Using console.log and firebug will show you what everything is.
console.log(data);
$.each(data.records, function(i,item){
console.log(i);
console.log(item);
console.log(this);
});
On Mar 2, 1:54 am, Alain Roger <raf.n...@gmail.com> wrote:
> On Mon, Mar 2, 2009 at 10:52 AM, Matt Quackenbush <quackfu...@gmail.com>wrote:
>
>
>
> > Yes.
>
> > On Mon, Mar 2, 2009 at 3:35 AM, Alain Roger wrote:
>
> >> Hi Wil,
>
> >> so if i understood well, having the following JSON:
>
> >>> {"records":[{"id":1,"abbreviation":"fre","description":"french"},{"id":2,"abbreviation":"eng","description"
> >>> :"english"},{"id":3,"abbreviation":"ger","description":"german"},{"id":4,"abbreviation":"svk","description"
>
> >>> :"slovak"},{"id":5,"abbreviation":"spa","description":"spanish"},{"id":6,"abbreviation":"usa","description"
> >>> :"american"},{"id":7,"abbreviation":"ita","description":"italian"},{"id":8,"abbreviation":"por","description"
>
> >>> :"portuguese"},{"id":9,"abbreviation":"cze","description":"czech"},{"id":10,"abbreviation":"rus","description"
> >>> :"russian"},null,null,null]}
>
> >>> i should write :
> >> $.each(data.records, function(i,item){...});
>
> >> where "data.records" is the name of my JSON result...so records (in my
> >> JSON).
> >> am I right ?
>
> thanks Matt.
