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 3943684
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Editorial Team
Asked: May 20, 20262026-05-20T00:48:39+00:00 2026-05-20T00:48:39+00:00

I have the following: console.log (a.time_ago() + ‘ ‘ + b.time_ago()); This is breaking

  • 0

I have the following:

console.log (a.time_ago() + ' ' + b.time_ago());

This is breaking in FireFox 3, meaning when FF hits that line in the JS, it goes no further. Strangely if I have Firebug open it doesn’t break and continues as normal. Some how firebug prevents this issue?

I’m puzzled on this one. Any thoughts as to why console.log would break firefox 3, but not if firebug is open?

Thanks

  • 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
    2026-05-20T00:48:40+00:00Added an answer on May 20, 2026 at 12:48 am

    This is not just Firefox. Your code will stop working in every browser (except Chrome and safari (in some instances) because they have console.log() built in along with their developer tools.)

    It is because when you don’t have firebug open, the object “console” is not defined. You should take care never too leave console.log() functions in your code, or it will break in every browser.


    I’d like to add that I have sometimes used this function:

    function log () {
        if (typeof console == 'undefined') {
            return;
        }
        console.log.apply(console, arguments);
    }
    

    Then you can simply call:

    log(somevar, anothervar);
    

    and it will work the same way as console.log, but will not fail if firebug is not loaded (and is shorter to type :P)

    Cheers

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

Sidebar

Related Questions

I have the following code $('#first').click(function() { myVar = $(this).next().val(); }); $('#second').blur(function() { console.log(myVar);
I'm using Jboss 7.1.1 and I have the following log definition: <subsystem xmlns=urn:jboss:domain:logging:1.1> <console-handler
I have the following code that allows my console app to go to the
I have following piece of code, which should be able to log to console
I have the following HTML: <input id=outer type=file onchange=console.log('No.');> <input id=inner type=file onchange=console.log('Yes!'); />
I have the following function: $(document).ready(function() { $(#dSuggest).keypress(function() { var dInput = $('input:text[name=dSuggest]').val(); console.log(dInput);
Possible Duplicate: Javascript infamous Loop problem? I have the following: function test(a,b,c){ console.log(a+b+c); }
I have following requirement, I have C#/.Net console application, which refers to 'System.Data.Sqlite.dll' 'System.Data.Sqlite.dll'
I have the following code: int i = 5000; Console.WriteLine(waiting + i + miliseconds);
I have the following syntax in the .cmd file, where PathList is console application

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.