如何在 spring 引导 JPA 中获得选择性列？

Question

I am new to spring boot.

I have two entities Invite and User.

@Entity
public class Invite {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private int id;
    
    @Column(name="email")
    private String email_id;
    private String mobile;
    private String pincode;
    private String name;
    private String status;
    private Date created_on;
    private Date updated_on;
 
    //gettes and setters
}


@Entity
public class User {
    @Id
    @GeneratedValue(strategy= GenerationType.IDENTITY)
    private int id;
    
    @OneToOne
    private Invite invite;
    private String firstName;
    private String LastName;
    @JsonIgnore
    private String password;
    private Date created_on;
    private Date updated_on;

    // getters and setters

}

I want the following data

+---------+---------------------------------------------+
|  id     |   firstName    email_id         mobile      |
+-------------------+------------+----------------------+
|         |         |            |                      |
| 1       |Ram      | s@mail.com |      1111111111      |
|         |         |            |                      |
+---------+---------+------------+----------------------+

So I have created the below query in UserRepository:

@Repository
public interface UserRepository extends JpaRepository<User, Integer> {

    
    @Query("select u.id,u.firstName,i.email_id, i.mobile from User u inner join u.invite i")
    public List<User> getUserDetails();
}

So I am getting the below error -

Failed to convert from type [java.lang.Object[]] to type [@org.springframework.data.jpa.repository.Query com.example.jpa.JPA.entity.User] for value

I know it's because the return value is of type Object and I am trying to get it into user.

Could you please give me a way to get any data without creating extra DTO classes or interface?

Answer 1

Its great you already know what the issue is.

@Query("select u.id,u.firstName,i.email_id, i.mobile from User u inner join u.invite i")
List<User> getUserDetails();

You are selecting a few columns but asking Data-Jpa to return you back with the a List of User Objects.

So you need to understand first, what JPA specification says about using JPQL.

• select u from User u where u.id =:id -> this returns a List of User type Objects This query in native is: SELECT * FROM USER WHERE ID = "{whatever value you passed}";

• select u.username from User u where u.id =:id" -> this returns just the username, which lets say is a String type. So the return object type is of the String type in case you using TypedQuery or Object is returned in case you using Query . The corresponding SQL generated is SELECT u.USERNAME FROM USER u WHERE u.id = {id}

• select u.id, u.username from User u where u.id =:id -> this returns Object[]. This is when you need the DTO projection to map it to a proper Type of what you want.

With native JPA implementations you can use DTO Projection or Query Projection

I mostly use DTO Projection from JPA spec ad not use Query Projection, so I am giving an example of DTO projection, but above documents if gone through gives a good reference how to work with each abstraction/feature.

1. Define a class to represent the DTO and package I am taking into consideration: com.example. Package name is important since you need the full class name for reference in JPQL

package com.example;


class UserDTO {
 int id;
 String firstName,
 String email_id;
 String mobile;

 //All args constructor is needed and madatory
public User(int id, String firstName, String email_id, String mobile) {
 // assignments
}
  // getters and setters
}

In the Repository:

@Query("select new com.example.UserDTO(u.id,u.firstName,i.email_id, i.mobile) from User u inner join u.invite i")
List<UserDTO> getUserDetails();

The most important thing being: new fully-qualifed-classname(all columns you are fetching)

This is how you use DTO projection with JPQL and Data Jpa Repository.

Answer 2

Write a constructor in User class for your selective column and use in query. Make sure you are also using a constructor in Invite class for email_id and mobile and use it inside constructor in User.

@Query("select new com.example.jpa.JPA.entity.User(u.id,u.firstName,i.email_id, i.mobile) from User u inner join u.invite i")
public List<User> getUserDetails();

Use this constructor in User and Invite class

public User(int id, String firstName, String email_id, String mobile) {
 // assign id and firstName
 // call invite class constuctor 
}

public Invite(String email_id, String mobile) {
 // assign email_id and mobile
}

Answer 3

If you want to have selected fields in query result or customize query result, you can simply use jpa Projection . create an interface with related property method

Answer 4

try this..this work for me: spring.jpa.hibernate.naming-strategy=org.hibernate.cfg.DefaultNamingStrategy

this is to be added in application.properties