is it possible to register live() an ajaxForm()?

Question

I'm doing like this:

$('.myforms').ajaxForm();

after I add some more forms to the html which have this .myforms class assigned, so I have to call the $('.myforms').ajaxForm() again

is it possible to register the ajaxForm live ?

score 2 · Accepted Answer · answered Dec 15 '10 at 02:04

2

checkout the Live Query plugin. It allows you to simulate the .live() but extends the behaviour to DOM elements, not just events.

answered Dec 15 '10 at 02:04

lomaxx

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I would like some peace of code from this plugin that would do this, without using the plugin – Omu Dec 15 '10 at 08:05

score 1 · Answer 2 · edited May 23 '17 at 11:55

1

Not with jQuery's live(), but you could monitor the form for changes to the DOM and trigger ajaxForm() from there. See this thread

edited May 23 '17 at 11:55

Community

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answered Dec 15 '10 at 01:47

Jake

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score 1 · Answer 3 · answered Feb 03 '12 at 08:41

You do not need to use jQuery live, all you need to do is put the ajaxForm code inside a function and then in the ajax success you just call the function again to rebind the new form in the view.

    function bindSubmitNewPost()  {

     var options = { 
        target:        '#new-post-message-1',  
        beforeSubmit:  showRequest, 
        success:       showResponse  

       }; 


    $('.FormNewPost').ajaxForm(options);  
}

is it possible to register live() an ajaxForm()?

3 Answers3