仅当字段在嵌套对象中时，Gson 才会序列化空值

Question

我面临的问题是，当仅涉及非顶级属性时，我想 ser/des null 值，而我不知道如何实现这一点。 所以假设我有一个 User 类：

Class User {
   String name;
   int id;
   Address address;
}

还有一个地址类：

Class Address{
   String street;
   String city;
   String country;
}

现在，我可以使用下面的 Gson 实例来 ser/des 空值：

Gson gson = new GsonBuilder().serializeNulls().create();
Address address = new Address(null, "New York", "US");
User user = new User("Adam", 123, address);  
String userJson = gson.toJson(user); 

输出是：

{
  "name": "Adam",
  "id": 123,
  "address": {
      "street": null,
      "city": "New York",
      "country": "US"
  }
}

不过，我不想SER / DES空当它涉及到用户的顶级属性。 例如以下用户：

User user = new User("Adam", 123, null);

我想要输出如下并且没有地址字段：

{
  "name": "Adam",
  "id": 123
}

我现在尝试使用自定义序列化程序对每个顶级属性进行硬编码，如果它们为空，则将其删除：

public class SerializerForUser implements JsonSerializer<ConfigSnapshot> {

    @Override
    public JsonElement serialize(User user, Type type, JsonSerializationContext jsc) {
        Gson gson = new GsonBuilder().serializeNulls().create();
        JsonObject jsonObject = gson.toJsonTree(user).getAsJsonObject();
        if (user.getAddress() == null) {
            jsonObject.remove("address");
        }
        // if... (same with other top-level attributes)
        return jsonObject;
    }
}

Gson gson = new GsonBuilder().serializeNulls().registerTypeAdapter(User.class, new SerializerForUser()).create();

但不知何故它不起作用，当例如地址为空时，我仍然会得到以下输出：

{
  "name": "Adam",
  "id": 123,
  "address: null
}

谁能给我一些关于我在这里做错了什么的提示？ 或者，如果有人能告诉我是否有更直接/一般的方法来实现这一点（因为我也想使用相同的 gson 实例来 ser/des 其他对象），那将是完美的？ 任何意见表示赞赏。

Answer 1

因为你正在使用

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

显示空值。

要跳过显示空值，让我们尝试

Gson gson = new Gson();

你可以在这里测试

public static void main(String[] args) {
    Gson yourGson = new GsonBuilder().serializeNulls().create(); // this is how you create your Gson object, which shows null value
    Address address = new Address(null, "New York", "US");
    User user = new User("Adam", 123, address);
    String userJson = yourGson.toJson(user);
    System.out.println(userJson);
    
    Gson newGson = new Gson(); // with this one, it doesn't show value
    System.out.println(newGson.toJson(user));
}

更新

我试图用几次覆盖方法serialize ，但它失败了，直到我尝试 #5

public class UserCustomSerializer implements JsonSerializer<User> {

    @Override
    public JsonElement serialize(User src, Type typeOfSrc, JsonSerializationContext context) {
        JsonObject obj = new JsonObject();

        if (src.name != null) {
            obj.addProperty("name", src.name);
        }
        obj.addProperty("id", src.id);

        if (src.address != null) {
            // try #1
            //JsonObject addressJsonObj = new JsonObject();
            //addressJsonObj.addProperty("street", src.address.street != null ? src.address.street : null);
            //addressJsonObj.addProperty("city", src.address.city != null ? src.address.city : null);
            //addressJsonObj.addProperty("country", src.address.country != null ? src.address.country : null);
            //obj.add("address", addressJsonObj);

            // try #2
            //Gson gson = new GsonBuilder().serializeNulls().create();
            //JsonElement jsonElement = gson.toJsonTree(src.address);
            //obj.add("address", jsonElement);

            // try #3
            //Gson gson2 = new GsonBuilder().serializeNulls().create();
            //obj.addProperty("address", gson2.toJson(src.address));

            // try #4
            //Gson gson = new GsonBuilder().serializeNulls().create();
            //JsonObject jsonObject = gson.toJsonTree(src.address).getAsJsonObject();
            //obj.add("address", jsonObject);
            
            // try #5
            JsonObject addressJsonObj = new JsonObject();
            addressJsonObj.addProperty("street", src.address.street != null ? src.address.street : "null");
            addressJsonObj.addProperty("city", src.address.city != null ? src.address.city : "null");
            addressJsonObj.addProperty("country", src.address.country != null ? src.address.country : "null");
            obj.add("address", addressJsonObj);
        }
        return obj;
    }
}

对于尝试 #3，我构建了不正确的字符串。

对于尝试#1、#2 和#4，我遇到了空值问题。 我在这里搜索并找到了原因和建议

在 JSON “对象”（又名字典）中，有两种方法可以表示缺失值：要么根本没有键/值对，要么使用 JSON 值为 null 的键。

因此，您要么使用具有适当值的 .add ，当您构建 JSON 时将转换为 null，要么您没有 .add 调用。

我的#5方法是检查子节点是否为空，我只是在字面上添加字符串“null”，然后在构建json字符串时替换它

private String parseToGson(User user){
    Gson gson = new GsonBuilder().registerTypeAdapter(User.class, new UserCustomSerializer()).create();
    return gson.toJson(user).replace("\"null\"", "null");
}

这是我定义的一些测试用例

@Test
public void children_attribute_is_null() throws Exception {
    String expected = "{\"name\":\"Adam\"," 
                     + "\"id\":123," 
                     + "\"address\":{" 
                                    + "\""+ "street\":null," 
                                    + "\"city\":\"New York\"," 
                                    + "\"country\":\"US" 
                                + "\"}" 
                     + "}";
    Address address = new Address(null, "New York", "US");
    User user = new User("Adam", 123, address);
    assertEquals(expected, parseToGson(user));
    
