We Keep Coding

Applying currency format using replace and a regular expression

I am trying to understand some code where a number is converted to a currency format. Thus, if you have 16.9 it converts to $16.90. The problem with the code is if you have an amount over $1,000, it just returns $1, an amount over $2,000 returns $2, etc. Amounts in the hundreds show up fine.
Here is the function:
var _formatCurrency = function(amount) {
return "$" + parseFloat(amount).toFixed(2).replace(/(\d)(?=(\d{3})+\.)/g, '$1,')
};
(The reason the semicolon is after the bracket is because this function is in itself a statement in another function. That function is not relevant to this discussion.)
I found out that the person who originally put the code in there found it somewhere but didn't fully understand it and didn't test this particular scenario. I myself have not dealt much with regular expressions. I am not only trying to fix it, but to understand how it is working as it is now.
Here's what I've found out. The code between the backslash after the open parenthesis and the backslash before the g is the pattern. The g means global search. The \d means digit, and the (?=\d{3})+\. appears to mean find 3 digits plus a decimal point. I'm not sure I have that right, though, because if that was correct shouldn't it ignore numbers like 5.4? That works fine. Also, I'm not sure what the '$1,' is for. It looks to me like it is supposed to be placed where the digits are, but wouldn't that change all the numbers to $1? Also, why is there a comma after the 1?

Regarding your comment
I was hoping to just edit the regex so it would work properly.
The regex you are currently using is obviously not working for you so I think you should consider alternatives even if they are not too similar, and
Trying to keep the code change as small as possible
Understandable but sometimes it is better to use a code that is a little bit bigger and MORE READABLE than to go with compact and hieroglyphical.
Back to business:
I'm assuming you are getting a string as an argument and this string is composed only of digits and may or may not have a dot before the last 1 or 2 digts. Something like
//input //intended output
1 $1.00
20 $20.00
34.2 $34.20
23.1 $23.10
62516.16 $62,516.16
15.26 $15.26
4654656 $4,654,656.00
0.3 $0.30
I will let you do a pre-check of (assumed) non-valids like 1. | 2.2. | .6 | 4.8.1 | 4.856 | etc.
Proposed solution:
var _formatCurrency = function(amount) {
amount = "$" + amount.replace(/(\d)(?=(\d{3})+(\.(\d){0,2})*$)/g, '$1,');
if(amount.indexOf('.') === -1)
return amount + '.00';
var decimals = amount.split('.')[1];
return decimals.length < 2 ? amount + '0' : amount;
};
Regex break down:
(\d): Matches one digit. Parentheses group things for referencing when needed.
(?=(\d{3})+(\.(\d){0,2})*$). Now this guy. From end to beginning:
$: Matches the end of the string. This is what allows you to match from the end instead of the beginning which is very handy for adding the commas.
(\.(\d){0,2})*: This part processes the dot and decimals. The \. matches the dot. (\d){0,2} matches 0, 1 or 2 digits (the decimals). The * implies that this whole group can be empty.
?=(\d{3})+: \d{3} matches 3 digits exactly. + means at least one occurrence. Finally ?= matches a group after the main expression without including it in the result. In this case it takes three digits at a time (from the end remember?) and leaves them out of the result for when replacing.
g: Match and replace globally, the whole string.
Replacing with $1,: This is how captured groups are referenced for replacing, in this case the wanted group is number 1. Since the pattern will match every digit in the position 3n+1 (starting from the end or the dot) and catch it in the group number 1 ((\d)), then replacing that catch with $1, will effectively add a comma after each capture.
Try it and please feedback.
Also if you haven't already you should (and SO has not provided me with a format to stress this enough) really really look into this site as suggested by Taplar

