Question

如果我有一个简单的对象，如下所示：

String name;
String email;
int age;
boolean isDeveloper;

然后假设接收到的JSON对象具有值：

{"name":null,"email":null,"age":26,"isDeveloper":true}

使用GSON作为默认值反序列化此JSON时，我有：

{"age":26,"isDeveloper":true}

但是电子邮件字段丢失会在以后导致我的应用程序失败，所以我想添加

email = null;

序列化回JSON，并且只有此字段值为null。也不要忽略任何非null字段。

其他空值不应添加到结果JSON中。

我尝试使用默认的GSON构建器进行反序列化，然后使用允许空值的GSON进行序列化：

Gson gson = new GsonBuilder().serializeNulls().create();

问题是：这将查找对象类中的所有null / empty值并将其全部设置

{"name":null,"email":null,"age":26,"isDeveloper":true}

如何设置电子邮件属性为null，然后序列化回仅包含此字段为JSON的JSON，而不忽略任何其他非null值？

我正在使用gson v2.2.4

这是示例代码：

import com.google.gson.Gson;
import com.google.gson.GsonBuilder;

public class App
{
    private class UserSimple {
        public String name;
        public String email;
        public int age;
        public boolean isDeveloper;


        UserSimple(String name, String email, int age, boolean isDeveloper) {
            this.name = name;
            this.email = email;
            this.age = age;
            this.isDeveloper = isDeveloper;
        }

    }

    public static void main( String[] args )
    {
        String userJson = "{'age':26,'email':'abc@dfg.gh','isDeveloper':true,'name':null}";
        Gson gson = new GsonBuilder()
                .setPrettyPrinting()
                .disableHtmlEscaping()
                .create();


        Gson gsonBuilder = new GsonBuilder().serializeNulls().create();

        UserSimple deserializedObj = gson.fromJson(userJson, UserSimple.class);
        System.out.println("\n"+gson.toJson(deserializedObj));
        deserializedObj.email = null;


        String serializedObj = gsonBuilder.toJson(deserializedObj, UserSimple.class);
        System.out.println("\n"+serializedObj);


    }
}

Answer 1

您可以创建自己的自定义适配器来解决当前的问题：

class MyCustomTypeAdapter extends TypeAdapter<UserSimple> {
    @Override
    public void write(JsonWriter writer, UserSimple userSimple) throws IOException {
        writer.beginObject();
        if(userSimple.getName() != null){
            writer.name("name");
            writer.value(userSimple.getName());
        }

        // you want to include email even if it's null
        writer.name("email");
        writer.value(userSimple.getEmail());

        if(userSimple.getAge() != null){
            writer.name("age");
            writer.value(userSimple.getAge());
        }
        if(userSimple.getDeveloper() != null){
            writer.name("isDeveloper");
            writer.value(userSimple.getDeveloper());
        }
        writer.endObject();
    }

    public UserSimple read(JsonReader reader) throws IOException {
    // you could create your own
        return null;
    }
}

输入：

String userJson = "{'age':null,'email':null,'isDeveloper':true,'name':'somename'}";

输出：

serializedObj :{"name":"somename","email":null,"isDeveloper":true}

输入：

String userJson = "{'age':20,'email':null,'isDeveloper':true,'name':'somename'}";

输出：

serializedObj :{"name":"somename","email":null,"age":20,"isDeveloper":true}

看看https://google.github.io/gson/apidocs/com/google/gson/TypeAdapter.html

Answer 2

您是否考虑过使用杰克逊？我认为它比Gson更强大，更快捷。但这只是我的意见，我并不是要冒犯任何人。我还遇到过杰克逊可以做gson无法做的事情。

以下是为您完成工作的代码：

import com.fasterxml.jackson.annotation.JsonInclude;
import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.databind.ObjectWriter;

public class Main {
    static class Person {
        @JsonInclude(JsonInclude.Include.ALWAYS)
        private String name;
        @JsonInclude(JsonInclude.Include.NON_NULL)
        private String email;
        @JsonInclude(JsonInclude.Include.NON_NULL)
        private Integer age;
        public Person(String name, String email, Integer age) {
            this.name = name;
            this.email = email;
            this.age = age;
        }

        public String getName() {
            return name;
        }

        public String getEmail() {
            return email;
        }

        public Integer getAge() {
            return age;
        }
    }
    public static void main(String[] args) throws Exception {
        ObjectWriter ow = new ObjectMapper().writer().withDefaultPrettyPrinter(); // pretty printing only enabled for demonstration purposes
        String test = ow.writeValueAsString(new Person(null, null, 2));
        System.out.println(test);
    }
}

输出为：

{
  "name" : null,
  "age" : 2
}

name也被输出，因为其包含策略为JsonInclude.Include.ALWAYS。

在Maven中，依赖项包括：

<dependency>
  <groupId>com.fasterxml.jackson.core</groupId>
  <artifactId>jackson-core</artifactId>
  <version>2.9.6</version>
</dependency>
<dependency>
  <groupId>com.fasterxml.jackson.core</groupId>
  <artifactId>jackson-annotations</artifactId>
  <version>2.9.6</version>
</dependency>
<dependency>
  <groupId>com.fasterxml.jackson.core</groupId>
  <artifactId>jackson-databind</artifactId>
  <version>2.9.6</version>
</dependency>

还有一个答案：Gson serialize null for specific class or field，但看起来还不是很好。

如何使用GSON仅返回一个特定的空字段？

2 个答案: