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));

JS Regex: how to remove first 2 zero's from "08/08/2017"

I'm new to regex, and have been researching all night how to remove the first 2 zeros from a string like "08/08/2017" (without removing 0 in "2017")
The 5+ regex tutorials I've reviewed do not seem to cover what I need here.
The date could be any sysdate returned from the system. So the regex also needs to work for "12/12/2017"
Here is the best I have come up with:
let sysdate = "08/08/2017"
let todayminuszero = str.replace("0","");
let today = todayminus0.replace("0","");
It works, but obviously it's unprofessional.
From the tutorials, I'm pretty sure I can do something along the lines of this:
str.replace(/\d{2}//g,""),);
This pattern would avoid getting the 3rd zero in str.
Replacement String would have to indicate 8/8/
Not sure how to write this though.

For date manipulation I would use other functions(best date related) but, this should do it, for the case that you stated. If you need other formats or so, I would suggest removing the zeros in an different way, but It all depends on you UseCase.
let sysdate = "08/08/2017";
let todayminuszero = sysdate.replace(/0(?=\d\/)/gi,"");
console.info(todayminuszero);
(?= ... ) is called Lookahead and with this you can see what is there, without replacing it
in this case we are checking for a number and a slash. (?=\d\/)
here some more information, if you want to read about lookahead and more http://www.regular-expressions.info/lookaround.html
A good place to test regex expressions is https://regex101.com/
I always use this for more advance expressions, since it displays all matching groups and so, with a great explaination. Great resource/help, if you are learning or creating difficult Expressions.
Info: as mentioned by Rajesh, the i flag is not needed for this Expression, I just use it out of personal preference. This flag just sets the expression-match to case insensitive.
-- Out of Scope, but may be interesting --
A longer solution without regex could look like this:
let sysdate = "08/08/2017";
let todayminuszero = sysdate.split("/").map(x => parseInt(x)).join("/");
console.info(todayminuszero);
Backside, this solution has many moving parts, the split function to make an array(´"08/08/2017"´ to ´["08", "08", "2017"]´), the map function, with a lambda function => and the parseInt function, to make out of each string item a nice integer (like: "08" to 8, ... ) and at last the join function that creates the final string out of the newly created integer array.

you should use this
let sysdate = "08/08/2017"
let todayminuszero = sysdate.replace(/(^|\/)0/g,"$1");
console.log(todayminuszero);

function stripLeadingZerosDate(dateStr){
return dateStr.split('/').reduce(function(date, datePart){
return date += parseInt(datePart) + '/'
}, '').slice(0, -1);
}
console.log(stripLeadingZerosDate('01/02/2016'));
console.log(stripLeadingZerosDate('2016/02/01'));
look at here
function stripLeadingZerosDate(dateStr){
return dateStr.split('/').reduce(function(date, datePart){
return date += parseInt(datePart) + '/'
}, '').slice(0, -1);
}
console.log(stripLeadingZerosDate('01/02/2016'));// 1/2/2016
console.log(stripLeadingZerosDate('2016/02/01'));// "2016/2/1"

By first 2 zeros, I understand you mean zero before 8 in month and in date.
You can try something like this:
Idea
Create a regex that captures group of number representing date, month and year.
Use this regex to replace values.
Use a function to return processed value.
var sysdate = "08/08/2017"
var numRegex = /(\d)+/g;
var result = sysdate.replace(numRegex, function(match){
return parseInt(match)
});
console.log(result)

Check dateformat with regex in JS

My question is simple but takes work. I tried lots of regex expressions to check my datetime is ok or not, but though I am sure my regex exprerssion is correct it always return to me isnotok with ALERT. Can you check my code?
validateForLongDateTime('22-03-1981')
function validateForLongDateTime(date){
var regex=new RegExp("/^\d{2}[.-/]\d{2}[.-/]\d{4}$/");
var dateOk=regex.test(date);
if(dateOk){
alert('ok');
}else{
alert('notok');
}
}

