Java Type Generic as Argument for GSON

JavaGenericsGson

Java Problem Overview


In GSON to get a list of objects you do

Gson gson = new Gson();
Type token = new TypeToken<List<MyType>>(){}.getType();
return gson.fromJson(json, token);

		

It works great, but I want to go further and have MyType parametrized so I can have a common function to parse list of objects with this code

// the common function 
public <T> List<T> fromJSonList(String json, Class<T> type) {
  Gson gson = new Gson();
  Type collectionType = new TypeToken<List<T>>(){}.getType();
  return gson.fromJson(json, collectionType);
}

// the call
List<MyType> myTypes = parser.fromJSonList(jsonString, MyType.class);

Sadly returns an array of StringMaps, not the type. T is being interpreted as another generic type, not my type. Any workaround ?

Java Solutions


Solution 1 - Java

Since gson 2.8.0, you can use TypeToken#getParametized((Type rawType, Type... typeArguments)) to create the typeToken, then getType() should do the trick.

For example:

TypeToken.getParameterized(List.class, myType).getType();

Solution 2 - Java

Generics work at compile-time. The reason super-type tokens work, is because (anonymous) inner classes can access the type arguments to their generic superclasses (superinterfaces), which in turn are stored directly in the bytecode metadata.

Once your .java source file is compiled, the type parameter <T> is obviously thrown away. Since it is not known at compile time, it cannot be stored in bytecode, so it's erased and Gson can't read it.

UPDATE

After newacct's answer, I tried to implement what he suggested in his option 2, ie implementing a ParameterizedType. The code looks like this (here is a basic test):

class ListOfSomething<X> implements ParameterizedType {

	private Class<?> wrapped;
	
	public ListOfSomething(Class<X> wrapped) {
		this.wrapped = wrapped;
	}
	
	public Type[] getActualTypeArguments() {
		return new Type[] {wrapped};
	}

	public Type getRawType() {
		return List.class;
	}

	public Type getOwnerType() {
		return null;
	}
	
}

the purpose of this code, is to be used inside getFromJsonList():

public List<T> fromJsonList(String json, Class<T> klass) {
	Gson gson = new Gson();
	return gson.fromJson(json, new ListOfSomething<T>(klass));
}

Even if the technique works and is indeed very clever (I didn't know it and I would have never thinked of it), this is the final accomplishment:

List<Integer> list = new Factory<Integer>()
         .getFromJsonList(text, Integer.class)

instead of

List<Integer> list = new Gson().fromJson(text,
         new TypeToken<List<Integer>>(){}.getType());

To me, all this wrapping in useless, even if I agree that TypeTokens make the code look nasty :P

Solution 3 - Java

public static final <T> List<T> getList(final Class<T[]> clazz, final String json)
{
	final T[] jsonToObject = new Gson().fromJson(json, clazz);

	return Arrays.asList(jsonToObject);
}

Example:

getList(MyClass[].class, "[{...}]");

Solution 4 - Java

Here is the full code base on great answer from @oldergod

public <T> List<T> fromJSonList(String json, Class<T> myType) {
    Gson gson = new Gson();
    Type collectionType = TypeToken.getParameterized(List.class, myType).getType();
    return gson.fromJson(json, collectionType);
}

Using

List<MyType> myTypes = parser.fromJSonList(jsonString, MyType.class);

Hope it help

Solution 5 - Java

I've taken Raffaele's approach one step further and generified the class, so that it works with every class A, where B is a non-parameterized class. Might be useful for Sets and other Collections.

    public class GenericOf<X, Y> implements ParameterizedType {

    private final Class<X> container;
    private final Class<Y> wrapped;

    public GenericOf(Class<X> container, Class<Y> wrapped) {
        this.container = container;
        this.wrapped = wrapped;
    }

    public Type[] getActualTypeArguments() {
        return new Type[]{wrapped};
    }

    public Type getRawType() {
        return container;
    }

    public Type getOwnerType() {
        return null;
    }

}

Solution 6 - Java

In Kotlin you can simply use this function:

inline fun <reified T> fromJson(json: String): T {
  return Gson().fromJson(json, object: TypeToken<T>(){}.type)
}

and use it like

val myTypes: List<MyType> = fromJson(jsonString);

It will parse any object including gereric types as List. Keyword inline and reified ensures that type will not be erased.

