I'm trying to load a local JSON file but it won't work. Here is my JavaScript code (using jQuery:  var json = $.getJSON("test.json"); var data = eval("(" +json.responseText + ")"); document.write(data["a"]);...
I need to debug a web application that uses jQuery to do some fairly complex and messy DOM manipulation. At one point, some of the events that were bound to particular elements, are not fired and simply stop working.  If I had a capability to edit...
I am using this little script to find out whether Firebug is open:  if (window.console && window.console.firebug) {     //is open };   And it works well. Now I was searching for half an hour to find a way to detect whether Google Chrome'...
What's a surefire way of detecting whether a user has Firebug enabled?...
I would like to change the context of the javascript executed in the webkit developer tool/firebug console to execute its code like it is running from inside an iframe on the page.  I know I could do this by opening the page in the iframe on a separa...
Are there any add-ons for Firefox that I can use to find out with part of the JavaScript causes memory leaks?...
I'm making a jQuery $.getJSON request to another domain, so am making sure that my GET URI ends with "callback=?" (i.e. using JSONP).  The NET panel of Firebug shows that I am receiving the data as expected, but for some reason the Consol...
I want to write a browser (Chrome/FF) extension that needs to select an element on a web page.  I would like it to behave like Firebug's element inspector does.  You click the inspect arrow and you can then hover/highlight elements.  When you cli...
Is it possible to access the previously-logged output of Firebug programmatically?  For example:  console.log('a'); console.log('b'); console.log('c');  for (var i = 0; i < console.output.length; ++i) {     alert(console.ou...
Which solution do you recommend, the second is simpler ( less code ), but there are drawbacks on using it ?  First: (Set a global debug flag)  // the first line of code var debug = true; try {     console.log } catch(e) {     if(e) {         debug=fa...

Tags

Recent Questions

Top Questions

Home Tags Terms of Service Privacy Policy DMCA Contact Us

©2020 All rights reserved.