The pattern is invalid, and your understanding of the function is incorrect. This function formats a number in a standard US currency, and here is how it works:
The parseFloat() function converts a string value to a decimal number.
The toFixed(2) function rounds the decimal number to 2 digits after the decimal point.
The replace() function is used here to add the thousands spearators (i.e. a comma after every 3 digits). The pattern is incorrect, so here is a suggested fix /(\d)(?=(\d{3})+\.)/g and this is how it works:
The (\d) captures a digit.
The (?=(\d{3})+\.) is called a look-ahead and it ensures that the captured digit above has one set of 3 digits (\d{3}) or more + followed by the decimal point \. after it followed by a decimal point.
The g flag/modifier is to apply the pattern globally, that is on the entire amount.
The replacement $1, replaces the pattern with the first captured group $1, which is in our case the digit (\d) (so technically replacing the digit with itself to make sure we don't lose the digit in the replacement) followed by a comma ,. So like I said, this is just to add the thousands separator.
Here are some tests with the suggested fix. Note that it works fine with numbers and strings:
var _formatCurrency = function(amount) {
return "$" + parseFloat(amount).toFixed(2).replace(/(\d)(?=(\d{3})+\.)/g, '$1,');
};
console.log(_formatCurrency('1'));
console.log(_formatCurrency('100'));
console.log(_formatCurrency('1000'));
console.log(_formatCurrency('1000000.559'));
console.log(_formatCurrency('10000000000.559'));
console.log(_formatCurrency(1));
console.log(_formatCurrency(100));
console.log(_formatCurrency(1000));
console.log(_formatCurrency(1000000.559));
console.log(_formatCurrency(10000000000.559));

Okay, I want to apologize to everyone who answered. I did some further tracing and found out the JSON call which was bringing in the amount did in fact have a comma in it, so it is just parsing that first digit. I was looking in the wrong place in the code when I thought there was no comma in there already. I do appreciate everyone's input and hope you won't think too bad of me for not catching that before this whole exercise. If nothing else, at least I now know how that regex operates so I can make use of it in the future. Now I just have to go about removing that comma.
Have a great day!

Assuming that you are working with USD only, then this should work for you as an alternative to Regular Expressions. I have also included a few tests to verify that it is working properly.
var test1 = '16.9';
var test2 = '2000.5';
var test3 = '300000.23';
var test4 = '3000000.23';
function stringToUSD(inputString) {
const splitValues = inputString.split('.');
const wholeNumber = splitValues[0].split('')
.map(val => parseInt(val))
.reverse()
.map((val, idx, arr) => idx !== 0 && (idx + 1) % 3 === 0 && arr[idx + 1] !== undefined ? `,${val}` : val)
.reverse()
.join('');
return parseFloat(`${wholeNumber}.${splitValues[1]}`).toFixed(2);
}
console.log(stringToUSD(test1));
console.log(stringToUSD(test2));
console.log(stringToUSD(test3));
console.log(stringToUSD(test4));

Need a RegExp to filter out all but one decimal point

I'm using the following code to negate the characters in the regexp. By checking the inverse, I can determine if the value entered is correctly formatted. Essentially, any digit can be allowed but only one decimal point (placed anywhere in the string.) The way I have it now, it catches all numerals, but allows for multiple decimal points (creating invalid floats.) How can I adjust this to catch more than one decimal points (since I only want to allow for one)?
var regex = new RegExp(/[^0-9\.]/g);
var containsNonNumeric = this.value.match(regex);
if(containsNonNumeric){
this.value = this.value.replace(regex,'');
return false;
}
Here is what I'm expecting to happen:
First, valid input would be any number of numerals with the possibility of only one decimal point. The current behavior: The user enters characters one by one, if they are valid characters they will show up. If the character is invalid (e.g. the letter A) the field will replace that character with ''(essentially behaving like a backspace immediately after filling the character in. What I need is the same behavior for the addition of one too many decimal points.

As I understand your question the code below might be what you are looking for:
var validatedStr=str.replace(/[^0-9.]|\.(?=.*\.)/g, "");
It replaces all characters other then numbers and dot (.), then it replaces all dots followed by any number of 0-9 characters followed by dot.
EDIT based on first comment - the solution above erases all dots but the last, the author wants to erase all but the first one:
Since JS does not support "look behind", the solution might be to reverse string before regex, then reverse it again or to use this regex:
var counter=0;
var validatedStr=str.replace(/[^0-9.]|\./g, function($0){
if( $0 == "." && !(counter++) ) // dot found and counter is not incremented
return "."; // that means we met first dot and we want to keep it
return ""; // if we find anything else, let's erase it
});
JFTR: counter++ only executes if the first part of condition is true, so it works even for strings beginning with letters

