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 6191869
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Editorial Team
Asked: May 24, 20262026-05-24T02:50:32+00:00 2026-05-24T02:50:32+00:00

var str =20110725; var dd = str.substring(6); var mm = str.substring(4,2); var yyyy =

  • 0
var str ="20110725";
var dd = str.substring(6);
var mm = str.substring(4,2);
var yyyy = str.substring(0,3);
alert(dd );//25
alert(mm);//11
alert(yyyy );//2011

Instead of the above output, I expected “25” as date, “07” as month and “2011” as year. Please correct me.

  • 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
    2026-05-24T02:50:32+00:00Added an answer on May 24, 2026 at 2:50 am

    I think you want substr(), not substring(). They’re different.

    • 0
    • Reply
    • Share
      Share
      • Share on Facebook
      • Share on Twitter
      • Share on LinkedIn
      • Share on WhatsApp
      • Report

Sidebar

Related Questions

Input: var str = 'text {{value1}} text {{value2}}text{{value1}}'; Regex: var result = str.match(/({{\S+}})+/ig); Output:
Given this string: var str = 'A1=B2;C3,D0*E9+F6-'; I would like to retrieve the substring
var str = '0.25'; How to convert the above to 0.25?
Why don't work? var str; $('table tr').each(function() { str = $(this).find('td').eq(6).html().trim().substring(10, 20); $(this).find('td').eq(6).text(str); });
var is:ImageSnapshot = myImagesnapshot; var str:String = ImageSnapshot.encodeImageAsBase64(is); As of now, I am sending
You can backreference like this in JavaScript: var str = 123 $test 123; str
HI I have a URL var str:String = conn=rtmp://server.com/service/&fileId=myfile.flv or var str:String = fileId=myfile.flv&conn=rtmp://server.com/service/
How can I quickly validate if a string is alphabetic only, e.g var str
I need to get all characters between '(' and ')' chars. var str =
For example, here is a string representing an expression: var str = 'total =

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.