Question

@Entity
public class Student implements Serializable {

    @Id
    @Column
    @GeneratedValue(strategy = GenerationType.AUTO)
    private Long id;


    @Column(name = "name")
    private String name;

    public Long getId() {
        return id;
    }

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


    public String getName() {
        return name;
    }


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




public interface StudentRepo extends JpaRepository<Student, Long> {


    @Query(value = "select id from student", nativeQuery = true)
    Student findCostum();
}



@RunWith(SpringJUnit4ClassRunner.class)
@ContextConfiguration(classes = {JpatestingApplication.class}, initializers = ConfigFileApplicationContextInitializer.class)
@Profile("default")
public class JpatestingApplicationTests {


    @Autowired
    StudentRepo repo;

    @Test
    public void contextLoads() {

        Student st = repo.findCostum();
        System.out.println(st);
        /*Student st = new Student();
        st.setName("nabeel");
        repo.save(st);*/
    }

}

引起：java.sql.SQLException：找不到列'name'。在 com.mysql.jdbc.SQLError.createSQLException（SQLError.java:964）at at com.mysql.jdbc.SQLError.createSQLException（SQLError.java:897）at at com.mysql.jdbc.SQLError.createSQLException（SQLError.java:886）at at com.mysql.jdbc.SQLError.createSQLException（SQLError.java:860）at at com.mysql.jdbc.ResultSetImpl.findColumn（ResultSetImpl.java:1077）at at com.mysql.jdbc.ResultSetImpl.getString（ResultSetImpl.java:5174）at at org.hibernate.type.descriptor.sql.VarcharTypeDescriptor $ 2.doExtract（VarcharTypeDescriptor.java:62）在 org.hibernate.type.descriptor.sql.BasicExtractor.extract（BasicExtractor.java:47）在 org.hibernate.type.AbstractStandardBasicType.nullSafeGet（AbstractStandardBasicType.java:238）在 org.hibernate.type.AbstractStandardBasicType.nullSafeGet（AbstractStandardBasicType.java:234）在 org.hibernate.type.AbstractStandardBasicType.nullSafeGet（AbstractStandardBasicType.java:224）在 org.hibernate.type.AbstractStandardBasicType.hydrate（AbstractStandardBasicType.java:300）在 org.hibernate.persister.entity.AbstractEntityPersister.hydrate（AbstractEntityPersister.java:2790）在org.hibernate.loader.Loader.loadFromResultSet（Loader.java:1729）在org.hibernate.loader.Loader.instanceNotYetLoaded（Loader.java:1655）在org.hibernate.loader.Loader.getRow（Loader.java:1544）at org.hibernate.loader.Loader.getRowFromResultSet（Loader.java:727）at org.hibernate.loader.Loader.processResultSet（Loader.java:972）at at org.hibernate.loader.Loader.doQuery（Loader.java:930）at org.hibernate.loader.Loader.doQueryAndInitializeNonLazyCollections（Loader.java:336）在org.hibernate.loader.Loader.doList（Loader.java:2617）... 72更多

Answer 1

您的查询：

@Query(value = "select id from student", nativeQuery = true)
Student findCostum();

没有任何意义。来自学生的查询选择ID将返回一个Longs列表（所有学生的ID）而不是单个学生本身。因此它抱怨名称列丢失，因为在此查询中只返回一列，即id。将此更改为：

@Query(value = "select id from student", nativeQuery = true)
List<Long> findCostum();

将修复您的错误，但我不相信这是您想要做的事情。我假设你只是想通过他们的身份找到一个给定的学生？在这种情况下，您可以在JpaRepository https://docs.spring.io/spring-data/jpa/docs/current/api/org/springframework/data/jpa/repository/JpaRepository.html#getOne-ID-

中使用getOne（id）方法

Student st = repo.getOne(id);

如何在spring jpa Repo

1 个答案: