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

  • Home
  • SEARCH
  • 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 1011067
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Editorial Team
Asked: May 16, 20262026-05-16T09:10:45+00:00 2026-05-16T09:10:45+00:00

This code doesn’t work: var number = $(this).find(‘.number’).text(); var current = 600; if (current

  • 0

This code doesn’t work:

var number = $(this).find('.number').text();
var current = 600;
if (current > number){
     // do something
}

HTML:

<div class="number">400</div>

Seems there is some problem with converting text() from text-like value to number.

What is the solution?

  • 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
    2026-05-16T09:10:45+00:00Added an answer on May 16, 2026 at 9:10 am

    Always use parseInt with a radix (base) as the second parameter, or you will get unexpected results:

    var number = parseInt($(this).find('.number').text(), 10);
    

    A popular variation however is to use + as a unitary operator. This will always convert with base 10 and never throw an error, just return zero NaN which can be tested with the function isNaN() if it’s an invalid number:

    var number = +($(this).find('.number').text());
    
    • 0
    • Reply
    • Share
      Share
      • Share on Facebook
      • Share on Twitter
      • Share on LinkedIn
      • Share on WhatsApp
      • Report

Sidebar

Ask A Question

Stats

  • Questions 491k
  • Answers 491k
  • Best Answers 0
  • User 1
  • Popular
  • Answers
  • Editorial Team

    How to approach applying for a job at a company ...

    • 7 Answers
  • Editorial Team

    What is a programmer’s life like?

    • 5 Answers
  • Editorial Team

    How to handle personal stress caused by utterly incompetent and ...

    • 5 Answers
  • Editorial Team
    Editorial Team added an answer They are the same. Primary key got NOT NULL automatically. May 16, 2026 at 10:10 am
  • Editorial Team
    Editorial Team added an answer Well, if you are asking how to catch different Exceptions… May 16, 2026 at 10:10 am
  • Editorial Team
    Editorial Team added an answer Why can't you modify your static method to return an… May 16, 2026 at 10:10 am

Trending Tags

analytics british company computer developers django employee employer english facebook french google interview javascript language life php programmer programs salary

Top Members

Related Questions

This code doesn't return data from table: var pom = from k in dataContext.student_gods
This code doesn't work jQuery.each([Alloggio,Macchina,Aereo,Treno], function(){ t = this; $(#ChkPrenotazione + t + Default).change(function(){
This code doesn't work as intended, what am I doing wrong? $(document).ready(function(){ $(a.hide-para).click(function(){ $('p').hide();
This code doesn't work: URL url = new URL( xmlPath ); InputSource input =
This code doesn't send the trailing null byte. How do I send the trailing
this code doesn't compile. I'm wondering what I am doing wrong: private static Importable
I am trying to access a REST webservice, however my code doesn't work and
I am using this code to extract a password protected RAR file. I am
When I write this code, I get an error on the Sort() Method. ArrayList
Possible Duplicate: or is not valid C++ : why does this code compile ?

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.