我是Spring(REST)的新手,我正在构建一个简单的REST Web服务。我使用RestController
来映射HTTP请求。我使用这个简单的方法来接受POST请求:
@RequestMapping(value = "/grade/create", method = RequestMethod.POST)
public ResponseEntity<Grade> createGrade(@RequestBody Grade grade)
{
dao.createGrade(grade);
return new ResponseEntity<Grade>(grade, HttpStatus.OK);
}
在这种情况下, dao
是一个带有@Repository
注释的类。请求shou; d看起来像这样:
{
"gradeType": "REGULAR",
"grade": "10",
"passed": 1,
"userId": 1,
"exam_id": 1,
"user_id": 3
}
问题是,Grade
有2个外键,用于用户和考试。我想能够只传递外国实体的ID,让Hibernate处理剩下的事情。但是,目前我将此作为回复:
{
"timestamp": 1484758525821,
"status": 500,
"error": "Internal Server Error",
"exception": "org.springframework.dao.InvalidDataAccessResourceUsageException",
"message": "could not execute statement; SQL [n/a]; nested exception is org.hibernate.exception.SQLGrammarException: could not execute statement",
"path": "/grade/create"
}
我该如何解决这个问题?我已经听说过关于JpaRepository
的一些事情,我可以用它来实现这个目标吗?
我的Grade
模式如下:
@Entity
@Table(name = "grades")
@NamedQueries(value = {
@NamedQuery(name = "Grade.get", query = "SELECT c FROM Grade c WHERE id = :id"),
@NamedQuery(name = "Grade.getAll", query = "SELECT c FROM Grade c"),
@NamedQuery(name = "Grade.getAllByUser", query = "SELECT g FROM Grade g INNER JOIN Exam e ON g.exam.id = e.id INNER JOIN Course c ON e.course.id = c.id WHERE g.user.id = :id"),
})
public class Grade {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private int id;
@Column(name = "type")
private String gradeType;
@Column(name = "grade")
private String grade;
@Column(name = "passed")
private int passed;
@JsonIgnore
@ManyToOne
@JoinColumn(name = "user_id")
private User user;
@ManyToOne
@JoinColumn(name = "exam_id")
private Exam exam;
private int examId;
private int userId;
public Grade(int id, String gradeType, String grade, int passed, User user, Exam exam)
{
setId(id);
setGradeType(gradeType);
setGrade(grade);
setPassed(passed);
setUser(user);
setExam(exam);
}
public Grade() {
}
我的存储库......
@Repository
public class GradeDao {
@PersistenceContext
private EntityManager em;
@Transactional
public List<Grade> getallGrades() {
return em.createNamedQuery("Grade.getAll", Grade.class)
.getResultList();
}
public List<Grade> getLimitedGradesByUser(int limit, int user_id) {
return em.createNamedQuery("Grade.getAllByUser", Grade.class)
.setParameter("id", user_id)
.setMaxResults(limit)
.getResultList();
}
@Transactional
public Grade getGrade(int id) {
return em.createNamedQuery("Grade.get", Grade.class).setParameter("id", id).getSingleResult();
}
@Transactional
public Grade createGrade(Grade grade) {
grade = em.merge(grade);
em.persist(grade);
return grade;
}
}
由于
答案 0 :(得分:2)
我该如何解决这个问题?我听说过一个关于JpaRepository的东西,我可以用它来完成这个吗?
是的,您可以使用JpaRepository执行此操作。您可以开始添加Interface
public interface GradeRepository extends PagingAndSortingRepository<Grade, Long>
,User
和Exam
相同
然后,当您在控制器中POST
Grade
时(我建议使用Service
)
1.创建new Grade()
2.使用setters
将您的User
和Exam
与@Autowire
和UserRepository
ExamRepository
联系起来findById
<登记/>
3.然后使用POST
来从setters
获取您的其他字段
4.从.save(grade)
@Autowired
GradeRepository
请注意,User
和Exam
应该已存在于DB
中,以便能够将其链接
此外,我强烈建议您不要通过HTTP
传递直接实体,而是使用DTOs
,而Grade
json永远不会传递exam id
。尝试通过其他字段查找考试,而不是table id
(例如:findByName)
我使用JPA和安全性实现了一个功能齐全的SpringBoot应用程序,你可能看看https://github.com/hodispk/internship