Re: [jQuery] form validation w/ simple ajax

[Blair McKenzie]

The problem is that you're returning a value from with the callback function (inside the post method). If you want isAjaxCheckEmail to return that value, you need to declare a variable at the start of isAjaxCheckEmail, set that variable to the result inside the callback function, then return that variable at the end of isAjaxCheckEmail.

Blair

On 11/2/06, bmsterling <[EMAIL PROTECTED]> wrote:

Blair,
Thanks for the response.

I was trying to return a true or false with:

function(json) {
                                                eval("var args = " + json);
                                                if ( args.success == "true"){
                                                  alert('true');
                                                 return true;
                                                }
                                                else
                                                {
                                                          alert('false');
                                                         return false;
                                                }
                                          }

but no luck, i also tried setting a variable outside the whole function and
then giving it a value in the same manner as above, and no luck there
either.
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