Re: json_encode/json_decode, take 2

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Christoph Boget wrote:
> The string used below in "$myArrEncoded" is generated in javascript, after
> creating the structure and spitting out:
> 
> var JSONVar = javascriptVar.toSource();
> 
> I can eval JSONVar and work with it as I would be working with the original
> javascriptVar so I know the transition back and forth from a structure to a
> string isn't causing problems.  The problem is when I try to work with the
> string in PHP.
> 
> Here is my code:

" is not what you think it is (but that could be my mail client),
and the exterior parentheses are also incorrect. your try/catch block is doing
nothing because neither function throws exceptions.

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