> For detail info I can recommend this Medium post

Solution 7 - Java

This has been answered in previous questions. Basically, there are 2 options:

  1. Pass the Type in from the calling site. The calling code will use TypeToken or whatever to construct it.
  2. Construct a Type corresponding to the parameterized type yourself. This will require you to write a class that implements ParameterizedType

Solution 8 - Java

If programming in kotlin, we can use reified type parameter in inline function

class GenericGson {

    companion object {
        inline fun <reified T : Any> Gson.fromJsonTokenType(jsonString: String): T {
            val type = object : TypeToken<T>() {}.type
            return this.fromJson(jsonString, type)
        }

        inline fun <reified T : Any> Gson.fromJsonType(jsonString: String): T = this.fromJson(jsonString, T::class.java)

        inline fun <reified T : Any> fromJsonTokenType(jsonString: String): T = Gson().fromJsonTokenType(jsonString)

        inline fun <reified T : Any> fromJsonType(jsonString: String): T = Gson().fromJsonType(jsonString)
    }
}

And use like below in your code

val arrayList = GenericGson.fromJsonTokenType<ArrayList<Person>>(json)

Solution 9 - Java

Kotlin "ListOfSomething" solution that worked for me:

fun <T: Any> getGsonList(json: String, kclass: KClass<T>) : List<T> {

    return getGsonInstance().fromJson<List<T>>(json, ListOfSomething<T>(kclass.java))
}


internal class ListOfSomething<X>(wrapped: Class<X>) : ParameterizedType {

    private val wrapped: Class<*>

    init {
        this.wrapped = wrapped
    }

    override fun getActualTypeArguments(): Array<Type> {
        return arrayOf<Type>(wrapped)
    }

    override fun getRawType(): Type {
        return ArrayList::class.java
    }

    override fun getOwnerType(): Type? {
        return null
    }
}

Solution 10 - Java

In kotlin simple use for example:

Get places function

fun getPlaces<T> (jsonString : String, clazz: Class<T>) : T { val places : T = Gson().fromJson(jsonString,clazz) return places }

Then you can use as:

val places = getPlaces(Array<Place>::class.java)

Solution 11 - Java

  public static <T> T getObject(String gsonStr) {
        Gson gson = new GsonBuilder()
                .setLenient()
                .create();
        Type collectionType = new TypeToken< T>(){}.getType();
        return gson.fromJson(gsonStr,
                collectionType);
    }

When use:

Class1 class1=  getObject(jsonStr);

Solution 12 - Java

This work for everything. e.g. map which has a key and value generic.

CustomType type = new CustomType(Map.class, String.class, Integer.class);

So no more TokenType.

class CustomType implements ParameterizedType {
    private final Class<?> container;
    private final Class<?>[] wrapped;

    @Contract(pure = true)
    public CustomType(Class<?> container, Class<?>... wrapped) {
        this.container = container;
        this.wrapped = wrapped;
    }

    @Override
    public Type[] getActualTypeArguments() {
        return this.wrapped;
    }

    @Override
    public Type getRawType() {
        return this.container;
    }

    @Override
    public Type getOwnerType() {
        return null;
    }
}

Solution 13 - Java

public <T> List<T> fromJSonList(String json) {
  Gson gson = new Gson();
  Type collectionType = new TypeToken<List<T>>(){}.getType();
  return gson.fromJson(json, collectionType);
}

//Just call
List<MyType> myTypes = parser.<MyType>fromJSonList(jsonString);

Attributions

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionRodrigo AsensioView Question on Stackoverflow
Solution 1 - JavaoldergodView Answer on Stackoverflow
Solution 2 - JavaRaffaeleView Answer on Stackoverflow
Solution 3 - Javakayz1View Answer on Stackoverflow
Solution 4 - JavaLinhView Answer on Stackoverflow
Solution 5 - JavaChristian BrüggemannView Answer on Stackoverflow
Solution 6 - JavagoofyView Answer on Stackoverflow
Solution 7 - JavanewacctView Answer on Stackoverflow
Solution 8 - JavaalijandroView Answer on Stackoverflow
Solution 9 - JavaMichael PetersonView Answer on Stackoverflow
Solution 10 - JavaCristian CardosoView Answer on Stackoverflow
Solution 11 - JavaAhmad AghazadehView Answer on Stackoverflow
Solution 12 - JavaEric ChanView Answer on Stackoverflow
Solution 13 - JavatangfuhaoView Answer on Stackoverflow