How to insert into db in spring-data?

Question

I want to make a request that inserts data into my database. The table has 4 columns: ID_DOCUMENT (PK), ID_TASK, DESCRIPTION, FILEPATH

Entity

... 
@Column(name = "ID_TASK")
private Long idTask;


@Column(name = "DESCRIPTION")
private String description;

@Column(name = "FILEPATH")
private String filepath;
...

Repository

@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

@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.

Remove your insertDocumentByTaskId. In your controller create an instance of TaskDocumentEntity set the properties and call the save method. The whole point of JPA is that you don't have to write your queries anymore... — M. Deinum
– M. Deinum, Commented Oct 7, 2015 at 10:23
Please post the code for your repository class not only the method declaration and also post your spring configuration class/xml. — dubonzi
– dubonzi, Commented Oct 7, 2015 at 14:32

pezetem · Accepted Answer · 2015-10-07 10:53:57Z

11

Spring data provides out of the box save method 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)

thus in your controller just create object TaskDocumentEntity and pass it to repository

@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);
    }

answered Oct 7, 2015 at 10:53

pezetem

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2 Comments

que1326 Over a year ago

I tried this also, but doesn`t persist, no changes the database (succesful test)

que1326 Over a year ago

It works now, I didnt inject the service in the test class, and didnt pass any params for the request, anyway I `ve rewritten the controller, thanks a lot guys !!!

user3448327 · Accepted Answer · 2017-08-17 22:42:22Z

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):

@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):

@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

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