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JpaRepository与不同表上的Native@Query

nativequery spring-boot spring

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Jakob · 技术社区 · 6 年前

我只是在spring数据中偶然发现了一些意想不到的行为。为了演示我用initializer设置了一些spring引导应用程序( https://start.spring.io/ )添加JPA、Web、H2。

应用程序包含两个表和一些数据:

数据.sql:

create table table1 (
  id int,
  name varchar(50)
);

create table table2 (
  id int,
  name varchar(50)
);

insert into table1 (id, name) values (1, 'First row from table 1');
insert into table1 (id, name) values (2, 'Second row from table 1');
insert into table1 (id, name) values (3, 'Third row from table 1');
insert into table1 (id, name) values (4, 'Fourth row from table 1');

insert into table2 (id, name) values (1, '** TABLE 2: 1st ROW UPPERCASE **');
insert into table2 (id, name) values (2, '** TABLE 2: 2nd ROW UPPERCASE **');

此表结构只有一个模型,因为两个表的结构相同。我为这个表创建了一个JpaRepository

@Repository
public interface SampleDAO extends JpaRepository<SampleModel, Integer> {

  @Query(value = "select id, name from table1", nativeQuery = true)
  List<SampleModel> findAllFromTable1();

  @Query(value = "select id, name from table2", nativeQuery = true)
  List<SampleModel> findAllFromTable2();
}

最后我添加了一个控制器(TestController):

@RestController
public class TestController {

  @Autowired
  private final SampleDAO sampleDAO;

  public TestController(SampleDAO sampleDAO) {
    this.sampleDAO = sampleDAO;
  }

  @GetMapping(path = "/")
  @ResponseBody
  public String testNativeQuery() {
    List<SampleModel> list1 = this.sampleDAO.findAllFromTable1();
    List<SampleModel> list2 = this.sampleDAO.findAllFromTable2();

    SampleModel m1 = list1.get(0);
    SampleModel m2 = list2.get(0);
    System.out.println("*****************************************");
    System.out.println("Data from findAllFromTable1():");
    list1.forEach(l -> {
      System.out.println(l.getName());
    });
    System.out.println("*****************************************");
    System.out.println("Data from findAllFromTable2():");
    list2.forEach(l -> {
      System.out.println(l.getName());
    });
    System.out.println("*****************************************");

    return "Done";
  }
}

我希望list1和list2包含我的两个表的数据,但令人惊讶的是,只获取了第一个表的结果:

*****************************************
Data from findAllFromTable1():
First row from table 1
Second row from table 1
Third row from table 1
Fourth row from table 1
*****************************************
Data from findAllFromTable2():
First row from table 1
Second row from table 1
*****************************************

https://github.com/steinmann321/nativequerydemo

这是预期的行为还是我做错了什么?

(请注意:这只是一个示例项目,实际查询要复杂得多,但结果是相同的)

1 回复 | 直到 6 年前

1

3

M. Rizzo 6 年前

您在持久实体之间的身份冲突方面遇到了问题。您可以看到,您从第一个查询中得到四条记录,从第二个查询中得到两条与数据库中的记录匹配的记录。但是当entitymanager正在处理第二个集合中的记录时,它看到已经有ID为1和2的持久实体,因此它不会创建新的实体。

创建JPA可嵌入密钥:

@Embeddable
public class MyKey implements Serializable {

    private static final long serialVersionUID = 1L;

    @Column(name = "id", nullable = false)
    private String id;

    @Column(name = "name", nullable = false)
    private String name;

    public String getId() {
      return id;
    }

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

    public String getName() {
      return name;
    }

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

    @Override
    public boolean equals(Object o) {
      if (this == o) {
        return true;
      }
      if (o == null || getClass() != o.getClass()) {
        return false;
      }
      MyKey that = (MyKey) o;
      return Objects.equals(id, that.getId()) && Objects.equals(name, that.getName());
    }

    @Override
    public int hashCode() {
      return Objects.hash(id, name);
    }
}

然后使用SampleModel实体中的嵌入键:

@Entity
public class SampleModel {

  @EmbeddedId
  private MyKey myKey;

  public MyKey getMyKey() {
    return myKey;
  }

  public void setMyKey(MyKey myKey) {
    this.myKey = myKey;
  }
}

System.out.println("*****************************************");
System.out.println("Data from findAllFromTable1():");
list1.forEach(l -> {
  System.out.println(l.getMyKey().getName());
});
System.out.println("*****************************************");
System.out.println("Data from findAllFromTable2():");
list2.forEach(l -> {
  System.out.println(l.getMyKey().getName());
});
System.out.println("*****************************************");

然后试试你的TestController。我试过了,结果是:

*****************************************
Data from findAllFromTable1():
First row from table 1
Second row from table 1
Third row from table 1
Fourth row from table 1
*****************************************
Data from findAllFromTable2():
** TABLE 2: 1st ROW UPPERCASE **
** TABLE 2: 2nd ROW UPPERCASE **
*****************************************