再看一下repository 接口的实现类SimpleJpaRepository的源码(只摘了部分源码): 1@Repository2@Transactional(3readOnly =true4)5publicclassSimpleJpaRepository<T, ID>implementsJpaRepositoryImplementation<T, ID>{67@Transactional8publicvoiddeleteById(ID id) {9Assert.notNull(id, "The given id must not be nu...
1@Repository2publicinterfaceUserRepositoryextendsJpaRepository<User,Long>{34//父类的保存方法5@Override6User save(User entity);78//按照JPA语法规则自定义的查询方法9List<User>findFirst10ByLastname(String lastName, Pageable pageable);10} 第四句话的意思是,当加上@Modifying注解时,JPA会以更新类语句来执行...
@Modifying@Query("update User u set u.firstname = ?1 where u.lastname = ?2")intsetFixedFirstnameFor(Stringfirstname,Stringlastname); 首先让人奇怪的是,repository method只能返回int或者转为void,因为这个操作只会把数据写入到数据库,但是不会select。 执行完modifying query, EntityManager可能会包含过时...
User u = repository.findOne(id); u.setFirstName("new first name"); u.setLastName("new last name"); If you have a detached entity and want to merge it, then use the save() method of CrudRepository: User attachedUser = repository.save(detachedUser); Share Improve this answer Follow...
4 Spring Data Repository: Detached entity passed to persist 5 Spring data jpa detached entity 2 jpa hibernate: detached entity passed to persist 0 Persist new JPA entity and detach 0 detached entity passed to persist Spring JPA 1 Spring boot/Spring data jpa - how to...
一、新增批量导入接口BatchSaveRepository isSave: true - save, false - update package com.easemob.oa.persistence.jpa; import org.springframework.data.repository.NoRepositoryBean; import java.util.List; @NoRepositoryBean public interface BatchSaveRepository<T> { ...
* 直接注解在repository方法上声明一个查找器 * */ @Retention(RetentionPolicy.RUNTIME) @Target({ ElementType.METHOD, ElementType.ANNOTATION_TYPE }) @QueryAnnotation @Documented public @interface Query { /** * 定义在执行带有@Query注解方法时,要执行的JPA查询。
Am成功地使用CDI注入jpa存储库。我想将自定义行为(软删除)添加到所有存储库中。当使用spring时,我可以通过指定存储库基类来启用客户行为 @EnableJpaRepositories(repositoryBaseClass = StagedRepositoryImpl.class) 如何在CDI中指定相同的内容?提前谢谢。 浏览1提问于2016-07-13得票数 3 回答已采纳 ...
return jpaUserRepository.findByUsername(request.username) } @PostMapping("/update") fun updateUserByUsername(@RequestBody request: Request): User { val user = jpaUserRepository.findByUsername(request.username) user.phone = request.phone user.email = request.email ...
findById(1); ProductCategory productCategory = repository.findById(1).get(); 2.数据更新 先查出来,然后对结果重新赋值。比插入多了一个查询的步骤。 // [2022-08-05 10:02:05] @Test public void testUpdate() { ProductCategory productCategory = repository.findById(1).get(); productCategory.set...