Building upon the original regex from #Jan Legner with a pair of string reversals to work around the look behind behavior. Succeeds at keeping the first decimal point.
Modified with an attempt to cover negatives as well. Can't handle negative signs that are out of place and special cases that should logically return zero.
let keep_first_decimal = function(s) {
return s.toString().split('').reverse().join('').replace(/[^-?0-9.]|\.(?=.*\.)/g, '').split('').reverse().join('') * 1;
};
//filters as expected
console.log(keep_first_decimal("123.45.67"));
console.log(keep_first_decimal(123));
console.log(keep_first_decimal(123.45));
console.log(keep_first_decimal("123"));
console.log(keep_first_decimal("123.45"));
console.log(keep_first_decimal("a1b2c3d.e4f5g"));
console.log(keep_first_decimal("0.123"));
console.log(keep_first_decimal(".123"));
console.log(keep_first_decimal("0.123.45"));
console.log(keep_first_decimal("123."));
console.log(keep_first_decimal("123.0"));
console.log(keep_first_decimal("-123"));
console.log(keep_first_decimal("-123.45.67"));
console.log(keep_first_decimal("a-b123.45.67"));
console.log(keep_first_decimal("-ab123"));
console.log(keep_first_decimal(""));
//NaN, should return zero?
console.log(keep_first_decimal("."));
console.log(keep_first_decimal("-"));
//NaN, can't handle minus sign after first character
console.log(keep_first_decimal("-123.-45.67"));
console.log(keep_first_decimal("123.-45.67"));
console.log(keep_first_decimal("--123"));
console.log(keep_first_decimal("-a-b123"));

JavaScript Regex to capture repeating part of decimal

Looking for the best way to take an arbitrary number with potentially repeating decimal part, and discover the repeating part (if present).
Ultimately, I need to decorate the number with overline notation (either with css text-decoration or MathML mline), so I need to know the index of where the repetition begins also.
So I need regex that will get me (or can be used in an algorithm to get) the following results:
1.333 // result: {"pattern": 3, index: 0}
1.5444 // result: {"pattern": 4, index: 1}
1.123123 // result: {"pattern": 123, index: 0}
1.5432121212 // result: {"pattern": 12, index: 4}
1.321 // result: null
1.44212 // result: null
Additional Example (from comments):
1.3333 // result: { "pattern": 3, index: 0}

function getRepetend(num) {
var m = (num+'').match(/\.(\d*?)(\d+?)\2+$/);
return m && {pattern: +m[2], index: m[1].length};
}
It works like this:
First, convert the number to string in order to be able to use regular expressions.
Then, match this regex: /\.(\d*?)(\d+)\2+$/:
\. matches the decimal dot.
(\d*?) matches the digits between the decimal dot and the repetend, and captures the result into backreference number 1.
(\d+?) matches the repetend, and captures it into backreference number 2.
\2+ matches repetitions of the repetend.
$ matches end of string.
Finally, if the match is null (i.e. there is no match), return null.
Otherwise, return an object that contains the repetend (backreference 2) converted to number, and the number of digits between the dot and the repetend (backreference 1).

You could try something like this:
(\d+?)\1+$
http://regex101.com/r/eX8eC3/3
It matched some number of digits and then uses a backreference to try and match the same set immediately afterwards 1 or more times. It's anchored at the end of the string because otherwise it'll be tripped up by, for example:
1.5432121212
It would see the 21 repeating instead of the 12.
Adding ? to the first group to make it non-greedy should fix the problem with 1.3333 as raised by Louis.

You can use this regex with RexExp#exec and use result.index in the resulting object:
var re = /(\d+)\1$/;
var s = '.5439876543212211211';
var result = re.exec( s );
console.log ( result.index );
//=> 14
console.log ( result[1] );
//=> 211
JsFiddle Demo

(.+)(?:\1)+$
Try this.See demo.
http://regex101.com/r/uH3tP3/10

The accepeted answer is OKish as per the given examples in the question are concerned. However if one day you find yourself here what you probably need is exactly what the topic says.
JavaScript Regex to capture repeating part of decimal
So you have the floating part of a string and you want to know if it is repeating or not. The accepted answer fails in practice. You can never guarantee that the string ends when the repeating part ends. Most of the time the string ends with a portion of the repeating part or sometimes *thanks* to the double precision errors jumps to irrelevant figures towards the end. So my suggestion is
/(\d+)\1+(?=\d*$)/g
Now this is not a silver bullet. It's helpful but won't protect you from vampires like 3.1941070707811985 which happens to have no repetend at all. In order to feel it you have to develop deeper mechanisms. However in most cases it's just fine in percieving the repend like in
3.1941070707811985 // 07 which is wrong so prove it later
7.16666666810468 // 666 but reduce it to 6 later
3.00000000000001 // 000000 but reduce it to "" later
0.008928571428571428 // 285714 just fine, do nothing
It is not an easy task to find if the floating part of a decimal has repetends or not in this environment. Most possibly you need to do further processing on the given string and the result of the regex for futher reduction / decision.

