.log('NumberFormat("85156",2)',NumberFormat("85156",2));
console.log('NumberFormat("85156.",2)',NumberFormat("85156.",2));
// NumberFormat(123.85,2) 123.85
// NumberFormat(123.851,2) 123.85
// NumberFormat(0.85,2) 0.85
// NumberFormat(0.851,2) 0.85
// NumberFormat(0.85156,2) 0.85
// NumberFormat(0.85156,4) 0.8515
// NumberFormat(0.85156,8) 0.85156
// NumberFormat(".85156",2) 0.85
// NumberFormat("0.85156",2) 0.85
// NumberFormat("1005.85156",2) 1005.85
// NumberFormat("0",2) 0
// NumberFormat("",2) 0
// NumberFormat(85156,8) 85156
// NumberFormat("85156",2) 85156
// NumberFormat("85156.",2) 85156
Already there are some suitable answer with regular expression and arithmetic calculation, you can also try this
function myFunction() {
var str = 12.234556;
str = str.toString().split('.');
var res = str[1].slice(0, 2);
document.getElementById("demo").innerHTML = str[0]+'.'+res;
}
// output: 12.23
Here is what is did it with string
export function withoutRange(number) {
const str = String(number);
const dotPosition = str.indexOf('.');
if (dotPosition > 0) {
const length = str.substring().length;
const end = length > 3 ? 3 : length;
return str.substring(0, dotPosition + end);
}
return str;
}

generate 4 digit random number using substring

I am trying to execute below code:
var a = Math.floor(100000 + Math.random() * 900000);
a = a.substring(-2);
I am getting error like undefined is not a function at line 2, but when I try to do alert(a), it has something. What is wrong here?
That's because a is a number, not a string. What you probably want to do is something like this:
var val = Math.floor(1000 + Math.random() * 9000);
console.log(val);
Math.random() will generate a floating point number in the range [0, 1) (this is not a typo, it is standard mathematical notation to show that 1 is excluded from the range).
Multiplying by 9000 results in a range of [0, 9000).
Adding 1000 results in a range of [1000, 10000).
Flooring chops off the decimal value to give you an integer. Note that it does not round.
General Case
If you want to generate an integer in the range [x, y), you can use the following code:
Math.floor(x + (y - x) * Math.random());
This will generate 4-digit random number (0000-9999) using substring:
var seq = (Math.floor(Math.random() * 10000) + 10000).toString().substring(1);
console.log(seq);
I adapted Balajis to make it immutable and functional.
Because this doesn't use math you can use alphanumeric, emojis, very long pins etc
const getRandomPin = (chars, len)=>[...Array(len)].map(
(i)=>chars[Math.floor(Math.random()*chars.length)]
).join('');
//use it like this
getRandomPin('0123456789',4);
$( document ).ready(function() {
var a = Math.floor(100000 + Math.random() * 900000);
a = String(a);
a = a.substring(0,4);
alert( "valor:" +a );
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.0/jquery.min.js"></script>
Your a is a number. To be able to use the substring function, it has to be a string first, try
var a = (Math.floor(100000 + Math.random() * 900000)).toString();
a = a.substring(-2);
You can get 4-digit this way .substring(startIndex, length), which would be in your case .substring(0, 4). To be able to use .substring() you will need to convert a to string by using .toString(). At the end, you can convert the resulting output into integer by using parseInt :
var a = Math.floor(100000 + Math.random() * 900000)
a = a.toString().substring(0, 4);
a = parseInt(a);
alert(a);
https://jsfiddle.net/v7dswkjf/
The problem is that a is a number. You cannot apply substring to a number so you have to convert the number to a string and then apply the function.
