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copied!<p>You have to make sure, that your return values are "really" json. Some browsers dont accept json which isn't syntactically correct.</p> <p>so make 100% sure you return something like:</p> <pre><code>{"foo": 1, "bar": "foobar"} </code></pre> <p>this is correct json.</p> <p>e.g. this is not:</p> <pre><code>{'foo': 1, 'bar': "foobar"} </code></pre> <p>this is also wrong:</p> <pre><code>{foo: 1, bar: "foobar"} </code></pre> <p>further some advices to your javascript code:</p> <pre><code> return { label: item.name, value: item.name } </code></pre> <p>this is not correct, wrap your keys in quotes. return { "label": item.name, "value": item.name }</p> <p>Generally I've read many time you should do ajax calls as GET Requests not POST, because POST produces more overhead and traffic. But don't nail me for that.</p> <p>Further, when you want to return e.g. an array from within php, use json_encode($var)</p> <p><a href="http://php.net/manual/de/function.json-encode.php" rel="nofollow">http://php.net/manual/de/function.json-encode.php</a></p> <p>For your concrete use case return:</p> <pre><code>{"name": "pizza", "point": 1} </code></pre> <p>from within your php script.</p> <p>Try this on your page in firebug:</p> <pre><code>var obj = jQuery.parseJSON('[{"name":"pizza","point":1}]'); for (var i = 0; i < obj.length; i++) { var object = obj[i]; for (property in object) { var s = property + "=" + object[property] + "<br>"; $("#console").after(s); console.log(s) } } </code></pre>

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