Jeff, Firebug and Web Developer extensions for Firefox will become very useful for you. There is also a jQuery extension for Firebug debugging: http://jquery.glyphix.com/ Good instructions on that page.
A JSON (or other Javascript) Object != a multidimensional array. Javascript objects can be accessed like an array, but are defined differently. One of the important differences is that Objects do not have a .length property by default. In Mike's example, you can access the length property of 'models' because models IS an array, defined by brackets ( [ ] ). This is more complex than I can properly describe, there is an excellent writeup at Quirksmode: http://www.quirksmode.org/js/associative.html The more you understand about Javascript, the easier jQuery will become for you. Though jQuery makes many things simpler and faster, it is a deep understanding of Javascript that will really open the power of jQuery to you. I recommend keeping Javascript reference guides open in browser tabs, such as http://www.w3schools.com/ W3C schools for the basics and http://www.quirksmode.org/ Quirksmode for the weird cases Google is also pretty good for searching code cases. Best of luck in your quest! Hope this is all helpful to you. Charles doublerebel.com On Aug 16, 9:51 am, "Erik Beeson" <[EMAIL PROTECTED]> wrote: > Like Mike said, you don't need eval() when using $.getJSON(). > > That console.* stuff is part of a FireFox extension called FireBug that's > fairly standard for doing development. > > --Erik > > On 8/16/07, jeff w <[EMAIL PROTECTED]> wrote: > > > > > 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
