CODEWITHSUNDEEP
javascriptarraysregexstringreplace
sen
sen

Reputation: 91

regex replace for multiple string array javascript

I have a array of string and the patterns like #number-number anywhere inside a string.

Requirements:

I got stuck and how to do in JavaScript. Here is the code I used:

let arrstr=["#12-1676","#02-8989898","#676-98908098","12-232","02-898988","676-98098","2-898988", "380100 6-764","380100 #6-764","380100 #06-764"]

for(let st of arrstr)
 console.log(st.replace(/#?(\d)?(\d-)/g ,replacer))
 
 function replacer(match, p1, p2, offset, string){
  let replaceSubString = p1 || "0";
  replaceSubString += p2;
  return replaceSubString;
 }

Upvotes: 1

Views: 150

Answers (4)

Wiktor Stribiżew
Wiktor Stribiżew

Reputation: 626699

I suggest matching # optionally at the start of string, and then capture one or more digits before - + a digit to later pad those digits with leading zeros and omit the leading # in the result:

st.replace(/#?\b(\d+)(?=-\d)/g, (_,$1) => $1.padStart(2,"0"))

See the JavaScript demo:

let arrstr=["#12-1676","#02-8989898","#676-98908098","12-232","02-898988","676-98098","2-898988", "380100 6-764","380100 #6-764","380100 #06-764"]

for(let st of arrstr)
 console.log(st,'=>', st.replace(/#?\b(\d+)(?=-\d)/g, (_,$1) => $1.padStart(2,"0") ))

The /#?\b(\d+)(?=-\d)/g regex matches all occurrences of

  • #? - an optional # char
  • \b - word boundary
  • (\d+) - Capturing group 1: one or more digits...
  • (?=-\d) - that must be followed with a - and a digit (this is a positive lookahead that only checks if its pattern matches immediately to the right of the current location without actually consuming the matched text).

Upvotes: 1

GenericUser
GenericUser

Reputation: 3230

Using the unary operator, here's a two liner replacer function.

const testValues = ["#162-7878", "#12-4598866", "#1-7878", "1-7878"];
const re = /#?(\d+?)-(\d+)/;

for(const str of testValues) {
  console.log(str.replace(re, replacer));
}

function replacer(match, p1, p2) {
  p1 = +p1 < 10 ? `0${p1}` : p1;
  return `${p1}-${p2}`; 
}

Upvotes: 1

user4308987
user4308987

Reputation:

I think a simple check is what you should do with the match function.

let arrstr=["#12-1676","#0-8989898","#676-98908098","12-232","02-898988","676-98098","2-898988"];
const regex = /#\d-/g;
for(i in arrstr){
    var found = arrstr[i].match(regex);
    if(found){
      arrstr[i]=arrstr[i].replace("#","0")
    }else{
      arrstr[i]=arrstr[i].replace("#","")
    }
}
console.log(arrstr);

or if you really want to stick with the way you have it.

let arrstr=["#12-1676","#02-8989898","#6-98908098","12-232","02-898988","676-98098","2-898988"]

for(let st of arrstr)
 console.log(st.replace(/#(\d)?(\d-)/g ,replacer))
 
 function replacer(match, p1, p2, offset, string){
  let replaceSubString = p1 || "0";
  replaceSubString += p2;
  return replaceSubString;
 }

remove the '?' from the regex so its not #? but just #

Upvotes: 0

Zachary
Zachary

Reputation: 1

let list_of_numbers =["#1-7878", "#162-7878", "#12-1676","#02-8989898","#676-98908098","12-232","02-898988","676-98098","2-898988"]

const solution = () => {
    let result = ''
    for (let number of list_of_numbers) {
        let nums = number.split('-')
        if (nums[0][0] == '#' && nums[0].length > 2) {
            result = `${nums[0].slice(1, number.length-1)}-${nums[1]}`
            console.log(result)
        } else if (nums[0][0] == '#' && nums[0].length == 2) {
            result = `${nums[0][0] = '0'}${nums[0][1]}-${nums[1]}`
            console.log(result)
        } else {
            console.log(number)
        }
    }
}

Upvotes: 0

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