CODEWITHSUNDEEP
javascriptregex
PengProgrammer
PengProgrammer

Reputation: 174

Regex add leading zeros to DD.MM.YYYY string

I was sure something like this had been answered before in some manner however I only managed to find reveresed versions like MM/DD/YYYY etc or special characters like / - and had little luck converting them as my regex knowledge is terrible.

The closest thread I found was this

I am struggling to find a clean approach using regex and javascript to add trailing zeros to the below string format?

dd.mm.yyyy

Eg 1.1.1998 => 01.01.1998 || 10.1.1998 => 10.01.1998

my current approach is along the lines of

const addLeadingZeros = (input: string) => {
   return input.replace(/.([1-9])(?!\d)/g, '.0$1');
}

Upvotes: 2

Views: 360

Answers (3)

Subham Bhattacharjee
Subham Bhattacharjee

Reputation: 11

You can also use this regex /^\d\.|(?<=\.)\d\./g it will select the first number and the number after the first dot, so it will only select date and month on your format.

You can use this like the code bellow

let regex = /^\d\.|(?<=\.)\d\./g;
console.log("8.8.2014".replace(regex, 
str => `0${str}`));
console.log("25.06.2014".replace(regex, 
str => `0${str}`));
console.log("9.06.2014".replace(regex, 
str => `0${str}`));
console.log("08.7.2014".replace(regex, 
str => `0${str}`));
console.log("1.1.1998".replace(regex, 
str => `0${str}`));

Now I'm gonna explain it part by part

regex

The regex is broken by a | which tells it to either match expression before it or after it.

The first part is ^\d\., if you know the basics of regex, you'll know that it means see if there is a number right after which there is a dot at the beginning, it selects the date if there is only one digit representing it.

The second part is (?<=\.)\d\., this part first sees if there is a dot before the string that is to be selected, then it sees if there is only one number after it then a dot. It will select the month if there is only one digit representing it.

Here the first part (?<=.) tells java script to seek for dot but not select it, if there is no dot before the number it won't select the string.

Now the last part g makes it select every instance of string which is matched.

So for the first example it selects the strings "8." and "8."

.replace

.replace is a string function taking two argument, first the string/regex to match, in my code it is a regex.

Second is a string or a callback that returns a string, in my code it's a callback.

The callback can get the an argument which is the matched string.

Then I add the "0" before the matched string with a template string.

So the ending output is

08.08.2014
25.06.2014
09.06.2014
08.07.2014
01.01.1998

Hope my explanation helped. I might be wrong at some places, as I'm still learning and use regex or even the callback of replace not very often. Hope there weren't many mistakes. If I'm wrong somewhere please let me know.

Upvotes: 1

kstepien
kstepien

Reputation: 1233

My solution:

const addLeadingZeros = (input: string) => {
   return input.replaceAll(/\d*(?<![\d])[\d](?=[.])/g, '0$&');
}

The regex: \d*(?<![\d])[\d](?=[.])

You can find the explanation about it here.

explanation

const addLeadingZeros = (input) => {
   return input.replaceAll(/\d*(?<![\d])[\d](?=[.])/g, '0$&');
}

console.log(addLeadingZeros('01.02.3456'))
console.log(addLeadingZeros('1.2.2001'))
console.log(addLeadingZeros('01.2.1234'))
console.log(addLeadingZeros('1.12.1998'))
console.log(addLeadingZeros('3.04.2021'))
console.log(addLeadingZeros('5.6.01'))
console.log(addLeadingZeros('5.06.01'))
console.log(addLeadingZeros('05.6.01'))
console.log(addLeadingZeros('7.8.1'))
console.log(addLeadingZeros('7.08.1'))
console.log(addLeadingZeros('07.8.1'))

Upvotes: 2

Amessihel
Amessihel

Reputation: 6424

How about:

input.replace(/(?<=^|\.)(\d)(?=\.|$)/g, '0$1');

It means: "replace all digits preceded by the beginning of a line or a dot and followed by a dot or the end of the line". This example works only for day and month parts.

Example (browser console outputs):

"01.01.1998".replace(/(?<=^|\.)(\d)(?=\.|$)/g, '0$1');
← "01.01.1998"
"1.01.1998".replace(/(?<=^|\.)(\d)(?=\.|$)/g, '0$1');
← "01.01.1998"
"1.1.1998".replace(/(?<=^|\.)(\d)(?=\.|$)/g, '0$1');
← "01.01.1998"

(Note: this is not a solution I particularly recommend, I just focused on the asked point.)

Upvotes: 2

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