使用 Spring Data JPA 选择一列
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Select one column using Spring Data JPA
提问by Alax
Does anyone have any idea how to get a single column using Spring Data JPA? I created a repository like below in my Spring Boot project, but always get the {"cause":null,"message":"PersistentEntity must not be null!"}
error when accessing the Restful URL.
有谁知道如何使用 Spring Data JPA 获取单个列?我在 Spring Boot 项目中创建了一个如下所示的存储库,但{"cause":null,"message":"PersistentEntity must not be null!"}
在访问 Restful URL 时总是出现错误。
@RepositoryRestResource(collectionResourceRel = "users", path = "users")
public interface UsersRepository extends CrudRepository<Users, Integer> {
@Query("SELECT u.userName FROM Users u")
public List<String> getUserName();
}
Then if I access the Restful URL like ../users/search/getUserName
, I get the error:
{"cause":null,"message":"PersistentEntity must not be null!"}
然后,如果我访问 Restful URL,例如../users/search/getUserName
,我会收到错误消息:
{"cause":null,"message":"PersistentEntity must not be null!"}
回答by Abhijeet Behare
This Works for me.
这对我有用。
public interface UserDataRepository extends JpaRepository<UserData, Long> {
@Query(value = "SELECT emp_name FROM user_data", nativeQuery = true)
public List<Object[]> findEmp_name();
}
System.out.println("data"+ userDataRepository.findEmp_name());
The above line gave me this result :
上面的行给了我这个结果:
data[abhijeet, abhijeet1, abhijeet2, abhijeet3, abhijeet4, abhijeet5]
数据[abhijeet, abhijeet1, abhijeet2, abhijeet3, abhijeet4, abhijeet5]
回答by Ethiraj
Concept is : In your entity class create a constructor with only required instant variables. And use that constructor in the repository method shown below.
概念是:在您的实体类中创建一个仅包含所需即时变量的构造函数。并在下面显示的存储库方法中使用该构造函数。
Lets say you have a interface Repository like below
假设您有一个如下所示的接口存储库
Repository implementation:
public interface UserRepository<User> extends JpaRepository<User,String> { @Query(value = "select new com.org.User(usr.userId) from User usr where usr.name(:name))") List<User> findUserIdAlone(@Param("name") String user); }
In Controller
@RestController public class UserController { @Autowired private UserRepository<User> userRepository; @Res public ResponseEntity<User> getUser(@PathVariable("usrname") String userName) { User resultUser = usrRepository.findUserIdAlone(userName); return ResponseEntity.ok(resultUser); } } public class User { private String userId,userName; public User(String userId) { this.userId=userId; } // setter and getters goes here }
存储库实现:
public interface UserRepository<User> extends JpaRepository<User,String> { @Query(value = "select new com.org.User(usr.userId) from User usr where usr.name(:name))") List<User> findUserIdAlone(@Param("name") String user); }
在控制器中
@RestController public class UserController { @Autowired private UserRepository<User> userRepository; @Res public ResponseEntity<User> getUser(@PathVariable("usrname") String userName) { User resultUser = usrRepository.findUserIdAlone(userName); return ResponseEntity.ok(resultUser); } } public class User { private String userId,userName; public User(String userId) { this.userId=userId; } // setter and getters goes here }
回答by Murder
If you need list all of the users, try select userName from Users
, if you need one user use "where"
look at spring data JPA http://docs.spring.io/spring-data/jpa/docs/current/reference/html/, try change CrudRepository to JpaRepository
如果您需要列出所有用户,请尝试select userName from Users
,如果您需要一个用户使用"where"
查看 spring 数据 JPA http://docs.spring.io/spring-data/jpa/docs/current/reference/html/,请尝试更改 CrudRepository到 JpaRepository
回答by Hatem Jaber
If you want to only return a single column you should look at Projections and Excerptswhich will allow you to filter specific columns and other things that are usefule.
如果您只想返回单个列,您应该查看Projections 和 Excerpts,这将允许您过滤特定的列和其他有用的内容。