DEMO: https://jsfiddle.net/L0dba54m/
var a = Math.floor(100000 + Math.random() * 900000);
a = a.toString();
a = a.substring(-2);
$(document).ready(function() {
var a = Math.floor((Math.random() * 9999) + 999);
a = String(a);
a = a.substring(0, 4);
});
// It Will Generate Random 5 digit Number & Char
const char = '1234567890abcdefghijklmnopqrstuvwxyz'; //Random Generate Every Time From This Given Char
const length = 5;
let randomvalue = '';
for ( let i = 0; i < length; i++) {
const value = Math.floor(Math.random() * char.length);
randomvalue += char.substring(value, value + 1).toUpperCase();
}
console.log(randomvalue);
function getPin() {
let pin = Math.round(Math.random() * 10000);
let pinStr = pin + '';
// make sure that number is 4 digit
if (pinStr.length == 4) {
return pinStr;
} else {
return getPin();
}
}
let number = getPin();
Just pass Length of to number that need to be generated
await this.randomInteger(4);
async randomInteger(number) {
let length = parseInt(number);
let string:string = number.toString();
let min = 1* parseInt( string.padEnd(length,"0") ) ;
let max = parseInt( string.padEnd(length,"9") );
return Math.floor(
Math.random() * (max - min + 1) + min
)
}
I've created this function where you can defined the size of the OTP(One Time Password):
generateOtp = function (size) {
const zeros = '0'.repeat(size - 1);
const x = parseFloat('1' + zeros);
const y = parseFloat('9' + zeros);
const confirmationCode = String(Math.floor(x + Math.random() * y));
return confirmationCode;
}
How to use:
generateOtp(4)
generateOtp(5)
To avoid overflow, you can validate the size parameter to your case.
Numbers don't have substring method. For example:
let txt = "123456"; // Works, Cause that's a string.
let num = 123456; // Won't Work, Cause that's a number..
// let res = txt.substring(0, 3); // Works: 123
let res = num.substring(0, 3); // Throws Uncaught TypeError.
console.log(res); // Error
For Generating random 4 digit number, you can utilize Math.random()
For Example:
let randNum = (1000 + Math.random() * 9000).toFixed(0);
console.log(randNum);
This is quite simple
const arr = ["one", "Two", "Three"]
const randomNum = arr[Math.floor(Math.random() * arr.length)];
export const createOtp = (): number => {
Number(Math.floor(1000 + Math.random() * 9000).toString());
}

Convert Fraction String to Decimal?

I'm trying to create a javascript function that can take a fraction input string such as '3/2' and convert it to decimal—either as a string '1.5' or number 1.5
function ratio(fraction) {
var fraction = (fraction !== undefined) ? fraction : '1/1',
decimal = ??????????;
return decimal;
});
Is there a way to do this?
Since no one has mentioned it yet there is a quick and dirty solution:
var decimal = eval(fraction);
Which has the perks of correctly evaluating all sorts of mathematical strings.
eval("3/2") // 1.5
eval("6") // 6
eval("6.5/.5") // 13, works with decimals (floats)
eval("12 + 3") // 15, you can add subtract and multiply too
People here will be quick to mention the dangers of using a raw eval but I submit this as the lazy mans answer.
Here is the bare bones minimal code needed to do this:
var a = "3/2";
var split = a.split('/');
var result = parseInt(split[0], 10) / parseInt(split[1], 10);
alert(result); // alerts 1.5
JsFiddle: http://jsfiddle.net/XS4VE/
Things to consider:
division by zero
if the user gives you an integer instead of a fraction, or any other invalid input
rounding issues (like 1/3 for example)
Something like this:
bits = fraction.split("/");
return parseInt(bits[0],10)/parseInt(bits[1],10);
I have a function I use to handle integers, mixed fractions (including unicode vulgar fraction characters), and decimals. Probably needs some polishing but it works for my purpose (recipe ingredient list parsing).
NPM
GitHub
Inputs "2 1/2", "2½", "2 ½", and "2.5" will all return 2.5. Examples:
var numQty = require("numeric-quantity");
numQty("1 1/4") === 1.25; // true
numQty("3 / 4") === 0.75; // true
numQty("¼" ) === 0.25; // true
numQty("2½") === 2.5; // true
numQty("¾") === 0.75; // true
numQty("⅓") === 0.333; // true
numQty("⅔") === 0.667; // true
One thing it doesn't handle is decimals within the fraction, e.g. "2.5 / 5".
