/*
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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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package cn.emay.sdk.util.json.gson;
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import java.lang.reflect.Type;
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/**
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* <p>
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* Interface representing a custom deserializer for Json. You should write a
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* custom deserializer, if you are not happy with the default deserialization
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* done by Gson. You will also need to register this deserializer through
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* {@link GsonBuilder#registerTypeAdapter(Type, Object)}.
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* </p>
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*
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* <p>
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* Let us look at example where defining a deserializer will be useful. The
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* {@code Id} class defined below has two fields: {@code clazz} and
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* {@code value}.
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* </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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*
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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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*
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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>
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* The default deserialization of {@code Id(com.foo.MyObject.class, 20L)} will
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* require the Json string to be
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* <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
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* hence just want to deserialize it from a Json string {@code 20}. You can
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* achieve that by writing a custom deserializer:
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* </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 deserialize(JsonElement json, Type typeOfT, JsonDeserializationContext context)
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* throws JsonParseException {
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* return new Id((Class)typeOfT, id.getValue());
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* }
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* </pre>
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*
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* <p>
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* You will also need to register {@code IdDeserializer} with Gson as follows:
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* </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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* <p>
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* New applications should prefer {@link TypeAdapter}, whose streaming API is
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* more efficient than this interface's tree API.
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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>
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* type for which the deserializer is being registered. It is
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* possible that a deserializer may be asked to deserialize a
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* 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
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* a field of the specified type.
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* <p>
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* In the implementation of this call-back method, you should consider invoking
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* {@link JsonDeserializationContext#deserialize(JsonElement, Type)} method to
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* create objects for any non-trivial field of the returned object. However, you
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* should never invoke it on the the same type passing {@code json} since that
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* will cause an infinite loop (Gson will call your call-back method again).
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*
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* @param json
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* The Json data being deserialized
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* @param typeOfT
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* The type of the Object to deserialize to
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* @return a deserialized object of the specified type typeOfT which is a
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* subclass of {@code T}
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* @throws JsonParseException
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* 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) throws JsonParseException;
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}
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