2011-03-04 17:46:41 +01:00
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/*
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* Copyright (C) 2008 Google Inc.
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*
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* Licensed under the Apache License, Version 2.0 (the "License");
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* you may not use this file except in compliance with the License.
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* You may obtain a copy of the License at
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*
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* http://www.apache.org/licenses/LICENSE-2.0
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*
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* Unless required by applicable law or agreed to in writing, software
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* distributed under the License is distributed on an "AS IS" BASIS,
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* WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
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* See the License for the specific language governing permissions and
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* limitations under the License.
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*/
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2011-04-08 15:51:07 +02:00
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package org.mcteam.factions.gson;
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2011-03-04 17:46:41 +01:00
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import java.lang.reflect.Type;
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/**
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* <p>Interface representing a custom deserializer for Json. You should write a custom
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* deserializer, if you are not happy with the default deserialization done by Gson. You will
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* also need to register this deserializer through
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* {@link GsonBuilder#registerTypeAdapter(Type, Object)}.</p>
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*
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* <p>Let us look at example where defining a deserializer will be useful. The {@code Id} class
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* defined below has two fields: {@code clazz} and {@code value}.</p>
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*
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* <pre>
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* public class Id<T> {
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* private final Class<T> clazz;
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* private final long value;
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* public Id(Class<T> clazz, long value) {
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* this.clazz = clazz;
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* this.value = value;
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* }
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* public long getValue() {
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* return value;
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* }
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* }
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* </pre>
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*
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* <p>The default deserialization of {@code Id(com.foo.MyObject.class, 20L)} will require the
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* Json string to be <code>{"clazz":com.foo.MyObject,"value":20}</code>. Suppose, you already know
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* the type of the field that the {@code Id} will be deserialized into, and hence just want to
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* deserialize it from a Json string {@code 20}. You can achieve that by writing a custom
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* deserializer:</p>
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*
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* <pre>
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* class IdDeserializer implements JsonDeserializer<Id>() {
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* public Id fromJson(JsonElement json, Type typeOfT, JsonDeserializationContext context)
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* throws JsonParseException {
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* return (Id) new Id((Class)typeOfT, id.getValue());
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* }
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* </pre>
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*
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* <p>You will also need to register {@code IdDeserializer} with Gson as follows:</p>
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*
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* <pre>
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* Gson gson = new GsonBuilder().registerTypeAdapter(Id.class, new IdDeserializer()).create();
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* </pre>
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*
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* @author Inderjeet Singh
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* @author Joel Leitch
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*
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* @param <T> type for which the deserializer is being registered. It is possible that a
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* deserializer may be asked to deserialize a specific generic type of the T.
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*/
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public interface JsonDeserializer<T> {
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/**
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* Gson invokes this call-back method during deserialization when it encounters a field of the
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* specified type.
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* <p>In the implementation of this call-back method, you should consider invoking
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* {@link JsonDeserializationContext#deserialize(JsonElement, Type)} method to create objects
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* for any non-trivial field of the returned object. However, you should never invoke it on the
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* the same type passing {@code json} since that will cause an infinite loop (Gson will call your
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* call-back method again).
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*
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* @param json The Json data being deserialized
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* @param typeOfT The type of the Object to deserialize to
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* @return a deserialized object of the specified type typeOfT which is a subclass of {@code T}
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* @throws JsonParseException if json is not in the expected format of {@code typeofT}
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*/
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public T deserialize(JsonElement json, Type typeOfT, JsonDeserializationContext context)
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throws JsonParseException;
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}
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