Simple Json to Java convertion using GSON

Question

I've recently started using GSON library for deserializing JSON that comes from web service but I can't make it work. I decide to test GSON on some simple input - can't make it any simpler and it still doesn't work. I've looked through similar issues like Converting JSON to Java, all of which suggest similar approach to solution. My guess is that I'm missing something really simple and obvious so a fresh view on the code would probably help. So here is what I have:

JSON

{"A":{"name":"qwrety","value1":1,"value2":2}}

Java class

import com.google.gson.Gson;

public class Main {

    public static void main(String[] args) {        
        String json = /*getting JSON from server*/       
        Gson gson = new Gson();
        A obj = gson.fromJson(json, A.class);
        System.out.println(obj);
    }
}

class A {
    private String name;
    private int value1;
    private int value2;

    public String getName() { return name; }
    public int getValue1() { return value1; }
    public int getValue2() { return value2; }

    public void setName(String name) { this.name = name; }
    public void setValue1(int value1) { this.value1 = value1; }
    public void setValue2(int value2) { this.value2 = value2; }

    public String toString() {
        return String.format("name: %s, value1: %d, value2: %d", name, value1, value2);
    }
}

What I get in return is

name: null, value1: 0, value2: 0

Anyone can tell what is wrong with this code?

UPD As @torbinsky and @Kevin-Dolan pointed out, the problem was because Java class structure didn't match the Json format. To fix this I added a new class

class Container {
    private A a;

    public A getA() { return a; }

    public void setA(A a) { this.a = a; }

}

and changed the deserialisation call to following

Gson gson = new Gson();
Container obj = gson.fromJson(json, Container.class);

System.out.println(obj.getA());

However I anyway get "null" printed out

torbinsky · Accepted Answer · 2011-02-13 00:58:37Z

4

I think your problem is your JSON string.

Try changing:

{"A":{"name":"qwrety","value1":1,"value2":2}}

to:

{"name":"qwrety","value1":1,"value2":2}

EDIT: I think the problem might be your JSON string has "A" and that is case-sensitive. Try changing your member variable to an uppercase "A". Sorry I am unable to test at the moment to confirm this.

If the JSON format is fixed then try changing your classes to something similar:

class Foo {
private String name;
private int value1;
private int value2;

public String getName() { return name; }
public int getValue1() { return value1; }
public int getValue2() { return value2; }

public void setName(String name) { this.name = name; }
public void setValue1(int value1) { this.value1 = value1; }
public void setValue2(int value2) { this.value2 = value2; }

public String toString() {
    return String.format("name: %s, value1: %d, value2: %d", name, value1, value2);
}
}

and:

class Container {
private Foo A;

public Foo getA() { return A; }

public void setA(Foo A) { this.A = A; }
}

edited Feb 13, 2011 at 0:58

answered Feb 13, 2011 at 0:08

torbinsky

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2 Comments

IvanR Over a year ago

Thanks, you're right, changing json to the one you suggested works, but as I said, I'm getting json from a web service, so I can't really change it's format. Your answer gave me an idea to introduce new class -> class Container { private A a; public A getA() { return a; } public void setA(A a) { this.a = a; } } and deserialize it instead -> Container obj = gson.fromJson(json, Container.class); but when I print obj.getA() I still get null

torbinsky Over a year ago

I think the problem might be your JSON string has "A" and that is case-sensitive. Try changing your member variable to an uppercase "A". Sorry I am unable to test at the moment to confirm this.

Kevin Dolan · Accepted Answer · 2011-02-13 00:08:26Z

2

The problem is, it's looking for a field in your object called "A", but instead it's finding fields "name", "value1" and "value2".

The json you want to be sending to gson is:

{"name":"qwrety","value1":1,"value2":2}

answered Feb 13, 2011 at 0:08

Kevin Dolan

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Comments

StaxMan · Accepted Answer · 2011-02-13 07:31:44Z

1

After changes, your field name is still "a" but JSON uses "A". So you need to either rename field, or add annotation to indicate that JSON uses "A" as name (@SerializedName("A") I think).

answered Feb 13, 2011 at 7:31

StaxMan

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