用于序列化List< User>的自定义序列化器在List< String>中 [英] Custom serializer for serializing a List<User> in List<String>

问题描述

我有一个Model对象组

I've a Model object Group

public class Group {
    String title;
    List<User> members;
    String createdBy;
}

我正在使用Jackson来序列化这个对象。我没有在列表members中序列化整个User对象，而是想要序列化 user.getTitle（）字段。
基本上我希望HashMap类似于

I'm using Jackson to serialize this Object. Instead of serializing the whole User object in list "members" I want to serializer just the user.getTitle() field. Basically I want a HashMap to be something like

{
  "title" : "sometitle"
  "members" : [user1.getTitle(), user2.getTitle()]
}

我为此编写了一个自定义序列化器

I've written a custom serializer for this

public class GroupSerializer extends JsonSerializer<Circle> {

    @Override
    public void serialize(Group value, JsonGenerator gen, SerializerProvider serializers) throws IOException, JsonProcessingException {

        if(value != null) {
            gen.writeStartObject();
            gen.writeStringField("title", value.getTitle());
            gen.writeStringField("createdBy", value.getCreatedBy());
            gen.writeFieldName("members");
            gen.writeStartArray();
            for(User user : value.getMembers()) {
                gen.writeString(user.getEmail());
            }
            gen.writeEndArray();
            gen.writeEndObject()
        }
   }
}

但它不起作用。如何仅序列化List的字段而不是整个用户对象？

But it's not working. How do I serialize only a field of List instead of whole User Object?

推荐答案

我建议您查看Jackson的 转换器 界面，它似乎比创建自定义序列化程序更适合任务。

I suggest that you look into Jackson's Converter interface, which seems more suited to the task than creating a custom serializer.

创建的一种方法转换器实例并将其添加到 ObjectMapper ，以便它将用于序列化所有用户实例。

One approach it to create a Converter instance and add it to the ObjectMapper, so that it will be used for the serialization of all User instances.

public class UserConverter extends StdConverter<User, String> {
    @Override
    public String convert(User user) {
        return user.getTitle();
    }
}

在 ObjectMapper上注册像这样：

SimpleModule simpleModule = new SimpleModule();
simpleModule.addSerializer(User.class, new StdDelegatingSerializer(new UserConverter()));

ObjectMapper om = new ObjectMapper().registerModule(simpleModule);

另一种方法，万一你不想转换所有用户实例到字符串，是用这样的转换器注释所选属性：

Another approach, in case you don't want to convert all User instances to String, is to annotate selected properties with a converter like this:

public class Group {
    String title;
    @JsonSerialize(converter = ListUserConverter.class)
    List<User> members;
    String createdBy;
}

并且有一个相应的转换器，如下所示：

And have a corresponding converter that looks something like this:

public class ListUserConverter extends StdConverter<List<User>, List<String>> {
    @Override
    public List<String> convert(List<User> users) {
        return users.stream().map(User::getTitle).collect(Collectors.toList());
    }
}

这篇关于用于序列化List< User>的自定义序列化器在List< String>中的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！