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I want to suppress before two zeros and include zeros before values with two digits

I want to suppress before two zeros and include zeros before values with two digits.
How to do the logic kindly help me on this.

This can be done in one line in C#:
string stringValue = "056";
string newStringValue = int.Parse(stringValue).ToString("000");
int.Parse() will automatically strip the leading zeroes and give you an int, and the "000" format in .ToString will ensure the resulting string is always three digits and pad with leading zeroes if necessary.

In javascript (or any other language), just add enough leading zeros to ensure you have a length of 3, and then take the rightmost 3 characters:
const n = '56';
const padded = n.padStart(3,'0').slice(-3) ;
// padded = '056'
It's not much different in C#, except that you need an explicit intermediate (because you need that string's length):
string n = '56';
string t = n.PadLeft(3,'0') ;
string padded = t.SubString( t.Length-3 ) ;

Javascript solution:
/^([0]{0,})/ regex replaces all leading contiguous zeros. Then we just pad it to length 3 with an initial zero.
const numbers = ['56', '056', '0056', '00056'];
const leadingZero = (nums) => {
return nums.map((num) => {
if (!num?.length) throw new Error();
return removeContigiousLeadingZeros(num);
});
};
const removeContigiousLeadingZeros = (num) => {
return num.replace(/^([0]{0,})/, '0')
};
console.log(leadingZero(numbers));

How to print out the even numbers in reverse order?

var num = prompt("Enter an integer : ");
var reversenum = num.reverse('');
document.write(reversenum);
I want to print out the even number of integers in reverse order after inputting an integer through the prompt, but I have no idea.
Even if I try to write split(), I don't think I can separate it because the letters are attached to each other. What should I do?
The result I want is,
Enter an integer : 8541236954
46248

Based on your updated question, I suppose what you want is to extract even-valued digits from a given integer, and display them in reverse order.
Since prompt() always returns a String, you can do one of the two ways to split it into digits and reverse their order:
Old-school JS way: num.split('').reverse()
ES6 array spread way: [...num].reverse()
Then, it is just a matter of using Array.prototype.filter() to return even numbers. Even numbers can be selected based on the criteria that their modulus of 2 is always 0 (i.e. when number is divided by 2, it has a remainder of 0).
Finally, join your filtered array so you get a string again.
See proof-of-concept order below:
const num = prompt("Enter an integer : "); // e.g. try '8541236954'
const digits = [...num].reverse();
const evenDigits = digits.filter(d => d % 2 ===0);
console.log(evenDigits.join('')); // e.g. '46248'

You need to first split the string into an array of characters, reverse the array and then join it again. I have given an easy to understand code which converts every character of the reversed string to an int and checks if that integer is even, followed by concatenating it in the answer.
var num = prompt("Enter an integer : ");
var reversenum = num.split('').reverse().join('');
var ans = "";
for (var i = 0; i < reversenum.length; i++)
{
var x = parseInt(reversenum[i]);
if(x%2 === 0)
ans = ans.concat(reversenum[i]);
}
console.log(ans);

Javascript replace regex to accept only numbers, including negative ones, two decimals, replace 0s in the beginning, except if number is 0

The question became a bit long, but it explains the expected behaviour.
let regex = undefined;
const format = (string) => string.replace(regex, '');
format('0')
//0
format('00')
//0
format('02')
//2
format('-03')
//-3
format('023.2323')
//23.23
format('00023.2.3.2.3')
//23.23
In the above example you can see the expected results in comments.
To summarize. I'm looking for a regex not for test, for replace which formats a string:
removes 0s from the beginning if it's followed by any numbers
allows decimal digits, but just 2
allows negative numbers
allows decimal points, but just one (followed by min 1, max 2 decimal digits)
The last one is a bit difficult to handle as the user can't enter period at the same time, I'll have two formatter functions, one will be the input in the input field, and one for the closest valid value at the moment (for example '2.' will show '2.' in the input field, but the handler will receive the value '2').
If not big favour, I'd like to see explanation of the solution, why it works, and what's the purpose of which part.
Right now I'm having string.replace(/[^\d]+(\.\[^\d{1,2}])+|^0+(?!$)/g, ''), but it doesn't fulfill all the requirements.

