score:3

Accepted answer

that datasource is ouputting jsonp, which is for cross-domain ajax requests. it's not valid 'raw' json because of that extra callback(...) wrapper.

read up about it here: http://api.jquery.com/jquery.ajax/ under the 'datatype' section.

score:2

as you say in your tags, it's not json, it's jsonp. do not parse it, catch it with a callback. use jquery.getscript to do it, and define function callback(data). inside that function, data should contain the (parsed) object. also, replace the ? in the url with callback (or whatever you named your function) - ? is not a valid identifier in javascript, so ?([....]) is nonsense.


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