Yes, I am using $.getJSON. eval() is a javascript function right? so
if I want more info on that, I should look at a javascript reference??
what is console.debug()? I've seen that in a bunch of posts. I am
guessing its a way to output results. Is it a cleaner way of using
something like alert(spit out results);?
Thanks for your patience if my questions are elementary, I'm just
trying to build some sort of foundation of basic concepts.
On Aug 16, 10:51 am, "Erik Beeson" <[EMAIL PROTECTED]> wrote:
> I assumed he was using $.getJSON or something similar that takes care of the
> eval'ing for you.
>
> --Erik
>
> On 8/16/07, Michael Geary <[EMAIL PROTECTED]> wrote:
>
>
>
> > > From: jeff w
>
> > > I am new to jQuery, and have started to play with JSON,but I
> > > need some info about how I refer to the JSON Object once it
> > > is returned from the server. I know I can loop through the
> > > contents of the object, and I can use json.count, but I am
> > > really unsure about the correct syntax to target the data
> > > that I need. Can anyone provide a link to a tutorial or some
> > > other help?
>
> > > Here is the JSON object that I need to return from the server:
>
> > > {"models": ["MDX SUV", "RDX SUV", "RL Sedan", "TL Sedan", "TSX Sedan"]}
>
> > > Thanks for your help.
>
> > > ************
>
> > > since writing this, I have made a guess at what might work. I
> > > confirmed that the data is returning as stated above (using
> > > Firebug), but when I echo json.count, i get 'undefined'. does
> > > that make sense?
>
> > I don't see a count property in your JSON data, and you didn't mention
> > eval'ing the JSON string (it's just a string until you do something with
> > it).
>
> > Assuming that you have received a string from your server in a variable
> > "json" containing the above JSON text:
>
> > json = eval( '(' + json + ')' );
> > var models = json.models;
> > for( var i = 0, n = models.length; i < n; ++i ) {
> > var model = models[i];
> > console.debug( model );
> > }
>
> > You can even use $.each if you like:
>
> > json = eval( '(' + json + ')' );
> > $.each( json.models, function( i, model ) {
> > console.debug( model );
> > });
>
> > -Mike
