Roe*_*ker 6 java rest spring json hibernate
我对 Spring (REST) 有点陌生,我正在构建一个简单的 REST web 服务。我使用 aRestController
来映射 HTTP 请求。我使用这个简单的方法来接受一个 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);
}
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dao
在这种情况下是一个带有@Repository
注释的类。一个请求应该是这样的:
{
"gradeType": "REGULAR",
"grade": "10",
"passed": 1,
"userId": 1,
"exam_id": 1,
"user_id": 3
}
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问题是,它Grade
有 2 个外键,用于用户和考试。我希望能够只传递外部实体的 ID,让 Hibernate 处理其余的事情。但是,目前我得到这个作为回应:
{
"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"
}
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我该如何解决这个问题?我听说过 a JpaRepository
,我可以用它来完成这个吗?
我的Grade
模态如下:
@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() {
}
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还有我的仓库...
@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;
}
}
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谢谢
我该如何解决这个问题?我听说过有关 JpaRepository 的信息,我可以用它来完成此任务吗?
是的,您可以使用 JpaRepository 来完成此操作。您可以开始添加一个Interface
public interface GradeRepository extends PagingAndSortingRepository<Grade, Long>
,同样适用于User
andExam
然后当您POST
在Grade
控制器中时(我建议Service
为此添加一个 )
1. 创建一个new Grade()
2.通过-ing 您的and并调用setters
来链接您的and
3. 然后从中获取其他字段您使用4. 从您的
电话User
Exam
@Autowire
UserRepository
ExamRepository
findById
POST
setters
.save(grade)
@Autowired
GradeRepository
请注意 和User
应该Exam
已经存在于 中DB
才能链接它们
另外,我强烈建议您永远不要通过HTTP
use传递直接实体DTOs
,并且在Grade
json 中永远不要传递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 Security 实现了一个功能齐全的 SpringBoot 应用程序,您可以看看https://github.com/hodispk/internship
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