如何在spring-data中插入db?

que*_*326 6 spring spring-data

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

实体

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


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

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

@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);
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调节器

@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);
}
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当我运行我的测试时,我收到此错误: 引起:org.hibernate.hql.ast.QuerySyntaxException:期待OPEN,在第1行第32列附近找到'c'[插入TaskDocumentEntity c(c.idTask,c.descriere) ,c.filepath)values(:id,:descriere,:filepath)] 我试图删除别名c,但仍然无法正常工作.

pez*_*tem 8

Spring数据提供了save用于插入数据库的开箱即用方法 - 无需使用@Query.看一下springData的核心概念(http://docs.spring.io/spring-data/jpa/docs/current/reference/html/#repositories.core-concepts)

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);
    }
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小智 5

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 中为我工作(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);
}
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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)
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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