我使用的是springboot 2.1.2,我遇到了一些仓库方面的问题。下面是我的实体类:
@Entity
@Getter
@Setter
@NoArgsConstructor
@Table(name = "orders")
public class Order {
@Id
@GeneratedValue
private long id;
@ManyToOne(targetEntity = User.class)
private User user;
//other fields and getters and setters are ignored
}
@Entity
@Getter
@Setter
@NoArgsConstructor
public class User {
@Id
private String email;
//other fields and getters and setters are ignored
}
以及我的OrderRepository:
@Repository
public interface OrderRepository extends JpaRepository<Order, Long> {
@Query("select o from Order o where o.user.email = ?1")
List<Order> findAllByUserId(String userId);
List<Order> findAllByUser(User user);
}
当我调用findAllByUserId或findAllByUser时,存储库返回一个null值而不是一个空列表,这很奇怪,因为我确信我的数据库有数据。
我读过其他类似的问题,它们似乎没有帮助。
我尝试使用调试器修复该问题,并跟踪到AsyncExecutionInterceptor类:
@Nullable
public Object invoke(MethodInvocation invocation) throws Throwable {
Class<?> targetClass = invocation.getThis() != null ? AopUtils.getTargetClass(invocation.getThis()) : null;
Method specificMethod = ClassUtils.getMostSpecificMethod(invocation.getMethod(), targetClass);
Method userDeclaredMethod = BridgeMethodResolver.findBridgedMethod(specificMethod);
AsyncTaskExecutor executor = this.determineAsyncExecutor(userDeclaredMethod);
if (executor == null) {
throw new IllegalStateException("No executor specified and no default executor set on AsyncExecutionInterceptor either");
} else {
Callable<Object> task = () -> {
try {
Object result = invocation.proceed();
if (result instanceof Future) {
return ((Future)result).get();
}
} catch (ExecutionException var4) {
this.handleError(var4.getCause(), userDeclaredMethod, invocation.getArguments());
} catch (Throwable var5) {
this.handleError(var5, userDeclaredMethod, invocation.getArguments());
}
return null;
};
return this.doSubmit(task, executor, invocation.getMethod().getReturnType());
}
}
我注意到,在这个方法的第13行,变量结果是带有适当Order对象的List,但是if子句失败,因此返回空值。
有人知道如何解决这个问题吗?
为了使它更清楚,我将显示我的db模式:
第一次
下面是Hibernate生成的sql:
Hibernate: select order0_.id as id1_8_, order0_.address_id as address_6_8_, order0_.date as date2_8_, order0_.deliver_time as deliver_3_8_, order0_.restaurant_id as restaura7_8_, order0_.status as status4_8_, order0_.total as total5_8_, order0_.user_email as user_ema8_8_ from orders order0_ where order0_.user_email=?
Hibernate: select address0_.id as id1_1_0_, address0_.location as location2_1_0_, address0_.name as name3_1_0_, address0_.phone as phone4_1_0_, address0_.user_email as user_ema5_1_0_, user1_.email as email1_14_1_, user1_.password as password2_14_1_, user1_.pts as pts3_14_1_, user1_.status as status4_14_1_, user1_.user_name as user_nam5_14_1_ from address address0_ left outer join user user1_ on address0_.user_email=user1_.email where address0_.id=?
Hibernate: select restaurant0_.id as id1_9_0_, restaurant0_.email as email2_9_0_, restaurant0_.location as location3_9_0_, restaurant0_.name as name4_9_0_, restaurant0_.password as password5_9_0_, restaurant0_.phone as phone6_9_0_, restaurant0_.status as status7_9_0_, restaurant0_.type as type8_9_0_, restaurant0_.vcode as vcode9_9_0_ from restaurant restaurant0_ where restaurant0_.id=?
Hibernate: select address0_.id as id1_1_0_, address0_.location as location2_1_0_, address0_.name as name3_1_0_, address0_.phone as phone4_1_0_, address0_.user_email as user_ema5_1_0_, user1_.email as email1_14_1_, user1_.password as password2_14_1_, user1_.pts as pts3_14_1_, user1_.status as status4_14_1_, user1_.user_name as user_nam5_14_1_ from address address0_ left outer join user user1_ on address0_.user_email=user1_.email where address0_.id=?
2条答案
按热度按时间jv4diomz1#
你的代码是模棱两可的。如果你想通过电子邮件获得订单列表,那么就写如下。没有必要的
@Query
和实体类,如下所示
我使用的是springboot 2.1.2,我遇到了一些仓库方面的问题。下面是我的实体类:
kmbjn2e32#
OrderRepository
接口中的以下方法签名应该有效:此外,您还可以添加下一个:
更新
尝试查询所有实体,尝试从存储库执行
findAll
,如果它也是null
,则代码库或配置中存在一些问题。请尝试更新类别Map定义: