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.