我有一个基本的SpringBoot应用程序。使用Spring Initializer,JPA,嵌入式Tomcat,Thymeleaf模板引擎并将其打包为可执行JAR文件。
我创建了这个Repository类:

@Repository
public interface MenuRepository extends CrudRepository<Menu, Long> {
..
}

和这个服务等级
@Service
@Transactional(readOnly = true)
public class MenuService {

     @Autowired
     protected MenuRepository menuRepository;

     @Transactional
     public void delete (Menu menu) {
         menuRepository.delete  (menu);
     }
     ..
}

这个Junit测试:
@ContextConfiguration(classes={TestSystemConfig.class})
@RunWith(SpringRunner.class)
@SpringBootTest(classes = MenuGestApplication.class)
public class MenuServiceTests {
...
@Test
    public void testDelete () {

        Menu menu = new menu();
        menu.setmenuId("bacalla-amb-tomaquet");
        menuService.save(menu);

        MenuPrice menuPrice = new menuPrice(menu);
        menuPrice.setPrice((float)20.0);
        menuPriceService.save(menuPrice);

        MenuPriceSummary menuPriceSummary = new menuPriceSummary(menu);
        menuPriceSummary.setFortnightlyAvgPrice((float)20.0);

        menuPriceSummaryService.save(menuPriceSummary);

        menu = menuService.findBymenuId("bacalla-amb-tomaquet");

        assertNotNull (menu);

        menuService.delete (menu);

        menu = menuService.findBymenuId("bacalla-amb-tomaquet");

        assertNull (menu);

    }
}

但是Junit失败了,因为没有删除对象并且没有抛出异常!

我建议这样做。
@OneToMany(mappedBy="menu", cascade = CascadeType.ALL,  orphanRemoval = true, fetch=FetchType.LAZY)
    private List<MenuPrice> price;

即使我在运行测试时在控制台中看到了这一点:
Caused by: com.mysql.jdbc.exceptions.jdbc4.MySQLIntegrityConstraintViolationException: Cannot delete or update a parent row: a foreign key constraint fails (`elcormenu`.`t_menu_price`, CONSTRAINT `FK19d0sljpshu4g8wfhrkqj7j7w` FOREIGN KEY (`menu_id`) REFERENCES `t_menu` (`id`))

和Menu类:
@Entity
@Table(name="t_menu")
public class Menu  implements Serializable {

    /**
     *
     */
    private static final long serialVersionUID = 1L;

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @JsonProperty("id")
    private Long id;

    @JsonProperty("MenuId")
    private String MenuId;

    @OneToMany(mappedBy = "Menu", cascade = CascadeType.ALL, orphanRemoval = true, fetch = FetchType.LAZY)
    @JsonIgnore
    private Set<MenuPrice> MenuPrice = new HashSet<>();

    @OneToOne(mappedBy = "Menu", cascade = CascadeType.ALL, fetch = FetchType.LAZY)
    @JsonIgnore
    private MenuPriceSummary summary;
...
}

最佳答案

您将同时使用@OneToMany@ManyToOne定义Menu和MenuPrice之间的双向关系。当您具有生物定向关系时,您需要设置双方。在测试中,您要在MenuPrice中设置Menu,但不将MenuPrice添加到Menu的MenuPrice Set中。创建menuPrice之后,请包含以下语句menu.MenuPrice.add(menuPrice);。由于您已经指定了save(),因此您也不需要多个Cascade.All。请尝试以下方法:

public void testDelete () {

    Menu menu = new menu();
    menu.setmenuId("bacalla-amb-tomaquet");

    MenuPrice menuPrice = new menuPrice(menu);
    menuPrice.setPrice((float)20.0);

    // Not needed
    // menuPriceService.save(menuPrice);

    // Add menuPrice to menu's menuPrice set
    menu.menuPrice.add(menuPrice);

    MenuPriceSummary menuPriceSummary = new menuPriceSummary(menu);
    menuPriceSummary.setFortnightlyAvgPrice((float)20.0);

    // Set menuPriceSummary in menu
    menu.summary = menuPriceSummary;

    // Not needed
    //menuPriceSummaryService.save(menuPriceSummary);

    // Saving menu will save it children too
    menuService.save(menu);

    menu = menuService.findBymenuId("bacalla-amb-tomaquet");

    assertNotNull (menu);

    menuService.delete (menu);

    menu = menuService.findBymenuId("bacalla-amb-tomaquet");

    assertNull (menu);

}

在某些情况下,您不需要双向关系,可以删除@OneToMany,但这取决于您的业务逻辑需要如何导航到子级或通过JPA查询它。

关于java - CrudRepository不会删除具有关系的对象,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/50818106/

10-10 17:22