You may use this code:
const arr = ['0', '00', '02', '-03', '023.2323', '00023.2.3.2.3', '-23.2.3.2.3']
var narr = []
// to remove leading zeroes
const re1 = /^([+-]?)0+?(?=\d)/
// to remove multiple decimals
const re2 = /^([+-]?\d*\.\d+)\.(\d+).*/
arr.forEach( el => {
el = el.replace(re1, '$1').replace(re2, '$1$2')
if (el.indexOf('.') >= 0)
el = Number(el).toFixed(2)
narr.push(el)
})
console.log(narr)
//=> ["0", "0", "2", "-3", "23.23", "23.23"]

If you aren't bound to the String#replace method, you can try this regex:
/^([+-])?0*(?=\d+$|\d+\.)(\d+)(?:\.(\d{1,2}))?$/
Inspect on regex101.com
It collects the parts of the number into capturing groups, as follows:
Sign: the sign of the number, +, - or undefined
Integer: the integer part of the number, without leading zeros
Decimal: the decimal part of the number, undefined if absent
This regex won't match if more then 2 decimal places present. To strip it instead, use this:
/^([+-])?0*(?=\d+$|\d+\.)(\d+)(?:\.(\d{1,2})\d*)?$/
Inspect on regex101.com
To format a number using one of the above, you can use something like:
let regex = /^([+-])?0*(?=\d+$|\d+\.)(\d+)(?:\.(\d{1,2}))?$/
const format = string => {
try{
const [, sign, integer, decimal = ''] = string.match(regex)
return `${(sign !== '-' ? '' : '-')}${integer}${(decimal && `.${decimal}`)}`
}catch(e){
//Invalid format, do something
return
}
}
console.log(format('0'))
//0
console.log(format('00'))
//0
console.log(format('02'))
//2
console.log(format('-03'))
//-3
console.log(format('023.23'))
//23.23
console.log(format('023.2323'))
//undefined (invalid format)
console.log(format('00023.2.3.2.3'))
//undefined (invalid format)
//Using the 2nd regex
regex = /^([+-])?0*(?=\d+$|\d+\.)(\d+)(?:\.(\d{1,2})\d*)?$/
console.log(format('0'))
//0
console.log(format('00'))
//0
console.log(format('02'))
//2
console.log(format('-03'))
//-3
console.log(format('023.23'))
//23.23
console.log(format('023.2323'))
//23.23
console.log(format('00023.2.3.2.3'))
//undefined (invalid format)

Another option is to use pattern with 3 capturing groups. In the replacement, use all 3 groups "$1$2$3"
If the string after the replacement is empty, return a single zero.
If the string is not empty, concat group 1, group 2 and group 3 where for group 3, remove all the dots except for the first one to keep it for the decimal and take the first 3 characters (which is the dot and 2 digits)
^([-+]?)0*([1-9]\d*)((?:\.\d+)*)|0+$
In parts
^ Start of string
( Capture group 1
[-+]? Match an optional + or -
) Close group
0* Match 0+ times a zero
( Capture group 2
[1-9]\d* Match a digit 1-9 followed by optional digits 0-9
) Close group
( Capture group 3
(?:\.\d+)* Repeat 0+ times matching a dot and a digit
) Close group
| Or
0+ Match 1+ times a zero
$ End of string
Regex demo
const strings = ['0', '00', '02', '-03', '023.2323', '00023.2.3.2.3', '-23.2.3.2.3', '00001234', '+0000100005.0001']
let pattern = /^([-+]?)0*([1-9]\d*)((?:\.\d+)*)|0+$/;
let format = s => {
s = s.replace(pattern, "$1$2$3");
return s === "" ? '0' : s.replace(pattern, (_, g1, g2, g3) =>
g1 + g2 + g3.replace(/(?!^)\./g, '').substring(0, 3)
);
};
strings.forEach(s => console.log(format(s)));

Splitting a string based on max character length, but keep words into account

So In my program I can receive strings of all kinds of lengths and send them on their way to get translated. If those strings are of a certain character length I receive an error, so I want to check & split those strings if necessary before that. BUT I can't just split the string in the middle of a word, the words themself also need to be intact & taken into account.
So for example:
let str = "this is an input example of one sentence that contains a bit of words and must be split"
let splitStringArr = [];
// If string is larger than X (for testing make it 20) characters
if(str.length > 20) {
// Split string sentence into smaller strings, keep words intact
//...
// example of result would be
// splitStringArr = ['this is an input', 'example of one sentence' 'that contains...', '...']
// instead of ['this is an input exa' 'mple of one senten' 'ce that contains...']
}
But I'm not sure how to split a sentence and still keep into account the sentence length.
Would a solution for this be to iterate over the string, add every word to it and check every time if it is over the maximum length, otherwise start a new array index, or are there better/existing methods for this?

