On 10/19/07, Jochem Maas <jochem@xxxxxxxxxxxxx> wrote: > > 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. i thought it might be something like that, but i didnt spend long enough at it to verify. here is a great tool to help: http://www.jslint.com/ -nathan
