Gson 序列化 HashMap<Teacher, List<Student>>

问题描述 投票:0回答:1

我有一个地图,其中键的值是对象列表。我能够通过

builder.enableComplexMapKeySerialization();
序列化键,但值未按预期序列化,因为它们在反序列化时返回字符串而不是对象。

下面是序列化的输出

[{"id":31001,"name":老师"]}, //这个是关键

[{"id":33033,"name":"student1"}, {"id":34001,"name":"student2"}]], //这是值列表

我使用了相关的 TypeToken,即

TypeToken<HashMap<Teacher, List<Student>>>
,但列表值在反序列化时仍然返回字符串而不是对象。

android serialization deserialization gson
1个回答
22
投票

JSON 由名称/值对组成(其中值方可以是事物列表)。其名称部分是一个字符串(请参阅:http://json.org

你想要做的是使用一个对象作为名称;你不能直接这样做。 JSON 对象不能是名称/值对的名称。

如果您阅读了 enableComplexMapKeySerialization 的文档,它会解释生成的 JSON 是什么。

它生成的 JSON(作为 JSON 数组的 Map)将完美反序列化回您的地图。以下是一个完整的工作示例(Java 7)。

请注意,一旦我从 JSON 反序列化回 Java,我就会迭代映射以获取密钥。这是因为如果没有在

equals()
中覆盖
hashCode()
Teacher
,就无法创建
Teacher
的新实例并让它作为键工作(仅比较参考值)。

import com.google.gson.Gson;
import com.google.gson.GsonBuilder;
import com.google.gson.reflect.TypeToken;
import java.lang.reflect.Type;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;


public class App 
{
    public static void main( String[] args )
    {
        HashMap<Teacher, List<Student>> map = new HashMap<>();
        Teacher t = new Teacher("12345", "Teacher");
        Teacher t2 = new Teacher("23456", "Teacher2");
        ArrayList<Student> list = new ArrayList<>();
        for (int i = 0; i < 3; i++) {
            list.add(new Student(String.valueOf(i), "Student" + String.valueOf(i)));
        }
        
        map.put(t, list);
        map.put(t2, list);
        
        GsonBuilder builder = new GsonBuilder();
        
        Gson gson =
            builder.enableComplexMapKeySerialization().setPrettyPrinting().create();
        Type type = new TypeToken<HashMap<Teacher,List<Student>>>(){}.getType();
        String json = gson.toJson(map, type);
        System.out.println(json);
        
        System.out.println("\nAfter deserialization:");
        HashMap<Teacher, List<Student>> map2 = gson.fromJson(json, type);
        
        for (Teacher t3 : map2.keySet()) {
            System.out.println(t3.name);
            for (Student s2 : map2.get(t3)) {
                System.out.println("\t" + s2.name);
            }
        }
    }
}

class Teacher {
    public String id;
    public String name;
    
    public Teacher(String id, String name) {
        this.id = id;
        this.name = name;
    }
}

class Student {
    public String id;
    public String name;
    
    public Student(String id, String name) {
        this.id = id;
        this.name = name;
    }
    
}

输出:

[
  [
    {
      "id": "12345",
      "name": "Teacher"
    },
    [
      {
        "id": "0",
        "name": "Student0"
      },
      {
        "id": "1",
        "name": "Student1"
      },
      {
        "id": "2",
        "name": "Student2"
      }
    ]
  ],
  [
    {
      "id": "23456",
      "name": "Teacher2"
    },
    [
      {
        "id": "0",
        "name": "Student0"
      },
      {
        "id": "1",
        "name": "Student1"
      },
      {
        "id": "2",
        "name": "Student2"
      }
    ]
  ]
]

After deserialization:
Teacher2
    Student0
    Student1
    Student2
Teacher
    Student0
    Student1
    Student2

如果您在

equals()
类中实现
hashCode()
Teacher
,那么您将能够使用
Teacher
的新实例从地图中检索内容:

class Teacher {

    public String id;
    public String name;
    
    public Teacher(String id, String name) {
        this.id = id;
        this.name = name;
    }
    
    @Override
    public int hashCode()
    {
        int hash = 3;
        hash = 37 * hash + Objects.hashCode(this.id);
        hash = 37 * hash + Objects.hashCode(this.name);
        return hash;
    }

    @Override
    public boolean equals(Object obj)
    {
        if (obj == null)
        {
            return false;
        }
        if (getClass() != obj.getClass())
        {
            return false;
        }
        final Teacher other = (Teacher) obj;
        if (!Objects.equals(this.id, other.id))
        {
            return false;
        }
        if (!Objects.equals(this.name, other.name))
        {
            return false;
        }
        return true;
    }
    
}

一旦你有了这个,你就可以:

...
HashMap<Teacher, List<Student>> map2 = gson.fromJson(json, type);
Teacher t = new Teacher("23456", "Teacher2");
List<Student> list = map2.get(t);
...
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