$document.ready Is Not A Function Break On This Error $document.readyfunction
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Uncaught Typeerror: $(...).ready Is Not A Function
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Uncaught Typeerror: Document.ready Is Not A Function
Sign up TypeError: jQuery(…).ready(…) is not a function up vote 16 down vote favorite OK, I know this has been asked before but none of the answers seems to apply to my case. I'm trying to get a very tiny piece of jQuery running (I'm just getting started on it). jQuery(document).ready(function(){ jQuery('.comtrig').on('click',function(){ $(this).next().animate({'display':'inline'},1000); }); })(); I get the error TypeError: jQuery(...).ready(...) is not typeerror $ is not a function wordpress a function in FF or Uncaught TypeError: object is not a function in Chrome. Solution 1 was to replace $ with jQuery but I obviously already did that as shown above I'm not in Wordpress either I'm using only jQuery and the above mini script, no other JS jQuery itself seems to load fine What am I missing here? javascript jquery share|improve this question asked Feb 13 '14 at 13:17 RubenGeert 1,22341434 1 What happens if you type $ or jQuery into console in browser? –Purple Hexagon Feb 13 '14 at 13:19 2 Just to be sure, are you loading jQuery before your jQuery(...).ready(...) call? –Niet the Dark Absol Feb 13 '14 at 13:19 what is the order of jquery file including? –ɹɐqʞɐ zoɹǝɟ Feb 13 '14 at 13:20 4 I think there is a typo })(); –Satpal Feb 13 '14 at 13:20 1 Replace })(); with });, This is incorrect {()) –Satpal Feb 13 '14 at 13:26 | show 3 more comments 5 Answers 5 active oldest votes up vote 23 down vote accepted try to remove this
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Stack Overflow the company Business Learn more about hiring developers or posting ads with us Stack Overflow Questions typeerror is not a function firefox Jobs Documentation Tags Users Badges Ask Question x Dismiss Join the Stack Overflow Community Stack Overflow is a community of 4.7 million programmers, just like you, helping each other. http://stackoverflow.com/questions/21755397/typeerror-jquery-ready-is-not-a-function Join them; it only takes a minute: Sign up $(document).ready(function() is not working up vote 15 down vote favorite I am using Jquery for getting a json object from a solr server. When I run my html file with Tomcat it is runns fine but when I embed it with my project which is running on weblogic it gets http://stackoverflow.com/questions/6341191/document-readyfunction-is-not-working this error: (debugging done through firebug) $ is not defined $(document).ready(function(){ Why do I get this error when I embed it in my project? This is the contents of my
tag, It is how I include jquery.js:here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow http://stackoverflow.com/questions/812149/is-not-defined-what-does-this-mean the company Business Learn more about hiring developers or posting ads with us Stack http://stackoverflow.com/questions/8220283/first-failing-jquery-ready-breaks-the-rest Overflow Questions Jobs Documentation Tags Users Badges Ask Question x Dismiss Join the Stack Overflow Community Stack Overflow is a community of 4.7 million programmers, just like you, helping each other. Join them; it only takes a minute: Sign up $ is not defined: What does this mean? up vote 3 down vote favorite 2 I get is not a JS error and I can't figure out how to fix it. When my page loads up, IE7 notifies me of a run time error. In addition, my Firebug on Firefox warns me of an error: $ is not defined (?) [Break on this error] $(document).ready(function() { $("a#sin...Out': 300, 'overlayShow': false }); }); When I go to the lines in question its this: Any help please. javascript jquery share|improve this question asked May 1 '09 at 16:39 Abs 13.1k68208356 add a comment| 11 Answers 11 active oldest votes up vote 18 down vote accepted You may have only included the dropdown part of jQuery, and not the whole thing. Try including just the JQuery.js file, without specification as to what part. share|improve this answer answered May 1 '09 at 16:42 Gordon Gustafson 20.3k1784140 add a comment| up vote 12 down vote Do you have a script reference to jQuery above the script block in question? The reason you are seeing this error is because you are using the jQuery function $ without referencing jQuery itself. You need to add a script reference to jQuery like this: if you have a local copy of jQuery.js. Otherwise you can use Google's hosted version like this: Just make sure that these script references live above the script block in question as that way your jQuery plugin will have $ defined for it. share|improve this answer answered May 1 '09 at 16:41 Andrew Hare 214k40497552 thanks I was having this problem and realized I declared the jsquery after my javascr
here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company Business Learn more about hiring developers or posting ads with us Stack Overflow Questions Jobs Documentation Tags Users Badges Ask Question x Dismiss Join the Stack Overflow Community Stack Overflow is a community of 4.7 million programmers, just like you, helping each other. Join them; it only takes a minute: Sign up First failing jQuery.ready() breaks the rest up vote 8 down vote favorite We allow users to write code which sometimes calls jQuery.ready(function(){..}) multiple times. It seems like the first function call that throws an error prevents execution of the rest of the functions. There was a discussion about this here with the solution wrapped jQuery.ready() in a delegate instead of wrapping the anonymous function parameter passed to jQuery.ready(..). How can I override jQuery.ready(fn) and wrap the passed-in function parameter in a delegate which wraps it in try/catch and the passes to jQuery.ready(delegate)? Here is an example:
script // here is some code from third-party developer that sometimes throws an error jQuery.ready(function() { if ((new Date()).getMonth() == 0) throw Error("It's January!"); }); script // here is my code which should run regardless of the error in the script jQuery.ready(function() { alert("I need to run even in January!"); }); What can I do to make code in run regardless of errors in ? jquery error-handling share|improve this question edited Nov 22 '11 at 17:24 asked Nov 22 '11 at 0:04 alecswan 1,40821218 We cannot enclose the code in try-catch block because that code is provided by third-party developers. It's a similar problem the poster described in the referenced discussion. –alecswan Nov 22 '11 at 2:16 possible duplicate of Handling errors in jQuery(document).ready –ripper234 Apr 23 '12 at 8:54 add a comment| 2 Answers 2 active oldest votes up vote 7 down vote accepted If you need to catch your own errors, then catch them with your own try/catch: $(document).ready(function() { try { // put your normal code here } catch (e) { // put any code you want to execute if there's an exception here } }); This will allow all subsequent code to continue wi