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[eluser]narkaT[/eluser]
you haven't solved the problem, you just worked around it ![]() your're probably misunderstanding the concept of AJAX. like Dready said, when the ajax-request is complete you'll have to do something with the result in order to see any changes. ps: Most frameworks prefer an object instead of a parameter-string when dealing with ajax: Quote:$.ajax({ |
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Load a view - by El Forum - 10-16-2008, 10:42 PM
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Load a view - by El Forum - 10-21-2008, 02:01 AM
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