There are at least 2 issues with the regex:
It has unescaped forward slashes
The hyphen in the character classes is unescaped and forms a range (matching only . and /) that is not what is necessary here.
The "fixed" regex will look like:
/^\d{2}[.\/-]\d{2}[.\/-]\d{4}$/
See demo
However, you cannot validate dates with it since it will also match 37-67-5734.
Here is an SO post with a comprehensive regex approach that looks viable
Here is my enahanced version with a character class for the delimiter:
^(?:(?:31([\/.-])(?:0?[13578]|1[02]))\1|(?:(?:29|30)([\/.-])(?:0?[1,3-9]|1[0-2])\2))(?:(?:1[6-9]|[2-9]\d)?\d{2})$|^(?:29([\/.-])0?2\3(?:(?:(?:1[6-9]|[2-9]\d)?(?:0[48]|[2468][048]|[13579][26])|(?:(?:16|[2468][048]|[3579][26])00))))$|^(?:0?[1-9]|1\d|2[0-8])([\/.-])(?:(?:0?[1-9])|(?:1[0-2]))\4(?:(?:1[6-9]|[2-9]\d)?\d{2})$
Here is an SO post showing another approach using Date.parse

this way you can validate date between 1 to 31 and month 1 to 12
var regex = /^(0[1-9]|[12][0-9]|3[01])[- \/.](0[1-9]|1[012])[- \/.](19|20)\d\d$/
see this demo here https://regex101.com/r/xP1bD2/1

Javascript - regular expressions with a prefix followed by 6 numbers

below is a my attempt at a function to validate a form field with the prefix ZHA or zha followed by 6 numbers. The prefix part seems to be working but if I enter 1 number it still validates. Any suggestions?
function checkHnum(hnumvalue){
var authTest = /^[ZHA]|[zha]+[\d]{6}$/;
return authTest.test(hnumvalue)
}
Thanks.

Your regex doesn't accept 1 digit only but it's buggy as it, for example, doesn't constraint the order of the letters ([ZHA] is "Z or H or A"). You seem to want
var ok = /^(ZHA|zha)\d{6}$/.test(yourString)
Note that if you also want to accept "Zha123456" then you can simply use a case insensitive regular expression :
var ok = /^zha\d{6}$/i.test(yourString)

Your regex should be:
/^ZHA\d{6}$/i
Note the i to make it case insensitive. The problem with yours was mainly the brackets. The brackets match one of the characters that inside of it.
For example
[ZHA] will match Z, or H, or A, but not the full ZHA
Hope this helps. Cheers

javascript regex for region code

I have a regex problem with validation for a region code.
My region code could be only one digit but it also could be a digits separated by '-'
for Example my region code could be one of the following:
6
6-66
77-7
As you can see I must have at least one digit or digits separated by '-' and if they are separated there should be a digits after the '-' sign (does not matter how many). So 6- must not be validated as legal region code. I try 2 hours to solve this, but I couldn't, so please help me! Thank you!

/\d+(-\d+)?$/
This will match 6, 6-66,77-7, but not6-`

If what you are looking for is the whole string:
/^\d+(?:-\d+)?$/
or something like that:
if (parseInt(yourstring.split(/-/)[0])>=eval(yourstring)) alert('true');
else alert('false');
But it is more complicated :) and less efficient! And if the condition is false you code will crash!

var data = ['6', '6-66', '77-7', '6-'];
var len = data.length;
for(var i=0; i<len; ++i) {
var current = data[i];
var result = data[i].match(/^(\d+|\d+[-]\d+)$/);
if(result != null) {
console.log(current);
}
}
--output:--
6
6-66
77-7

For a quick answer you can try following:
/^([0-9])|([0-9]\-[0-9][0-9])|([0-9][0-9]\-[0-9])$/
or in case your engine support perl-styled character classes:
/^(\d)|(\d\-\d\d)|(\d\d\-\d)$/
here what it does:
between / and / resides as string defining a regular expression
\d stands for one digit it coudl also be writen as [0-9]
() defines a sub-expression, so (\d) matches your first one-digit, (\d-\d\d) second three digits style, and last (\d\d-\d) third variant of three-digit region code
| goes as "OR" like (A)|(B)|(C), so by combining previous three we will get:
/(\d)|(\d-\d\d)|(\d\d-\d)/
Finally ^ means start of string, and $ - end of string.
also there is so called BRE mode (in which you have to add "\" symbol before each parentheses), but I think it is not the case. However if you would have some free time, please consider any quick tutorial like this one.