Re: json_encode/json_decode, take 2

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.