Getting javascript wrapper error in Qcodo PHP Framework?
Sounds like you are trying to reference a control you didn't render. This is often done by accident when people tell it to render a QDialog, but have the dialog set to Visible=false (don't send to browser) rather than Display=false (set CSS style to display:none).
– Angela Feb 10 '10 at 19:32 Or just set the Display property on the object in the underlying PHP code that instantiates the control – Rob Cowell Feb 16 '10 at 18:47.
From the sample code you give, it makes more sense to define that class outside of the if / else statements entirely, which will both reduce the amount of code to maintain and resolve your error. However, I suspect you may be mismatching the braces when reading the code; from the snippet as you paste it, the error shouldn't occur. If the second class definition is occurring as well as the first (because they aren't part of the same conditional) then you would expect to see that error.
If you do use this approach, it would make sense to check carefully that the class is defined as you'd expect it in all cases. Is this code sample from the QCodo codebase? If so, you probably ought to report the issue via their "Bugs and Issues" forum.
See this and this ...
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