Sign Up

Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.

Have an account? Sign In

Have an account? Sign In Now

Sign In

Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.

Sign Up Here

Forgot Password?

Don't have account, Sign Up Here

Forgot Password

Lost your password? Please enter your email address. You will receive a link and will create a new password via email.

Have an account? Sign In Now

You must login to ask a question.

Forgot Password?

Need An Account, Sign Up Here

Please briefly explain why you feel this question should be reported.

Please briefly explain why you feel this answer should be reported.

Please briefly explain why you feel this user should be reported.

Sign InSign Up

The Archive Base

The Archive Base Logo The Archive Base Logo

The Archive Base Navigation

  • SEARCH
  • Home
  • About Us
  • Blog
  • Contact Us
Search
Ask A Question

Mobile menu

Close
Ask a Question
  • Home
  • Add group
  • Groups page
  • Feed
  • User Profile
  • Communities
  • Questions
    • New Questions
    • Trending Questions
    • Must read Questions
    • Hot Questions
  • Polls
  • Tags
  • Badges
  • Buy Points
  • Users
  • Help
  • Buy Theme
  • SEARCH
Home/ Questions/Q 9272941
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Editorial Team
Asked: June 18, 20262026-06-18T15:59:01+00:00 2026-06-18T15:59:01+00:00

Forgive me if this is a double post, but I couldn’t find a similair

  • 0

Forgive me if this is a double post, but I couldn’t find a similair one that worked for me.

I have a domain name ex. google.co.uk (but this can also be google.com). I need to split this on the first period so I get an array object like this: ["google", "co.uk"] or ["google", "com"]

In antoher post I found this: 'google.co.uk'.split(/.(.+)?/)[1]; but that doesn’t seem to work…

Can anyone help me? Thanks in advance!

  • 1 1 Answer
  • 0 Views
  • 0 Followers
  • 0
Share
  • Facebook
  • Report

Leave an answer
Cancel reply

You must login to add an answer.

Forgot Password?

Need An Account, Sign Up Here

1 Answer

  • Voted
  • Oldest
  • Recent
  • Random
  1. Editorial Team
    Editorial Team
    2026-06-18T15:59:02+00:00Added an answer on June 18, 2026 at 3:59 pm

    Replace the first . with something else that will never turn up in the string (such as |), then split off that character.

    As str.replace only replaces the first instance of the character it finds by default, the code is pretty simple:

    str = "google.co.uk";
    str.replace(".", "|").split("|");
    
    • 0
    • Reply
    • Share
      Share
      • Share on Facebook
      • Share on Twitter
      • Share on LinkedIn
      • Share on WhatsApp
      • Report

Sidebar

Related Questions

Forgive me if this is a dupe but I couldn't find anything that hit
Forgive the double post but I am keen on an answer to this. I
Forgive me if this post is too naive-sounding. I have a Twitter app that
Forgive me if this has already been asked, I couldn't find it. I have
Forgive my ignorance if this is 'a basic question'..... I have a method that
Forgive me if this is a repeated question but I am having trouble finding
Forgive me for this unbeautiful code, but I'm trying to get it to work
Forgive me if this has been asked before, but I assure you I've scoured
Forgive me if this is a dumb beginners problem, but I really don't get
I didnt find anything directly related in searching, so please forgive if this is

Explore

  • Home
  • Add group
  • Groups page
  • Communities
  • Questions
    • New Questions
    • Trending Questions
    • Must read Questions
    • Hot Questions
  • Polls
  • Tags
  • Badges
  • Users
  • Help
  • SEARCH

Footer

© 2021 The Archive Base. All Rights Reserved
With Love by The Archive Base

Insert/edit link

Enter the destination URL

Or link to existing content

    No search term specified. Showing recent items. Search or use up and down arrow keys to select an item.