How to check if a string contains a number in JavaScript?

I don't get how hard it is to discern a string containing a number from other strings in JavaScript.
Number('') evaluates to 0, while '' is definitely not a number for humans.
parseFloat enforces numbers, but allow them to be tailed by abitrary text.
isNaN evaluates to false for whitespace strings.
So what is the programatically function for checking if a string is a number according to a simple and sane definition what a number is?

By using below function we can test whether a javascript string contains a number or not. In above function inplace of t, we need to pass our javascript string as a parameter, then the function will return either true or false
function hasNumbers(t)
{
var regex = /\d/g;
return regex.test(t);
}

If you want something a little more complex regarding format, you could use regex, something like this:
var pattern = /^(0|[1-9][0-9]{0,2}(?:(,[0-9]{3})*|[0-9]*))(\.[0-9]+){0,1}$/;
Demo
I created this regex while answering a different question awhile back (see here). This will check that it is a number with atleast one character, cannot start with 0 unless it is 0 (or 0.[othernumbers]). Cannot have decimal unless there are digits after the decimal, may or may not have commas.. but if it does it makes sure they are 3 digits apart, etc. Could also add a -? at the beginning if you want to allow negative numbers... something like:
/^(-)?(0|[1-9][0-9]{0,2}(?:(,[0-9]{3})*|[0-9]*))(\.[0-9]+){0,1}$/;

There's this simple solution :
var ok = parseFloat(s)==s;
If you need to consider "2 " as not a number, then you might use this one :
var ok = !!(+s==s && s.length && s.trim()==s);

You can always do:
function isNumber(n)
{
if (n.trim().length === 0)
return false;
return !isNaN(n);
}

Let's try
""+(+n)===n
which enforces a very rigid canonical way of the number.
However, such number strings can be created by var n=''+some_number by JS reliable.
So this solution would reject '.01', and reject all simple numbers that JS would stringify with exponent, also reject all exponential representations that JS would display with mantissa only. But as long we stay in integer and low float number ranges, it should work with otherwise supplied numbers to.

No need to panic just use this snippet if name String Contains only numbers or text.
try below.
var pattern = /^([^0-9]*)$/;
if(!YourNiceVariable.value.match(pattern)) {//it happen while Name Contains only Charectors.}
if(YourNiceVariable.value.match(pattern)) {//it happen while Name Contains only Numbers.}

This might be insane depending on the length of your string, but you could split it into an array of individual characters and then test each character with isNaN to determine if it's a number or not.

A very short, wrong but correctable answer was just deleted. I just could comment it, besides it was very cool! So here the corrected term again:
n!=='' && +n==n'
seems good. The first term eliminates the empty string case, the second one enforces the string interpretataion of a number created by numeric interpretation of the string to match the string. As the string is not empty, any tolerated character like whitespaces are removed, so we check if they were present.

Regex x(?!n) with optional characters in the x selector

So I have a javascript program that solves for 1 variable. I'm coming to a roadblock when selecting numbers that DON'T have a variable associated with them.
Here is my current regex expression:
(\+|-)?([0-9]+)(\.[0-9]+)?(?![a-z])
takes input like 15000.53=1254b+21
and returns [15000.53, 125, +21], when it should return [15000.53, +21] (yes, the + is supposed to be there)
I know why it is happening. The number of digits is optional so the function can handle large numbers and floats, but they are optional, so it is hard to make sure the entire number is selected. The result of this is selecting all the digits of the number EXCEPT the one directly next to the variable.
Anyone know of a way for the number of digits to stay optional, yet still make sure a variable doesn't follow the number? Thanks!
var reg = (\+|-)?([0-9]+)(\.[0-9]+)?(?![a-z]);
var numbers = [];
var equation = '15000.53=1254b+21';
while (aloneInt = reg.exec(side[0])) {
numbers.push(aloneInt[0]);
}

Try the following expression:
(?![+-]?[0-9.]+[a-z])(\+|-)?([0-9]+)(\.[0-9]+)?
The added negative lookahead (?![+-]?[0-9.]+[a-z]) makes sure there isn't one or more optionally signed floating point numbers that are followed by a letter from the alphabet.
In other words, it makes sure there isn't a number followed by a variable name, then it matches the number.
Regex101 Demo