如何用Spring RestController只对对象的子对象应用@JsonView--序列化?

问题描述 投票:0回答:1

我有这样的案例。

public class Project{

    private long id;

    private List<User> users;

    // other properties and getter setter etc
}

public class User{

   @JsonView(MinimalUser.class) 
   private long id;

   @JsonView(MinimalUser.class)
   private String name;

   private ComplexObject anything;

}

现在是RestMethod:


    @JsonView(MinimalUser.class)
    @GetMapping("/client/{id}")
    public List<Project> findProjectsByClientId(@PathVariable long id) {
        return projectService.findProjectsByClientId(id);
    }

这里我只是想用最小的Users对象初始化一个项目,但是什么都不会被初始化,因为在我的项目中没有 "MinimalUser.class "JsonView,因此它没有被初始化。

我不想把 @JsonView Annotation 放在 Project 中的所有变量上,因为这将是矫枉过正。

我如何告诉Controller只对Project的User(child)应用@JsonView-FilteringSerialization?

因为这个调用就很好用(我只得到了必填字段)。


    @JsonView(MinimalUser.class)
    @GetMapping("/minimal")
    public List<User> findAllUsers() {
        return userService.findAllUsers();
    }

java spring serialization jackson json-view
1个回答
1
投票

如果你扩展 MinimalUser 视图,并将此新视图应用于整个项目,您可以得到以下结果。

public class MinimalProject extends MinimalUser {}
public class DetailedProject extends MinimalProject {}

// apply view to entire class
@JsonView(MinimalProject.class)
public class Project {
    // keep existing fields as is

    // limit view for project
    @JsonView(DetailedProject.class)
    private String details;
}

// test
ObjectMapper mapper = new ObjectMapper().disable(MapperFeature.DEFAULT_VIEW_INCLUSION);

User u = new User();
u.setId(1);
u.setName("John Doe");
u.setEmail("[email protected]");
u.setDetails("some details");

System.out.println("user default view: " + mapper.writeValueAsString(u));

System.out.println("user minimal view: " + mapper.writerWithView(MinimalUser.class).writeValueAsString(u));

Project p = new Project();

p.setId(1000);
p.setName("MegaProject");
p.setUsers(Arrays.asList(u));
p.setDetails("Project details: worth $1M");

System.out.println("project default view: " + mapper.writeValueAsString(p));

System.out.println("project minimal view: " + mapper.writerWithView(MinimalProject.class).writeValueAsString(p));

输出:

user default view: {"id":1,"name":"John Doe","details":"some details","email":"[email protected]"}
user minimal view: {"id":1,"name":"John Doe"}
project default view: {"id":1000,"users":[{"id":1,"name":"John Doe","details":"some details","email":"[email protected]"}],"name":"MegaProject","details":"Project details: worth $1M"}
project minimal view: {"id":1000,"users":[{"id":1,"name":"John Doe"}],"name":"MegaProject"}
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