Expected BEGIN_ARRAY but was BEGIN_OBJECT at line 1 column 2

Question

Hi i need to find out value of one key(i.e type) in the Json before parsing it to actual Java object, but i doing so leads to following exception:

java.lang.IllegalStateException: Expected BEGIN_ARRAY but was BEGIN_OBJECT at line 1 column 2 

Json

{
    "type":1,
    "password":"ac@123",
    "role":"normaluser",
    "name":"Archana Chatterjee",
    "username":"a.chatterjee",
    "designation":"Teacher",
    "id":"T_02",
    "age":42
}

Code

Type listType = new TypeToken<ArrayList<Pair>>() {}.getType();
ArrayList<Pair> list = gson.fromJson(json, listType);

Class

class Pair {
    String property;
    Object value;
    // has getters and setters
}

Answer 1

After reading this, i found a better approach of solving my problem. Here i need a low-level streaming parser to get the JsonObject.

        String json3 = "{\"type\":1,\"password\":\"ac@123\",\"role\":\"normaluser\",\"name\":\"Archana Chatterjee\",\"username\":\"a.chatterjee\",\"designation\":\"Teacher\",\"id\":\"T_02\",\"age\":42}";
        JsonParser jParser = new JsonParser();
        JsonObject jObject = (JsonObject) jParser.parse(json3);

Now JsonObject has capability of returning the value of a key. javadoc for same

        JsonElement elem = jObject.get("type");

JsoElement's toString() returns the value of the key. (i.e value of type here)

        System.out.println(elem);

Output

1

Answer 2

Firstly I suggest you to take a look at Gson docs and JSON specs because it looks like you're rather lost...

With Gson, the normal and the easiest way to parse your JSON is creating a structure of Java classes that matches the structure of your JSON.

In this case, your JSON is an object (it's surrounded by { }) and contains a number of properties (i.e., type, password, role, etc.).

So, what you need is to create a Java class with those properties (matching the names):

class User
  String type
  String password
  String role
  // ...

Then you just need to call Gson, give it some JSON string and tell it in which class you want that JSON to be parsed into:

User someUser = gson.fromJson( yourJson, User.class );

---

EDIT: If your JSON may change, you cannot use a single class to parse it, so you have a couple of options. Probably the easiest one is to try to parse your JSON into a Map structure.

A Map is an object (like your JSON is) that contains pairs Key=>Value (e.g., 'type'=>1). So you can try to parse it likt this:

Type mapType = new TypeToken< Map<String, String> >() {}.getType();
HashMap<String, String> map = gson.fromJson( yourJson, mapType );

Note: the problem with this approach is that, as Java is a strongly-typed language, your JSON has to have pairs of the same type (e.g., ). In this case, your fields 'type' and 'age' are not strings, but integers, but I'm pretty sure it can automatically convert them into strings...

If it can't, or if it's really important to parse 'type' and 'age' into integers, your only option is to write a Custom Deserializer...