Switch out everything except for the initial one

Question

Switch out everything except for the initial one

Can all instances be replaced except for the first one? For example, 123.45.67..89.0 should turn into 123.4567890.

Edit: Specifically seeking a regex solution. I am aware of how to achieve it through concatenation or using the index.

Answer 1

Answer №1

To accomplish this, consider implementing a positive lookbehind:

(?<=\..*)\.

Modify your code to resemble the following:

"123.45.67..89.0".replace(/(?<=\..*)\./g, '');

Answer 2

To accomplish this, consider implementing a positive lookbehind:

(?<=\..*)\.

Modify your code to resemble the following:

"123.45.67..89.0".replace(/(?<=\..*)\./g, '');

Answer 3

Answer №2

JavaScript Solution:

let ipAddress = "123.45.67.89.0";

let firstDotIndex = ipAddress.search(/\./) + 1;

let resultIPAddress = ipAddress.substr(0, firstDotIndex) + ipAddress.slice(firstDotIndex).replace(/\./g, '');

console.log(resultIPAddress);

I hope you find this JS code helpful :)

Answer 4

JavaScript Solution:

let ipAddress = "123.45.67.89.0";

let firstDotIndex = ipAddress.search(/\./) + 1;

let resultIPAddress = ipAddress.substr(0, firstDotIndex) + ipAddress.slice(firstDotIndex).replace(/\./g, '');

console.log(resultIPAddress);

I hope you find this JS code helpful :)

Answer 5

Answer №3

Although the OP requested solutions with regex, I have an alternative approach that utilizes String.prototype.split() and Array.prototype.reduce():

function replaceAllExceptFirst(str, search, replace) {
  return str
    .split(search)
    .reduce((prev, curr, i) => prev + (i == 1 ? search : replace) + curr);
}

To use this function, one would do something like:

replaceAllExceptFirst('123.45.67..89.0', '.', '')

In my situation, I needed to replace all instances of a specific substring except for the last occurrence. With a minor tweak, it can be achieved as follows:

function replaceAllExceptLast(str, search, replace) {
  return str
    .split(search)
    .reduce(
      (prev, curr, i, substrs) =>
        prev + (i !== substrs.length - 1 ? replace : search) + curr
    );
}

I understand that using regex might offer better performance. However, I thought this method could be easier to grasp for individuals like myself who struggle with regex concepts.

PS: This marks my debut post on StackOverflow, so please show some kindness :)

Answer 6

Although the OP requested solutions with regex, I have an alternative approach that utilizes String.prototype.split() and Array.prototype.reduce():

function replaceAllExceptFirst(str, search, replace) {
  return str
    .split(search)
    .reduce((prev, curr, i) => prev + (i == 1 ? search : replace) + curr);
}

To use this function, one would do something like:

replaceAllExceptFirst('123.45.67..89.0', '.', '')

In my situation, I needed to replace all instances of a specific substring except for the last occurrence. With a minor tweak, it can be achieved as follows:

function replaceAllExceptLast(str, search, replace) {
  return str
    .split(search)
    .reduce(
      (prev, curr, i, substrs) =>
        prev + (i !== substrs.length - 1 ? replace : search) + curr
    );
}

I understand that using regex might offer better performance. However, I thought this method could be easier to grasp for individuals like myself who struggle with regex concepts.

PS: This marks my debut post on StackOverflow, so please show some kindness :)

Answer 7

Answer №4

To eliminate the initial character of your string, employ slice method followed by substituting all instances of - with empty spaces.

let example = "-22-2---222----2--"
example.slice(0, 1) + example.slice(1).replaceAll('-', '')

Answer 8

To eliminate the initial character of your string, employ slice method followed by substituting all instances of - with empty spaces.

let example = "-22-2---222----2--"
example.slice(0, 1) + example.slice(1).replaceAll('-', '')

Answer 9

Answer №5

Is this considered cheating:

"123.45.67.89.0".replace(/\..*/, newC => "." + newC.replace(/\./g, () => ""))

Answer 10

Is this considered cheating:

"123.45.67.89.0".replace(/\..*/, newC => "." + newC.replace(/\./g, () => ""))

Answer 11

Answer №6

To exclude the first occurrence when using the replace method, we can pass a function like this:

givenString.replace(/{pattern_here}/g, (item) => (!index++ ? item : ""));

const givenString = '123.45.67..89.0'
let index = 0

let result = givenString.replace(/\./g, (item) => (!index++ ? item : ""));

console.log(result)

Answer 12

To exclude the first occurrence when using the replace method, we can pass a function like this:

givenString.replace(/{pattern_here}/g, (item) => (!index++ ? item : ""));

const givenString = '123.45.67..89.0'
let index = 0

let result = givenString.replace(/\./g, (item) => (!index++ ? item : ""));

console.log(result)

Answer 13

Answer №7

Give this a shot

let ipAddress = '123.45.67.89.0';

const modifiedIP = ipAddress.split('.')[0].concat('.'+ipAddress.split('.')[1]+ipAddress.split('.')[2]+ipAddress.split('.')[3]+ipAddress.split('.')[4]);

Answer 14

Give this a shot

let ipAddress = '123.45.67.89.0';

const modifiedIP = ipAddress.split('.')[0].concat('.'+ipAddress.split('.')[1]+ipAddress.split('.')[2]+ipAddress.split('.')[3]+ipAddress.split('.')[4]);

Answer 15

Answer №8

Maybe using splice() is a better option in this scenario.

var str = "123.45.67.89.0";
var arr = str.split(".");
var ans = arr.splice(0,2).join('.') + arr.join('');

alert(ans);
// 123.4567890

Answer 16

Maybe using splice() is a better option in this scenario.

var str = "123.45.67.89.0";
var arr = str.split(".");
var ans = arr.splice(0,2).join('.') + arr.join('');

alert(ans);
// 123.4567890

Answer 17

Answer №9

To locate the initial instance of the ., you can utilize the indexOf method, followed by using the replace function to retrieve and substitute the matched string.

const inputString = "123.45.67..89.0";
const firstDotIndex = inputString.indexOf(".");
const modifiedString = inputString.replace(/\./g, (...args) => {
  if (args[1] === firstDotIndex) return args[0];
  else return "";
});

console.log(modifiedString);

Answer 18

To locate the initial instance of the ., you can utilize the indexOf method, followed by using the replace function to retrieve and substitute the matched string.

const inputString = "123.45.67..89.0";
const firstDotIndex = inputString.indexOf(".");
const modifiedString = inputString.replace(/\./g, (...args) => {
  if (args[1] === firstDotIndex) return args[0];
  else return "";
});

console.log(modifiedString);

Answer 19

Answer №10

const ipAddress = "123.45.67.89.0";
const array = ipAddress.split(".");
const result = [];
result = array.map(function(value, index){
return value + (index ? "" : ".");
})
console.log(result.join().replace(/,/g, ""));

Answer 20

const ipAddress = "123.45.67.89.0";
const array = ipAddress.split(".");
const result = [];
result = array.map(function(value, index){
return value + (index ? "" : ".");
})
console.log(result.join().replace(/,/g, ""));

Switch out everything except for the initial one

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

Answer №6

Answer №7

Answer №8

Answer №9

Answer №10

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