I created a nice function to do just that, everything was based off of this question and answers but it will take the string and output the decimal value but will also output whole numbers as well with out errors
https://gist.github.com/drifterz28/6971440
function toDeci(fraction) {
fraction = fraction.toString();
var result,wholeNum=0, frac, deci=0;
if(fraction.search('/') >=0){
if(fraction.search('-') >=0){
wholeNum = fraction.split('-');
frac = wholeNum[1];
wholeNum = parseInt(wholeNum,10);
}else{
frac = fraction;
}
if(fraction.search('/') >=0){
frac = frac.split('/');
deci = parseInt(frac[0], 10) / parseInt(frac[1], 10);
}
result = wholeNum+deci;
}else{
result = fraction
}
return result;
}
/* Testing values / examples */
console.log('1 ',toDeci("1-7/16"));
console.log('2 ',toDeci("5/8"));
console.log('3 ',toDeci("3-3/16"));
console.log('4 ',toDeci("12"));
console.log('5 ',toDeci("12.2"));
Too late, but can be helpful:
You can use Array.prototype.reduce instead of eval
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/Reduce
ES6
const fractionStrToDecimal = str => str.split('/').reduce((p, c) => p / c);
console.log(fractionStrToDecimal('1/4/2')); // Logs 0.125
console.log(fractionStrToDecimal('3/2')); // Logs 1.5
CJS
function fractionStrToDecimal(str) {
return str.split('/').reduce((p, c) => p / c);
}
console.log(fractionStrToDecimal('1/4')); // Logs 0.25
[EDIT] Removed reducer initial value and now the function works for numerators greater than 1. Thanks, James Furey.
Function (ES6):
function fractionToDecimal(fraction) {
return fraction
.split('/')
.reduce((numerator, denominator, i) =>
numerator / (i ? denominator : 1)
);
}
Function (ES6, condensed):
function fractionToDecimal(f) {
return f.split('/').reduce((n, d, i) => n / (i ? d : 1));
}
Examples:
fractionToDecimal('1/2'); // 0.5
fractionToDecimal('5/2'); // 2.5
fractionToDecimal('1/2/2'); // 0.25
fractionToDecimal('10/5/10'); // 0.2
fractionToDecimal('0/1'); // 0
fractionToDecimal('1/0'); // Infinity
fractionToDecimal('cat/dog'); // NaN
With modern destructuring syntax, the best/safest answer can be simplified to:
const parseFraction = fraction => {
const [numerator, denominator] = fraction.split('/').map(Number);
return numerator / denominator;
}
// example
parseFraction('3/2'); // 1.5
In other words, split the faction by its / symbol, turn both resulting strings into numbers, then return the first number divided by the second ...
... all with only two (very readable) lines of code.
If you don't mind using an external library, math.js offers some useful functions to convert fractions to decimals as well as perform fractional number arithmetic.
console.log(math.number(math.fraction("1/3"))); //returns 0.3333333333333333
console.log(math.fraction("1/3") * 9) //returns 3
<script src="https://cdnjs.cloudflare.com/ajax/libs/mathjs/3.20.1/math.js"></script>
const fractionStringToNumber = s => s.split("/").map(s => Number(s)).reduce((a, b) => a / b);
console.log(fractionStringToNumber("1/2"));
console.log(fractionStringToNumber("1/3"));
console.log(fractionStringToNumber("3/2"));
console.log(fractionStringToNumber("3/1"));
console.log(fractionStringToNumber("22/7"));
console.log(fractionStringToNumber("355 / 113"));
console.log(fractionStringToNumber("8/4/2"));
console.log(fractionStringToNumber("3")); // => 3, not "3"
From a readability, step through debugging perspective, this may be easier to follow:
// i.e. '1/2' -> .5
// Invalid input returns 0 so impact on upstream callers are less likely to be impacted
function fractionToNumber(fraction = '') {
const fractionParts = fraction.split('/');
const numerator = fractionParts[0] || '0';
const denominator = fractionParts[1] || '1';
const radix = 10;
const number = parseInt(numerator, radix) / parseInt(denominator, radix);
const result = number || 0;
return result;
}
To convert a fraction to a decimal, just divide the top number by the bottom number. 5 divided by 3 would be 5/3 or 1.67. Much like:
function decimal(top,bottom) {
return (top/bottom)
}
Hope this helps, haha
It works with eval() method but you can use parseFloat method. I think it is better!