You can use match and lookahead and word boundaries, |.+ to take care string at the end which are less then max length at the end
let str = "this is an input example of one sentence that contains a bit of words and must be split"
console.log(str.match(/\b[\w\s]{20,}?(?=\s)|.+$/g))

Here's an example using reduce.
const str = "this is an input example of one sentence that contains a bit of words and must be split";
// Split up the string and use `reduce`
// to iterate over it
const temp = str.split(' ').reduce((acc, c) => {
// Get the number of nested arrays
const currIndex = acc.length - 1;
// Join up the last array and get its length
const currLen = acc[currIndex].join(' ').length;
// If the length of that content and the new word
// in the iteration exceeds 20 chars push the new
// word to a new array
if (currLen + c.length > 20) {
acc.push([c]);
// otherwise add it to the existing array
} else {
acc[currIndex].push(c);
}
return acc;
}, [[]]);
// Join up all the nested arrays
const out = temp.map(arr => arr.join(' '));
console.log(out);

What you are looking for is lastIndexOf
In this example, maxOkayStringLength is the max length the string can be before causing an error.
myString.lastIndexOf(/\s/,maxOkayStringLength);
-- edit --
lastIndexOf doesn't take a regex argument, but there's another post on SO that has code to do this:
Is there a version of JavaScript's String.indexOf() that allows for regular expressions?

I would suggest:
1) split string by space symbol, so we get array of words
2) starting to create string again selecting words one by one...
3) if next word makes the string exceed the maximum length we start a new string with this word
Something like this:
const splitString = (str, lineLength) => {
const arr = ['']
str.split(' ').forEach(word => {
if (arr[arr.length - 1].length + word.length > lineLength) arr.push('')
arr[arr.length - 1] += (word + ' ')
})
return arr.map(v => v.trim())
}
const str = "this is an input example of one sentence that contains a bit of words and must be split"
console.log(splitString(str, 20))

JavaScript split string by regex

I will have a string never long than 8 characters in length, e.g.:
// represented as array to demonstrate multiple examples
var strs = [
'11111111',
'1RBN4',
'12B5'
]
When ran through a function, I would like all digit characters to be summed to return a final string:
var strsAfterFunction = [
'8',
'1RBN4',
'3B5'
]
Where you can see all of the 8 single 1 characters in the first string end up as a single 8 character string, the second string remains unchanged as at no point are there adjacent digit characters and the third string changes as the 1 and 2 characters become a 3 and the rest of the string is unchanged.
I believe the best way to do this, in pseudo-code, would be:
1. split the array by regex to find multiple digit characters that are adjacent
2. if an item in the split array contains digits, add them together
3. join the split array items
What would be the .split regex to split by multiple adajcent digit characters, e.g.:
var str = '12RB1N1'
=> ['12', 'R', 'B', '1', 'N', '1']
EDIT:
question:
What about the string "999" should the result be "27", or "9"
If it was clear, always SUM the digits, 999 => 27, 234 => 9

You can do this for the whole transformation :
var results = strs.map(function(s){
return s.replace(/\d+/g, function(n){
return n.split('').reduce(function(s,i){ return +i+s }, 0)
})
});
For your strs array, it returns ["8", "1RBN4", "3B5"].

var results = string.match(/(\d+|\D+)/g);
Testing:
"aoueoe34243euouoe34432euooue34243".match(/(\d+|\D+)/g)
Returns
["aoueoe", "34243", "euouoe", "34432", "euooue", "34243"]

George... My answer was originally similar to dystroy's, but when I got home tonight and found your comment I couldn't pass up a challenge
:)
Here it is without regexp. fwiw it might be faster, it would be an interesting benchmark since the iterations are native.
function p(s){
var str = "", num = 0;
s.split("").forEach(function(v){
if(!isNaN(v)){
(num = (num||0) + +v);
} else if(num!==undefined){
(str += num + v,num = undefined);
} else {
str += v;
}
});
return str+(num||"");
};
// TESTING
console.log(p("345abc567"));
// 12abc18
console.log(p("35abc2134mb1234mnbmn-135"));
// 8abc10mb10mnbmn-9
console.log(p("1 d0n't kn0w wh#t 3153 t0 thr0w #t th15 th1n6"));
// 1d0n't0kn0w0wh#t12t0thr0w0#t0th6th1n6
// EXTRY CREDIT
function fn(s){
var a = p(s);
return a === s ? a : fn(a);
}
console.log(fn("9599999gh999999999999999h999999999999345"));
// 5gh9h3
and here is the Fiddle & a new Fiddle without overly clever ternary