Question

我正在使用Spring API和连接到MySQL数据库的JPA构建一个带有REST API的应用程序。要在User类中搜索名称，我实现了一种查询方法：List< User > findByFirstnameLike( String name );

此方法仅返回完全匹配的结果，这不是我想要的。我误解了它的用法吗？

存储库界面：

public interface UserRepository extends JpaRepository<User, Long> {

    public List< User > findByFirstname( String name );

    public List< User > findByFirstnameLike( String name );
}

服务bean方法：

@Override
public List<User> findByFirstNameLike(String name) {
    logger.info( "searching for first name: {}", name);
    List< User > ret = userRepo.findByFirstnameLike(name);

    if( null == ret ){
        logger.info("No list returned from search");
    }
    else{
        logger.info( "List size = {}", ret.size() );
    }

    return( ret );
}

REST接口方法：

@RequestMapping(
        value="/{firstName}",
        method=RequestMethod.GET,
        produces=MediaType.APPLICATION_JSON_VALUE )
public ResponseEntity< List< User > > SearchForUserByFirstName( @PathVariable( "firstName" ) String firstName ){
    return( new ResponseEntity< List< User > >( userService.findByFirstNameLike( firstName ), HttpStatus.OK) );
}

实体类：

@Entity
public class User {

    public enum Department {
        BS, BA, BT, BD, UX, SALES
    }

    @Id
    @GeneratedValue
    private Long id;

    private String firstname;

    private String lastname;

    private String email;

    private String phone;

    private Department department;

所以......在数据库中有一个名为“Adam”的用户，查询方法返回除“Adam”之外的任何字符串的空列表。

编辑2：在打开show SQL并将对象插入数据库之后，我搜索了第一个名称的一部分并产生了这个输出和一个空列表：

searching for first name: dam <-- This is the log print
Hibernate: select user0_.id as id1_6_, user0_.department as 
departme2_6_, user0_.email as email3_6_, user0_.firstname as 
firstnam4_6_, user0_.lastname as lastname5_6_, user0_.phone as 
phone6_6_ from user user0_ where user0_.firstname like ?

精确匹配的搜索，返回带有一个对象的数组：

searching for first name: Adam
Hibernate: select user0_.id as id1_6_, user0_.department as 
departme2_6_, user0_.email as email3_6_, user0_.firstname as 
firstnam4_6_, user0_.lastname as lastname5_6_, user0_.phone as 
phone6_6_ from user user0_ where user0_.firstname like ?

Answer 1

尝试使用findByFirstnameContaining代替findByFirstnameLike。

我不确定如何解释like。但是docs说：

Keyword     | Sample                    | JPQL snippet

Containing  | findByFirstnameContaining | … where x.firstname like ?1 (parameter bound wrapped in %)

我想你想搜索%dam%。

如果您想在开头或结尾处绑定通配符，也可以：findByFirstnameStartingWith或findByFirstnameEndingWith。

Answer 2

来自documentation：

关键字|样品JPQL代码段
喜欢| findByFirstnameLike…|其中x.firstname像？1

作为参数传递的值必须包含'％'，例如：

列表ret = userRepo.findByFirstnameLike（name +“％”）;

使用spring查询方法

2 个答案: