Spring boot join query Validation failed for query for public abstract in using three entity

Question

我使用的是弹簧靴。我曾经加入三个实体来获取存储库中的数据。但它显示以下错误..

Caused by: org.springframework.beans.factory.BeanCreationException: Error creating bean with name 'menuRightRepo': Invocation of init method failed; nested exception is java.lang.IllegalArgumentException: Validation failed for query for method public abstract java.util.List...  

Caused by: java.lang.IllegalArgumentException: Validation failed for query for method public abstract java.util.List

我使用了下面的代码..

1. 我的第一个名为 MenuNameEntity 的实体如下所示

   @Entity
   @Table(name = "MenuName")
   public class MenuNameEntity {
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name = "id")
    private long id;
   
    private String parentId;
    private String menuName;
    private String status;
    // getter and setter
    }
   

2. 我的第二个实体在下面

   @Entity
   @Table(name = "MenuChild")
   public class MenuChildEntity {
   
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name = "id")
    private long id;   
    private String parentId;   
    private String childMenuName;  
    private String url;    
    private String status;
    //getter and setter
   }
   

3. 我的第三个实体在下面

   @Entity
   @Table(name = "MenuRight")
   public class MenuRightEntity {
     @Id
     @GeneratedValue(strategy = GenerationType.IDENTITY)
     @Column(name = "id")
     private long id;
   
     private String companyId; 
     private String userId;    
     private String url;   
     private String status;    
     private String enqMode;   
     private String insertMode;    
     private String updateMode;    
     private String deleteMode;
    //getter and setter
    }
   

4. 我的 MenuRightRepository 在下面

    public interface MenuRightRepo extends JpaRepository<MenuRightEntity, Long> {
   
     @Query("select new com.rms.info.MenuRightResponse(m.companyId, m.userId,m.enqMode,m.insertMode,m.updateMode,m.deleteMode, p.parentId,c.childId,c.childMenuName,c.url) from MenuNameEntity p,MenuChildEntity c,MenuRightEntity m where p.parentId = c.parentId  and p.status =1 and c.status =1 and c.url= m.url ")
    List<MenuRightResponse> getMenuMenuRights();   
   
   }
   

5. 我已经使用 MenuRightResponse 来绑定从存储库中获取的值

     package com.rms.info;
     public class MenuRightResponse {
   
   private String companyId= "";
   private String userId= "";
   private String enqMode= "";
   private String insertMode= "";
   private String updateMode= "";
   private String deleteMode= "";
   private String parentId= "";
   private String childId= "";
   private String childMenuName= "";
   private String url= "";     
   
     public MenuRightResponse(String companyId, String userId, String enqMode, String insertMode, String updateMode,
       String deleteMode, String parentId, String childId, String childMenuName, String url) {
   super();
   this.companyId = companyId;
   this.userId = userId;
   this.enqMode = enqMode;
   this.insertMode = insertMode;
   this.updateMode = updateMode;
   this.deleteMode = deleteMode;
   this.parentId = parentId;
   this.childId = childId;
   this.childMenuName = childMenuName;
   this.url = url;
     }
   //getter and setter
    }
   

当我在 MenuRightRepo 中调用 getMenuMenuRights() 时，它会显示上述错误。我没有使用任何主键和外键。代码有什么问题。请帮帮我

Answer 1

我已经解决了下面的问题...

1. 我已将我的 MenuNameEntity 更改为

   @Entity
   @Table(name = "MenuName")
   public class MenuNameEntity {
   
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name = "id")
    private long id;
   
   private String parentId;
   private String menuName;
   private String status;
   
    @OneToMany(fetch = FetchType.LAZY, mappedBy = "MenuName")  
    private Set<MenuChildEntity> menuChildEntities;
   
   // getter and setter
   }
   

2. 我已将我的 MenuChildEntity 更改为

   @Entity
   @Table(name = "MenuChild")
   public class MenuChildEntity {
   
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name = "id")
    private long id; 
   
    @Column(name = "parentId", nullable = false,insertable = false,updatable = false) 
    private String parentId;   
    private String childMenuName;  
    private String url;    
    private String status;
   
    @OneToMany(fetch = FetchType.LAZY, mappedBy = "MenuChild") 
    private Set<MenuRightEntity> menuRightEntities;
   
    @ManyToOne
    @JoinColumn(name = "parentId")
    private MenuNameEntity MenuName;
   
    //getter and setter
    }
   

3. 我已将我的 MenuRightEntity 更改为

   @Entity
   @Table(name = "MenuRight")
   public class MenuRightEntity {
     @Id
     @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name = "id")
    private long id;
   
    private String companyId; 
    private String userId;  
   
   @Column(name = "url", nullable = false,insertable = false,updatable = false)  
    private String url;   
    private String status;    
    private String enqMode;   
    private String insertMode;    
    private String updateMode;    
    private String deleteMode;        
   
    @ManyToOne
    @JoinColumn(name = "url")
    private MenuChildEntity MenuChild;
   
     //getter and setter
   
    }
   

我的存储库查询然后工作正常...我只是加入三个表无条件 一对多和多对一 并加入列 nullable false 工作很好