Parse JsonString to JsonObject in Java

Question

I have the following String :

{
    "response": true,
    "model_original_id": "5acea0b5:1431fde5d6e:-7fff",
    "model_new_id": 500568,
    "model_new_version": 1,
    "reload": true,
    "idsModelProperties": [{
        "key": "creation_date",
        "value": "2013-12-23"
    },
    {
        "key": "state",
        "value": 1,
        "displayValue": "Analisi"
    }],
    "idsNodes": [],
    "idsConnectors": [],
    "idsNodesProperties": []
}

and i need to parse it as a JSONObject. I tried to use quickjson but it gives me an exception when it tries to parse an emty string . This is what i tried :

JsonParserFactory factory=JsonParserFactory.getInstance();
JSONParser parser=factory.newJsonParser();
Map jsonData=parser.parseJson(response_output);

Exception: Exception in thread "main" com.json.exceptions.JSONParsingException: @Key-Hierarchy::root/idsNodes[0]/ @Key:: Value is expected but found empty...@Position::256

Any idea?

Is the string empty or null ? If it fails on "empty" string only parse if non-empty string is given. — PeterMmm
– PeterMmm, Commented Dec 23, 2013 at 15:32
It looks like quickjson cannot handle empty json arrays. Try using a different parsing library. — Sotirios Delimanolis
– Sotirios Delimanolis, Commented Dec 23, 2013 at 15:36

Sotirios Delimanolis · Accepted Answer · 2013-12-23 15:41:02Z

3

I'll give you an alternative since it looks like quick-json has trouble parsing empty json arrays. Check out Gson.

String json = "{ \"response\": true, \"model_original_id\": \"5acea0b5:1431fde5d6e:-7fff\", \"model_new_id\": 500568, \"model_new_version\": 1, \"reload\": true, \"idsModelProperties\": [{ \"key\": \"creation_date\", \"value\": \"2013-12-23\" }, { \"key\": \"state\", \"value\": 1, \"displayValue\": \"Analisi\" }], \"idsNodes\": [], \"idsConnectors\": [], \"idsNodesProperties\": []}";
JsonParser jsonParser = new JsonParser();
JsonElement jsonElement = jsonParser.parse(json);

JsonElement is an abstract class. Its sub types are JsonArray, JsonNull, JsonObject and JsonPrimitive. In the example above, the actual instance is a JsonObject because your json String is a json object. It internally contains a LinkedTreeMap but you don't really need access to it. You can access the different json objects directly on the JsonElement.

answered Dec 23, 2013 at 15:41

Sotirios Delimanolis

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2 Comments

Blerta Dhimitri Over a year ago

In fact I get the json string dinamically... So I cannot do escape of " manually ...

Sotirios Delimanolis Over a year ago

@user3107262 That's just an example to make it run. You don't need to escape anything if it's already in the String. Escaping is just for String literals.

noamik · Accepted Answer · 2014-08-14 08:58:45Z

2

While the author of the question has already accepted an answer, someone else might be searching for the same issue. I've fixed the issues I know of with quick-json in https://code.google.com/p/quick-json/issues/detail?id=11

answered Aug 14, 2014 at 8:58

noamik

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