CODEWITHSUNDEEP
javajsontypesjson-simple
Z2VvZ3Vp
Z2VvZ3Vp

Reputation: 7613

How to tell if return is JSONObject or JSONArray with JSON-simple (Java)?

I am hitting a service and sometimes getting back something like this:

{ "param1": "value1", "param2": "value2" }

and sometimes getting return like this:

[{ "param1": "value1", "param2": "value2" },{ "param1": "value1", "param2": "value2" }]

How do I tell which I'm getting? Both of them evaluate to a String when I do getClass() but if I try to do this:

json = (JSONObject) new JSONParser().parse(result);

on the second case I get an exception

org.json.simple.JSONArray cannot be cast to org.json.simple.JSONObject

How to avoid this? I would just like to know how to check which I'm getting back. (The first case will sometimes have [] in it so I can't do index of and I'd like a cleaner way than just checking the first character.

There has got to be some sort of method that checks this?

Upvotes: 12

Views: 15151

Answers (2)

Stephen C
Stephen C

Reputation: 719249

Simple Java:

Object obj = new JSONParser().parse(result); 
if (obj instanceof JSONObject) {
    JSONObject jo = (JSONObject) obj;
} else {
    JSONArray ja = (JSONArray) obj;
}

You could also test if the (purported) JSON starts with a [ or a { if you wanted to avoid the overhead of parsing the wrong kind of JSON. But be careful with leading whitespace.

For what it is worth, an API that returns either a JSON object or an array of JSON objects depending on the result set size is ... badly designed. It would be better to always return a JSON array; e.g. an array containing zero, one or many objects. That avoids the client-side messiness we are having to deal with in this question.

Upvotes: 25

Saurabh Mittal
Saurabh Mittal

Reputation: 73

Though it's similar to above one, but their is not default constructor of JSONParser. Error coming was: The constructor JSONParser() is undefined

Use this instead

JsonElement jsonElement = new JsonParser().parse(jsonString);
if (jsonElement.isJsonArray()) {
    //Your Code
} else {
    //Your Code
}

Upvotes: 1

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