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java使用userId创建对象列表

我有一个应用程序,用户可以将电影标题添加到列表中。目前我使用以下代码返回列表:

@Override 
public List<Movie> getAllmovies() {

    List<Movie> movies = new ArrayList<Movie>();
    Iterator<Movie> iterator = movieRepository.findAll().iterator();
    while (iterator.hasNext()) {
        movies.add(iterator.next());
    }

    return movies;
}

正如您所料,这将返回所有电影,而不是当前用户的电影。所以我改为使用findAllByUserId()而不是findAll()的代码:

@Override 
public List<Movie> getAllmovies() {

    User current_user = userService.getUser();  
    List<Movie> movies = new ArrayList<Movie>();
    movies = movieRepository.findAllByUserId(current_user.getId());

    return movies;
}

我在我的电影记录片中加入了这个方法:

@Repository
public interface MovieRepository extends JpaRepository<Movie, Serializable> {
    List<Movie> findAllByUserId(Long userId);
}

但现在当我编译代码时,我得到了一个错误:

Caused by: org.springframework.data.mapping.PropertyReferenceException: No property userId found for type Movie! Did you mean 'users'?

这个错误来自哪里?这不是电影故事里的userId论点

//编辑。在混乱了一会儿之后,我发现userIdfindAllByUserId()方法有关。所以这个错误意味着我不能使用userId,因为它不是类型为Movie的属性。那么,如何按用户id返回电影列表呢

//编辑。添加了电影模型:

package com.movieseat.models;

import com.movieseat.model.security.User;
import java.util.HashSet;
import java.util.Set;
import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.ManyToMany;
import javax.persistence.Table;

@Entity(name = "Movie")
@Table(name = "movie")
public class Movie {

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    @Column(name = "id")
    private Integer id;

    private String name;

    public Movie(){}

    public Movie(String name) {
        this.name = name;
    }

    public Movie(Integer id, String name ) {
        this.id = id;
        this.name = name;
    }

    public Movie(String name, Set<User> users){
        this.name = name;
        this.users = users;
    }    

    @ManyToMany(mappedBy = "movies")
    private Set<User> users = new HashSet<>(); 

    public Set<User> getUsers() {
        return users;
    }

    public void setUsers(Set<User> users) {
        this.users = users;
    }

    public Integer getId() {
        return id;
    }

    public void setId(Integer id) {
        this.id = id;
    }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    @Override
    public String toString(){
        return "id: " + id + "name: " + name;
    }

}

共 (1) 个答案

  1. # 1 楼答案

    以下方法签名将获得您想要的:

    List<Movie> findByUsers_Id(Long id)
    

    这是使用Spring数据JPA的属性表达式功能。签名用户Id将被翻译成JPQL x.users.id

    More info here