Hi Gafa, You cant return it like this. You have to make your processing into the onsuccess callback.
So... create a function and use it inside your onSuccess callback, or make your hole logic into the callback method (maybe bad). -- Fábio Miranda Costa Solucione Sistemas Front-End Engineer http://meiocodigo.com On Fri, Aug 7, 2009 at 4:00 PM, Gafa <[email protected]> wrote: > > How do I return the onsuccess resonsetext back to the function? > > getDataJSON always returns undefined. > > function getDataJSON(psType,psParams) > { > var sParms="someurl"+psType+psParams > > var jsonRequest = new Request.JSON( > { > url: sParms, > noCache: true, > async: false, > onSuccess: function(searchResults){ > return searchResults; > }, > onFailure: function(errfail) { > alert(errfail); > return ""; > } > }).send(); > > } > > thanks for any help