    Gson g = new Gson(); 
    User usr = g.fromJson( parseToGson(user), User.class);
    assertEquals("Adam", usr.name);
    assertEquals(123, usr.id);
    assertEquals(null, usr.address.street);
    assertEquals("New York", usr.address.city);
    assertEquals("US", usr.address.country);
}

@Test
public void parent_attribute_is_null() throws Exception {
    String expected = "{\"name\":\"Adam\"," 
                     + "\"id\":123" + "}";
    User user = new User("Adam", 123, null);
    assertEquals(expected, parseToGson(user));
    
    Gson g = new Gson(); 
    User usr = g.fromJson( parseToGson(user), User.class);
    assertEquals("Adam", usr.name);
    assertEquals(123, usr.id);
    assertEquals(null, usr.address);
}

@Test
public void parent_attribute_and_children_attribute_are_null() throws Exception {
    String expected = "{\"id\":123," 
                     + "\"address\":{" 
                                + "\"street\":null," 
                                + "\"city\":\"New York\","
                                + "\"country\":\"US" 
                                + "\"}" 
                     + "}";
    Address address = new Address(null, "New York", "US");
    User user = new User(null, 123, address);
    assertEquals(expected, parseToGson(user));
    
    Gson g = new Gson(); 
    User usr = g.fromJson( parseToGson(user), User.class);
    assertEquals(null, usr.name);
    assertEquals(123, usr.id);
    assertEquals(null, usr.address.street);
    assertEquals("New York", usr.address.city);
    assertEquals("US", usr.address.country);
}

更新 #2

由于以前的版本不是通用版本，我想更新答案。

对于通用，我创建了MyCustomSerializer如下

public class MyCustomSerializer<T> implements JsonSerializer<T> {

    private final Class<T> type;

    public MyCustomSerializer(Class<T> type) {
        this.type = type;
    }

    public Class<T> getMyType() {
        return this.type;
    }

    @Override
    public JsonElement serialize(T src, Type typeOfSrc, JsonSerializationContext context) {
        JsonObject obj = new JsonObject();
        try {
            Field[] declaredFields = this.type.getDeclaredFields();
            for (Field field : declaredFields) {
                Object object = field.get(src);
                if (object != null) {
                    // Here, we check for 4 types of JsonObject.addProperty
                    if (object instanceof String) {
                        obj.addProperty(field.getName(), (String) object);
                        continue;
                    }
                    if (object instanceof Number) {
                        obj.addProperty(field.getName(), (Number) object);
                        continue;
                    }
                    if (object instanceof Boolean) {
                        obj.addProperty(field.getName(), (Boolean) object);
                        continue;
                    }
                    if (object instanceof Character) {
                        obj.addProperty(field.getName(), (Character) object);
                        continue;
                    }
                    // This is where we check for other types
                    // The idea is if it is an object, we need to care its child object as well, so parse it into json string and replace the null value.
                    Gson gson = new GsonBuilder().serializeNulls().create();
                    String json = gson.toJson(object);
                    json = json.replace("null", "\"null\""); // We have to build the string first, then replace it with our special keys. In this case, I use the string "null"
                    JsonObject convertedObject = new Gson().fromJson(json, JsonObject.class); // Then convert it back to json object
                    obj.add(field.getName(), convertedObject);
                }
            }
        } catch (IllegalAccessException e) {
            e.printStackTrace();
        }
        return obj;
    }
}

主要思想仍然与以前的版本相同，但我把它变成了一个通用的。

我还添加了一些额外的属性来测试此代码使用结果构建的字符串

{
   "id":123,
   "address":{
      "street":null,
      "city":"New York",
      "country":"US",
      "info":{
         "zipcode":null,
         "address2":"stackoverflow",
         "workPlaceAddress":{
            "street":null,
            "companyName":"google"
         }
      }
   }
}

要调用这个，我们需要做

private String parseToGson(User user) {
    Gson gson = new GsonBuilder().registerTypeAdapter(User.class, new MyCustomSerializer<>(User.class)).create();
    return gson.toJson(user).replace("\"null\"", "null"); 
}

更新 #3

由于您仍然担心您的解决方案，因此我也尝试对其进行调整

public class YourSerializer <T> implements JsonSerializer<T>{
    
    private final Class<T> type;

    public YourSerializer(Class<T> type) {
        this.type = type;
    }

    public Class<T> getMyType() {
        return this.type;
    }

    @Override
    public JsonElement serialize(T src, Type typeOfSrc, JsonSerializationContext context) {
           Gson gson = new GsonBuilder().serializeNulls().create();
            JsonObject jsonObject = gson.toJsonTree(src).getAsJsonObject();
            
            Field[] declaredFields = this.type.getDeclaredFields();
            for (Field field : declaredFields) {
                try {
                    if(field.get(src) == null) {
                        jsonObject.remove(field.getName());
                    }
                } catch (IllegalAccessException e) {
                    e.printStackTrace();
                }
            }
            return jsonObject;
    }
}

原因是您使用serializeNulls()不正确，这使您的输出不正确。 要纠正它，您应该先registerTypeAdapter创建您的自定义 json，然后调用serializeNulls

private String parseToGson(User user) {
    Gson gson = new GsonBuilder().registerTypeAdapter(User.class, new YourSerializer<>(User.class)).serializeNulls().create();  
    return gson.toJson(user);
}

我用 update#2 测试并得到了相同的结果

{
   "id":123,
   "address":{
      "street":null,
      "city":"New York",
      "country":"US",
      "info":{
         "zipcode":null,
         "address2":"aaa",
         "workPlaceAddress":{
            "street":null,
            "companyName":"google"
         }
      }
   }
}