Unfortunately it will work only with that kind of values - "12.2" not with "5/8", but since you can handle with calculation I think this is good approach!
If you want to use the result as a fraction and not just get the answer from the string, a library like https://github.com/infusion/Fraction.js would do the job quite well.
var f = new Fraction("3/2");
console.log(f.toString()); // Returns string "1.5"
console.log(f.valueOf()); // Returns number 1.5
var g = new Fraction(6.5).div(.5);
console.log(f.toString()); // Returns string "13"
Also a bit late to the party, but an alternative to eval() with less security issues (according to MDN at least) is the Function() factory.
var fraction = "3/2";
console.log( Function("return (" + fraction + ");")() );
This would output the result "1.5" in the console.
Also as a side note: Mixed fractions like 1 1/2 will not work with neither eval() nor the solution with Function() as written as they both stumble on the space.
safer eval() according to MDN
const safeEval = (str) => {
return Function('"use strict";return (' + str + ")")();
}
safeEval("1 1/2") // 1.5
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/eval#Do_not_ever_use_eval!
This too will work:
let y = "2.9/59"
let a = y.split('')
let b = a.splice(a.indexOf("/"))
console.log(parseFloat(a.join('')))
a = parseFloat(a.join(''))
console.log(b)
let c = parseFloat(b.slice(1).join(''))
let d = a/c
console.log(d) // Answer for y fraction
I developed a function to convert a value using a factor that may be passed as a fraction of integers or decimals. The user input and conversion factor might not be in the correct format, so it checks for the original value to be a number, as well as that the conversion can be converted to a fraction assuming that /number means 1/number, or there are a numerator and a denominator in the format number/number.
/**
* Convert value using conversion factor
* #param {float} value - number to convert
* #param {string} conversion - factor
* #return {float} converted value
*/
function convertNumber(value, conversion) {
try {
let numberValue = eval(value);
if (isNaN(numberValue)) {
throw value + " is not a number.";
}
let fraction = conversion.toString();
let divider = fraction.indexOf("/");
let upper = 1;
let denominator = 1;
if (divider == -1) {
upper = eval(fraction);
} else {
let split = fraction.split("/");
if (split.length > 2) {
throw fraction + " cannot be evaluated to a fraction.";
} else {
denominator = eval(split[1]);
if (divider > 0) {
upper = eval(split[0]);
}
}
}
let factor = upper/denominator;
if (isNaN(factor)) {
throw fraction + " cannot be converted to a factor.";
}
let result = numberValue * factor;
if (isNaN(result)) {
throw numberValue + " * " + factor + " is not a number.";
}
return result
} catch (err) {
let message = "Unable to convert '" + value + "' using '" + conversion + "'. " + err;
throw message;
}
}
You can use eval() with regex to implement a secure method to calculate fraction
var input = "1/2";
return input.match(/^[0-9\/\.]+$/) != null ? eval(input) : "invalid number";

How to format a float in javascript?

In JavaScript, when converting from a float to a string, how can I get just 2 digits after the decimal point? For example, 0.34 instead of 0.3445434.
There are functions to round numbers. For example:
var x = 5.0364342423;
print(x.toFixed(2));
will print 5.04.
EDIT:
Fiddle
var result = Math.round(original*100)/100;
The specifics, in case the code isn't self-explanatory.
edit: ...or just use toFixed, as proposed by Tim Büthe. Forgot that one, thanks (and an upvote) for reminder :)
Be careful when using toFixed():
First, rounding the number is done using the binary representation of the number, which might lead to unexpected behaviour. For example
(0.595).toFixed(2) === '0.59'
instead of '0.6'.
