【Gson四】范型POJO的反序列化

编程技术  /  houtizong 发布于 3年前   288

在下面这个例子中，POJO(Data类)是一个范型类，在Tests中，指定范型类为PieceData，POJO初始化完成后，通过

String str = new Gson().toJson(data);

得到范型化的POJO序列化得到的JSON串，然后将这个JSON串反序列化为POJO

 

import com.google.gson.Gson;import java.util.ArrayList;import java.util.List;class PieceData {    private String name;    private int weight;    public String getName() {        return name;    }    public void setName(String name) {        this.name = name;    }    public int getWeight() {        return weight;    }    public void setWeight(int weight) {        this.weight = weight;    }}class Data<T> {    List<T> pieces;    public List<T> getPieces() {        return pieces;    }    public void setPieces(List<T> pieces) {        this.pieces = pieces;    }}public class Tests {    public static void main(String[] args) {        PieceData data1 = new PieceData();        data1.setName("Cat");        data1.setWeight(6);        PieceData data2 = new PieceData();        data2.setName("Dog");        data2.setWeight(36);        List<PieceData> pieces = new ArrayList<PieceData>();        pieces.add(data1);        pieces.add(data2);        Data<PieceData> data = new Data<PieceData>();        data.setPieces(pieces);        String str = new Gson().toJson(data);        System.out.println(str);        Data<PieceData> result = new Gson().fromJson(str, Data.class);        for (int i = 0; i < result.getPieces().size(); i++) {            Object o = result.getPieces().get(i);             System.out.println(o.getClass() + "\t" + o);            // System.out.println(piece.getName() + "\t" + piece.getWeight());        }    }}

 

针对上面的代码，有如下分析：

1. POJO序列化的JSON串是{"pieces":[{"name":"Cat","weight":6},{"name":"Dog","weight":36}]}

2. Data<PieceData> result = new Gson().fromJson(str, Data.class);为什么可以编译通过，直观的感觉是因为参数是Data.class,因此返回值只能是Data result = new Gson().fromJson(str, Data.class):通过察看Gson的fromJson方法 public <T> T fromJson(String json, Class<T> classOfT)，看不出来上面的代码为什么可以编译通过

3. 上面的代码输出是

{"pieces":[{"name":"Cat","weight":6},{"name":"Dog","weight":36}]}
class com.google.gson.internal.LinkedTreeMap    {name=Cat, weight=6.0}
class com.google.gson.internal.LinkedTreeMap    {name=Dog, weight=36.0}

为什么result.getPieces()返回的List，其中的数据类型为TreeMap,明明是PieceData的，原因我猜想，虽然代码中，result类型由于是Data<PieceData>，即result.getPieces()应该是PieceData的集合，可实际运行时，范型有个特性称为Generics Erasure,也就是说实际运行中result.getPieces()未必是PieceData的集合，在这里是TreeMap的集合，TreeMap保存的是PieceData数据的key-value对。

4.可否把

Data<PieceData> result = new Gson().fromJson(str, Data.class);

改为

Data<PieceData> result = new Gson().fromJson(str, Data<PieceData>.class);

 不可以！编译不通过，为什么Data<PieceData>编译出错

5. 使用Type包装Data<PieceData>,代码如下

 

 

package generic;import com.google.gson.Gson;import com.google.gson.reflect.TypeToken;import java.lang.reflect.Type;import java.util.ArrayList;import java.util.List;class PieceData {    private String name;    private int weight;    public String getName() {        return name;    }    public void setName(String name) {        this.name = name;    }    public int getWeight() {        return weight;    }    public void setWeight(int weight) {        this.weight = weight;    }}class Data<T> {    List<T> pieces;    public List<T> getPieces() {        return pieces;    }    public void setPieces(List<T> pieces) {        this.pieces = pieces;    }}public class Tests {    public static void main(String[] args) {        PieceData data1 = new PieceData();        data1.setName("Cat");        data1.setWeight(6);        PieceData data2 = new PieceData();        data2.setName("Dog");        data2.setWeight(36);        List<PieceData> pieces = new ArrayList<PieceData>();        pieces.add(data1);        pieces.add(data2);        Data<PieceData> data = new Data<PieceData>();        data.setPieces(pieces);        String str = new Gson().toJson(data);        System.out.println(str);        Type type = new TypeToken<Data<PieceData>>() {        }.getType();        Data<PieceData> result = new Gson().fromJson(str, type);        for (int i = 0; i < result.getPieces().size(); i++) {            Object o = result.getPieces().get(i);            System.out.println(o.getClass() + "\t" + o);            PieceData p = result.getPieces().get(i);            System.out.println(p.getName() + "\t" + p.getWeight());        }    }}

 

 

 此时结果符合预期

{"pieces":[{"name":"Cat","weight":6},{"name":"Dog","weight":36}]}
class generic.PieceData    generic.PieceData@674a93a6
Cat    6
class generic.PieceData    generic.PieceData@5123ac44
Dog    36

 

总结

1.当要反序列化的POJO是范型POJO时，如例子中的Data，那么必须使用TypeToken对范型进行包装，即调用public <T> T fromJson(String json, Type typeOfT)

2.如果POJO本身不是范型类，但是POJO中包含范型集合时，可以不使用TypeToken进行包装，即调用 public <T> T fromJson(String json, Class<T> classOfT)即可完成反序列化，参见http://bit1129.iteye.com/blog/2101301

3.public <T> T fromJson(String json, Type typeOfT)和public <T> T fromJson(String json, Class<T> classOfT)的javadoc明确说明了它们的应用场景

public <T> T fromJson(String json, Class<T> classOfT)

 

   /* public <T> T fromJson(String json, Class<T> classOfT) deserializes the specified Json into an object of the specified class. It is not   * suitable to use if the specified class is a generic type since it will not have the generictype information because of the Type Erasure feature of Java. Therefore, this method should not   * be used if the desired type is a generic type. Note that this method works fine if the any of the fields of the specified object are generics, just the object itself should not be a   * generic type. For the cases when the object is of generic type, invoke {@link #fromJson(String, Type)}. If you have the Json in a {@link Reader} instead of   * a String, use {@link #fromJson(Reader, Class)} instead.   */

public <T> T fromJson(String json, Type typeOfT)

 

 

  /**   * This method deserializes the specified Json into an object of the specified type. This method   * is useful if the specified object is a generic type. For non-generic objects, use   * {@link #fromJson(String, Class)} instead. If you have the Json in a {@link Reader} instead of   * a String, use {@link #fromJson(Reader, Type)} instead.   *   * @param <T> the type of the desired object   * @param json the string from which the object is to be deserialized   * @param typeOfT The specific genericized type of src. You can obtain this type by using the   * {@link com.google.gson.reflect.TypeToken} class. For example, to get the type for   * {@code Collection<Foo>}, you should use:   * <pre>   * Type typeOfT = new TypeToken&lt;Collection&lt;Foo&gt;&gt;(){}.getType();   * </pre>   * @return an object of type T from the string   * @throws JsonParseException if json is not a valid representation for an object of type typeOfT   * @throws JsonSyntaxException if json is not a valid representation for an object of type   */

 

 

 

 

 

 

 

 

上一篇：【Gson三】Gson解析{"data":{"IM":["MSN","QQ","Gtalk"]}}
下一篇：【Gson五】日期对象的序列化和反序列化

请勿发布不友善或者负能量的内容。与人为善，比聪明更重要！

留言需要登陆哦

技术博客集 - 网站简介：
前后端技术：
后端基于Hyperf2.1框架开发,前端使用Bootstrap可视化布局系统生成
网站主要作用：
1.编程技术分享及讨论交流，内置聊天系统;
2.测试交流框架问题，比如：Hyperf、Laravel、TP、beego;
3.本站数据是基于大数据采集等爬虫技术为基础助力分享知识，如有侵权请发邮件到站长邮箱，站长会尽快处理;
4.站长邮箱：[email protected];

      订阅博客周刊 去订阅

文章归档

文章标签

友情链接

首页
关于我们
Auther ·HouTiZong
侯体宗的博客
© 2020 zongscan.com
版权所有ICP证 : 粤ICP备20027696号
PHP交流群 也可以扫右边的二维码
侯体宗的博客