插入子项和父项 JSON Spring 启动

Question

我的 spring 启动应用程序数据中有 3 个实体，应用程序用户、教师和学生

1. 应用用户

   @Id
   @GeneratedValue(strategy = GenerationType.AUTO) 
   @Column(name= "appuserId", updatable=false)
   private Long appuserId;     
   @Column(name = "username")
   private String username;
   
   @Column(name = "fullname")
   private String fullName;
   
   @Column(name = "email")
   private String email;   
   
   @Column(name = "password")
   private String password;
   
   @OneToOne(mappedBy = "appuser")
   private Teacher teacher;
   
   @OneToOne(mappedBy = "appuser")
   private Student student;
   

2. 老师

   @Id
   @GeneratedValue(strategy=GenerationType.AUTO)
   @Column(name = "teacherId" , updatable = false)
   private Long teacherId;
   
   @Column(name= "firstname")
   private String firstName;
   
   @Column(name = "lastname")
   private String lastName;
   
   @Column(name="designation")
   private String designation;
   
   @OneToOne(cascade = CascadeType.ALL)
   @JoinColumn(name = "appuserId", nullable = true)
   private Appuser appuser;
   
   @OneToMany(cascade = CascadeType.ALL, orphanRemoval=true ,mappedBy="teacher")
   @JsonIgnore
   private List<Course> courses;
   

3. 学生

   @Id
   @GeneratedValue(strategy = GenerationType.AUTO) 
   @Column(name= "studentId", updatable=false)
   private Long studentId;
   
   @Column(name = "firstName")
   private String firstName;
   
   @Column(name = "lastName")
   private String lastName;
   
   @Column(name = "enrolledSince")
   private String enrolledSince;
   
   @OneToOne(cascade = CascadeType.ALL)
   @JoinColumn(name = "appuserId", nullable = false)
   private Appuser appuser;
   
   @OneToMany(cascade = CascadeType.ALL,orphanRemoval=true, mappedBy="student")
   @JsonIgnore
   private List<CourseStudent> courseStudents;
   

我可以在 appusers table 中使用 json 格式在 postman 中插入并且运行良好。但是当我尝试插入教师或学生 table 时，appusers 中的结果为空。它不应该为空，因为应用用户的教师和学生外键。

Answer 1

这不应该发生。当您保存教师或学生时，您应该指定数据库中已经存在的 appuser。并使用appuserId代替Appuser，这足以识别它属于哪个appuser。

保存教师或学生并请求加入数据库后，您可以获取您的appuser。

Answer 2

当您尝试插入教师或学生 table 时，请确保您在坚持时将值设置为 appuser。

Student std =new Student();
// create an object of appuser,set its vaue and assign it to student object
 Appuser ap = new Appuser();
 // assigning values to the appuser object as ap.setfullname="...";... so on
 std.setAppuser=ap;

现在保留这个学生对象，条目将反映在映射中 table 或者您可以在已经持久化的 std 对象中设置 appuser 的 id。