spring 如何在spring-data中插入db？
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How to insert into db in spring-data?
提问by que1326
I want to make a request that inserts data into my database. The table has 4 columns: ID_DOCUMENT (PK), ID_TASK, DESCRIPTION, FILEPATH
我想发出一个将数据插入我的数据库的请求。该表有 4 列：ID_DOCUMENT (PK)、ID_TASK、DESCRIPTION、FILEPATH
Entity
实体
...
@Column(name = "ID_TASK")
private Long idTask;
@Column(name = "DESCRIPTION")
private String description;
@Column(name = "FILEPATH")
private String filepath;
...
Repository
存储库
@Modifying
@Query("insert into TaskDocumentEntity c (c.idTask, c.description, c.filepath) values (:id,:description,:filepath)")
public void insertDocumentByTaskId(@Param("id") Long id,@Param("description") String description,@Param("filepath") String filepath);
Controller
控制器
@RequestMapping(value = "/services/tasks/addDocument", method = RequestMethod.POST)
@ResponseBody
public void set(@RequestParam("idTask") Long idTask,@RequestParam("description") String description,@RequestParam("filepath") String filepath){
//TaskDocumentEntity document = new TaskDocumentEntity();
taskDocumentRepository.insertDocumentByTaskId(idTask,descriere,filepath);
}
When I run my test, I get this error: Caused by: org.hibernate.hql.ast.QuerySyntaxException: expecting OPEN, found 'c' near line 1, column 32 [insert into TaskDocumentEntity c (c.idTask, c.descriere, c.filepath) values (:id,:descriere,:filepath)]I tried to remove the alias c, and still doesn`t work.
当我运行我的测试时，我得到这个错误： 由：org.hibernate.hql.ast.QuerySyntaxException：期望打开，在第 1 行，第 32 列附近找到“c”[插入到 TaskDocumentEntity c (c.idTask, c.descriere , c.filepath) values (:id,:descriere,:filepath)]我试图删除别名 c，但仍然不起作用。
回答by pezetem
Spring data provides out of the box savemethod used for insertion to database - no need to use @Query. Take a look at core concepts of springData (http://docs.spring.io/spring-data/jpa/docs/current/reference/html/#repositories.core-concepts)
Spring data 提供了save用于插入数据库的开箱即用的方法 - 无需使用@Query. 查看 springData 的核心概念（http://docs.spring.io/spring-data/jpa/docs/current/reference/html/#repositories.core-concepts）
thus in your controller just create object TaskDocumentEntityand pass it to repository
因此在您的控制器中只需创建对象TaskDocumentEntity并将其传递给存储库
@RequestMapping(value = "/services/tasks/addDocument", method = RequestMethod.POST)
@ResponseBody
public void set(@RequestParam("idTask") Long idTask,@RequestParam("description") String description,@RequestParam("filepath") String filepath){
// assign parameters to taskDocumentEntity by constructor args or setters
TaskDocumentEntity document = new TaskDocumentEntity(idTask,descriere,filepath);
taskDocumentRepository.save(document);
}
回答by A.P
There is a way to do this but it depends on the db you're using. Below worked for me in Oracle (using Dual table):
有一种方法可以做到这一点，但这取决于您使用的数据库。下面在 Oracle 中对我来说有效（使用双表）：
@Repository
public interface DualRepository extends JpaRepository<Dual,Long> {
@Modifying
@Query("insert into Person (id,name,age) select :id,:name,:age from Dual")
public int modifyingQueryInsertPerson(@Param("id")Long id, @Param("name")String name, @Param("age")Integer age);
}
So in your case, it would be (if Oracle):
因此，在您的情况下，它将是（如果是 Oracle）：
@Modifying
@Query("insert into TaskDocumentEntity (idTask,description,filepath) select :idTask,:description,:filepath from Dual")
public void insertDocumentByTaskId(@Param("idTask") Long id,@Param("description") String description,@Param("filepath") String filepath)
I'm not sure which db you're using, here's a link which shows at the bottom which db's support select stmts without a from clause : http://modern-sql.com/use-case/select-without-from
我不确定您使用的是哪个 db，这里有一个链接，它在底部显示哪个 db 支持 select stmts 没有 from 子句：http://modern-sql.com/use-case/select-without-from