Second, there's an IE bug with toFixed(). In IE (at least up to version 7, didn't check IE8), the following holds true:
(0.9).toFixed(0) === '0'
It might be a good idea to follow kkyy's suggestion or to use a custom toFixed() function, eg
function toFixed(value, precision) {
var power = Math.pow(10, precision || 0);
return String(Math.round(value * power) / power);
}
One more problem to be aware of, is that toFixed() can produce unnecessary zeros at the end of the number.
For example:
var x=(23-7.37)
x
15.629999999999999
x.toFixed(6)
"15.630000"
The idea is to clean up the output using a RegExp:
function humanize(x){
return x.toFixed(6).replace(/\.?0*$/,'');
}
The RegExp matches the trailing zeros (and optionally the decimal point) to make sure it looks good for integers as well.
humanize(23-7.37)
"15.63"
humanize(1200)
"1200"
humanize(1200.03)
"1200.03"
humanize(3/4)
"0.75"
humanize(4/3)
"1.333333"
var x = 0.3445434
x = Math.round (x*100) / 100 // this will make nice rounding
The key here I guess is to round up correctly first, then you can convert it to String.
function roundOf(n, p) {
const n1 = n * Math.pow(10, p + 1);
const n2 = Math.floor(n1 / 10);
if (n1 >= (n2 * 10 + 5)) {
return (n2 + 1) / Math.pow(10, p);
}
return n2 / Math.pow(10, p);
}
// All edge cases listed in this thread
roundOf(95.345, 2); // 95.35
roundOf(95.344, 2); // 95.34
roundOf(5.0364342423, 2); // 5.04
roundOf(0.595, 2); // 0.60
roundOf(0.335, 2); // 0.34
roundOf(0.345, 2); // 0.35
roundOf(551.175, 2); // 551.18
roundOf(0.3445434, 2); // 0.34
Now you can safely format this value with toFixed(p).
So with your specific case:
roundOf(0.3445434, 2).toFixed(2); // 0.34
There is a problem with all those solutions floating around using multipliers. Both kkyy and Christoph's solutions are wrong unfortunately.
Please test your code for number 551.175 with 2 decimal places - it will round to 551.17 while it should be 551.18 ! But if you test for ex. 451.175 it will be ok - 451.18. So it's difficult to spot this error at a first glance.
The problem is with multiplying: try 551.175 * 100 = 55117.49999999999 (ups!)
So my idea is to treat it with toFixed() before using Math.round();
function roundFix(number, precision)
{
var multi = Math.pow(10, precision);
return Math.round( (number * multi).toFixed(precision + 1) ) / multi;
}
If you want the string without round you can use this RegEx (maybe is not the most efficient way... but is really easy)
(2.34567778).toString().match(/\d+\.\d{2}/)[0]
// '2.34'
function trimNumber(num, len) {
const modulu_one = 1;
const start_numbers_float=2;
var int_part = Math.trunc(num);
var float_part = String(num % modulu_one);
float_part = float_part.slice(start_numbers_float, start_numbers_float+len);
return int_part+'.'+float_part;
}
There is no way to avoid inconsistent rounding for prices with x.xx5 as actual value using either multiplication or division. If you need to calculate correct prices client-side you should keep all amounts in cents. This is due to the nature of the internal representation of numeric values in JavaScript. Notice that Excel suffers from the same problems so most people wouldn't notice the small errors caused by this phenomen. However errors may accumulate whenever you add up a lot of calculated values, there is a whole theory around this involving the order of calculations and other methods to minimize the error in the final result. To emphasize on the problems with decimal values, please note that 0.1 + 0.2 is not exactly equal to 0.3 in JavaScript, while 1 + 2 is equal to 3.
Maybe you'll also want decimal separator? Here is a function I just made:
function formatFloat(num,casasDec,sepDecimal,sepMilhar) {
if (num < 0)
{
num = -num;
sinal = -1;
} else
sinal = 1;
var resposta = "";
var part = "";
if (num != Math.floor(num)) // decimal values present
{
part = Math.round((num-Math.floor(num))*Math.pow(10,casasDec)).toString(); // transforms decimal part into integer (rounded)
while (part.length < casasDec)
part = '0'+part;
if (casasDec > 0)
{
resposta = sepDecimal+part;
num = Math.floor(num);
} else
num = Math.round(num);
} // end of decimal part
while (num > 0) // integer part
{
part = (num - Math.floor(num/1000)*1000).toString(); // part = three less significant digits
num = Math.floor(num/1000);
if (num > 0)
while (part.length < 3) // 123.023.123 if sepMilhar = '.'
part = '0'+part; // 023
resposta = part+resposta;
if (num > 0)
resposta = sepMilhar+resposta;
}
if (sinal < 0)
resposta = '-'+resposta;
return resposta;
}
/** don't spend 5 minutes, use my code **/
function prettyFloat(x,nbDec) {
if (!nbDec) nbDec = 100;
var a = Math.abs(x);
var e = Math.floor(a);
var d = Math.round((a-e)*nbDec); if (d == nbDec) { d=0; e++; }
var signStr = (x<0) ? "-" : " ";
var decStr = d.toString(); var tmp = 10; while(tmp<nbDec && d*tmp < nbDec) {decStr = "0"+decStr; tmp*=10;}
var eStr = e.toString();
return signStr+eStr+"."+decStr;
}
prettyFloat(0); // "0.00"
prettyFloat(-1); // "-1.00"
prettyFloat(-0.999); // "-1.00"
prettyFloat(0.5); // "0.50"
I use this code to format floats. It is based on toPrecision() but it strips unnecessary zeros. I would welcome suggestions for how to simplify the regex.
function round(x, n) {
var exp = Math.pow(10, n);
return Math.floor(x*exp + 0.5)/exp;
}
Usage example:
function test(x, n, d) {
var rounded = rnd(x, d);
var result = rounded.toPrecision(n);
result = result.replace(/\.?0*$/, '');
result = result.replace(/\.?0*e/, 'e');
result = result.replace('e+', 'e');
return result;
}
document.write(test(1.2000e45, 3, 2) + '=' + '1.2e45' + '<br>');
document.write(test(1.2000e+45, 3, 2) + '=' + '1.2e45' + '<br>');
document.write(test(1.2340e45, 3, 2) + '=' + '1.23e45' + '<br>');
document.write(test(1.2350e45, 3, 2) + '=' + '1.24e45' + '<br>');
document.write(test(1.0000, 3, 2) + '=' + '1' + '<br>');
document.write(test(1.0100, 3, 2) + '=' + '1.01' + '<br>');
document.write(test(1.2340, 4, 2) + '=' + '1.23' + '<br>');
document.write(test(1.2350, 4, 2) + '=' + '1.24' + '<br>');
countDecimals = value => {
if (Math.floor(value) === value) return 0;
let stringValue = value.toString().split(".")[1];
if (stringValue) {
return value.toString().split(".")[1].length
? value.toString().split(".")[1].length
: 0;
} else {
return 0;
}
};
formatNumber=(ans)=>{
let decimalPlaces = this.countDecimals(ans);
ans = 1 * ans;
if (decimalPlaces !== 0) {
let onePlusAns = ans + 1;
let decimalOnePlus = this.countDecimals(onePlusAns);
if (decimalOnePlus < decimalPlaces) {
ans = ans.toFixed(decimalPlaces - 1).replace(/\.?0*$/, "");
} else {
let tenMulAns = ans * 10;
let decimalTenMul = this.countDecimals(tenMulAns);
if (decimalTenMul + 1 < decimalPlaces) {
ans = ans.toFixed(decimalPlaces - 1).replace(/\.?0*$/, "");
}
}
}
}
I just add 1 to the value and count the decimal digits present in the original value and the added value. If I find the decimal digits after adding one less than the original decimal digits, I just call the toFixed() with (original decimals - 1). I also check by multiplying the original value by 10 and follow the same logic in case adding one doesn't reduce redundant decimal places.
A simple workaround to handle floating-point number rounding in JS. Works in most cases I tried.

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