I'm downloading jQuery and specifying it using the src attribute on the scropt tag from http://code.jquery.com/jquery-1.6.2.min.js and i'm getting that error.I've tried downloading jQuery and saving it locally and that works. However if i use the CDN it's giving me that error on firebug.
I'm coding a simple hover-opacity thing for some images on my site on the navigation bar. Although I'm receiving this message in my Firebug. uncaught exception: Syntax error, unrecognized expression: . Here's the navigation code:
For some strange reason, none of the scripts get loaded when I load the page. However, once I open firebug to try to see what's happening, it works fine.
When I go to the site using FF5 or Safari 5, everything pulls up great. When I try to open the page using FF3.6 or IE8 none of the classes show up, I then up Firebug or IE Developer toolbar and refresh and the xml appears without failing.
I've a technical problem, i use jQuery Form Plugin and it's working well except for upload file. Effectively after uploading file we can't see any response in firebug. (You can see an example in the official website of jquery form plugin [URL] And in the response i call a script with
The profiler in firebug is great for figuring out what calls are taking up the most time. I have already optimized a lot of my selectors and it is much faster.
However the profile shows a lot of jquery error() calls. In the attached image of the firebug profile window you can see it was called 52 times, accounting for 15.4 of the processing time.
Is that normal for jquery to call its error() like that? My code works flawlessy, and there are no error messages in the firefox error console. It seems like that is a significant performance hit. Is there anyway to get more info on what the errors are?
I trying to make a call to an external domain using $.ajax() and it WORKS, the server receives the call, but the response in firebug errors out in jquery.js line 7760. I've been beating my head at this all day and don't feel like I've made it much further.
I suspect it has something to do with the dataType or type of the request. But I've tried all kinds of things from POST to GET to JSONP in the type. For dataType, I've also tried "html", "text", "xml", "json", and even some combos of "text html" but no success.
I'm looking to have a lightbox pop up when a user clicks the Back button in their browser rather than just navigating back. The purpose is to ask a question with a Yes/No answer, and if they click No, I allow them to go back. The only thing I've found anything like this is the onUnload event, but that doesn't prevent them from going back. How should this be handled?
I am trying to write a script that uses the IF statement to see wether or not a user clicked the back button to come to a page, and then if it's true to not let the page load and kick them back X number of pages (say 4) This is what I have so far:
I have a application that uses jquery-1.6.4.jsandjquery-ui-1.8.14.custom.min.js. A call is made by a js file that works if I break on the call and single step using firebug but when run w/o the single step the call to the php in the server never seems to occure. IE doesn't seem to have this problem
I am trying to capture the back button and redirect if it is a certain URL, if not just go back like a normal back button.I've never really messed with the history except for something like this: <a href="#" onClick="history.go(-1)">Anyone have an example using this plugin: [URL]r any other plugin that might achieve this
I have firebug installed in my firefox browser, but I can't get console.log, console.debug or console.trace to work. My console window remains empty despite all my efforts. For example I have some code which throws up an alert box ( alert ('here is an alert') so I know 100% that it is being executed. If I now place the line console.log("please log this to the console");next to it, nothing gets written to the console. I also have firephp installed and this is logging to the console for my php code absolutely fine
An old co-worker wrote some scripts based on prototype to show input fields on a form when certain options were selected on the previous input fields. I'm trying to duplicate his old code but for some reason it's just not working. I can see that the reason those input fields aren't showing up is because display:none is being added on to the element by the js on the second step after the breakpoint (break on attribute change in firebug). This happens on the working page as well but the working page has about 10 more steps afterwards that eventually remove that style.
I have a webpage popup (lets call it POPUP) which refreshes the opener window (this one we call PARENT) when we close it, saying we want to save data. For this, we use
When we don't want to save the POPUP data, we just close the popup and don't refresh the PARENT. In the PARENT we have a "Back" link which executes a simple
The problem is: If we refresh parent, we need to go back 2 pages, because the refresh method adds another page to the history. But we have no (easy) way of knowing in the PARENT if it had been refreshed. With this, our users are forced to click two times in the Back link.
Is there anyway of going back to the previous page, no matter how many refreshes happened in the current one ?