Php :: PHP To Jquery Dialog() Interaction ?

Jan 10, 2012

I have an html button:

[code]...

This would load flash content:

[code]...

I want to use the jquery dialog to open this content. Everything passed in seems to be perfect (according to the GET response from firebug) but I still get a jquery error. missing ; before statement jquery.js line 612

[code]...

If I paste this url into my browser the flash app starts in the browser as expected but not with the jquery dialog obviously. Must be something wrong with my jquery code?

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