[jQuery] Re: JSON + variable

Hi mueller,

The response is look like a function call.

/*
yasearch(
{
"q": "uboot",
"f": ["k"],
"r": [["uboote"], ["uboot chat"], ["ubootpc spiel"], ["uboot
simulation"], ["modell uboot"], ["uboot hamburg"], ["uboot museum"],
["atom-uboot"], ["uboot ping"], ["rc uboot"]]
}
)

*/

try this,

function yasearch(a) {
alert(a.q);

}

$(function() {
$.ajax({
type: 'GET',
url: 'js.html',
data: "format=json&id=123",
success: function(data) {
eval(data);
},
dataType: 'json'
});
});

On Feb 3, 5:44 pm, weidc <mueller.juli...@googlemail.com> wrote:
> Hello,
>
> well i got a problem with using JSON.
>
> My code looks like this:
>
> $(function()
> {
>
>         $.ajax({
>                 type:'GET',
>                 url: url,
>                 data:"format=json&id=123",
>                 success:function(data) {
>                         alert(data); // doesn't work
>                 },
>                 dataType:'jsonp'
>         });
>
> });
>
> The URL is an external one.
> The answer looks something like this:
>
> yasearch({"q":"uboot","f":["k"],"r":[["uboote"],["uboot chat"],["uboot
> pc spiel"],["uboot simulation"],["modell uboot"],["uboot hamburg"],
> ["uboot museum"],["atom-uboot"],["uboot ping"],["rc uboot"]]})
>
> I always get this error msg in firebug:
> yasearch is not defined
>
> I tried to define it in any way but it doesn't seem to work. I use
> JSON for the first time now and don't know how to get an useable
> answer.
>
> I hope someone can help me.