我有3个实体。
他们每个人之间都有一对多的关系。我需要检索单张票的记录。但是,当我随身携带数据时,就会映射到员工的数据。在即将到来的员工数据中,我不希望将密码字段数据与其他字段一起检索。因此,对此的标准查询
员工班
@Entity
@NamedQuery(name = "getUserByEmail", query = "from Employee where emaillAddress = :emailAddress")
public class Employee implements Serializable {
/**
*
*/
private static final long serialVersionUID = 1L;
@JsonIgnore
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Column(name = "employee_id", updatable = false)
private int empId;
@JsonIgnore
@Column(name ="emp_code" ,unique = true, nullable = false)
private long employeeCode;
@Column(name = "full_name", nullable = false)
private String fullName;
@JsonIgnore
@Column(name = "email_address", nullable = false, unique = true)
private String emaillAddress;
@JsonIgnore
@Column(name = "password", nullable = false)
private String password;
@Column(name = "employee_role", nullable = false)
private int role;
@JsonIgnore
@OneToMany(mappedBy = "owner", cascade = CascadeType.ALL, orphanRemoval = true, fetch = FetchType.EAGER)
private Collection<Ticket> tickets = new ArrayList<>();
public Employee() {
this.fullName = "";
this.password = "";
this.emaillAddress = "";
this.role = 2;
}
}
机票类
@Entity
public class Ticket {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private int ticketId;
private String title;
private String message;
@Enumerated(EnumType.STRING)
private TicketPriority priority;
@Enumerated(EnumType.STRING)
private TicketStatus status;
@Enumerated(EnumType.STRING)
private TicketType type;
@JsonFormat(shape = JsonFormat.Shape.STRING,pattern = "dd-MM-yyyy | HH:mm",timezone="Asia/Kolkata")
@Temporal(TemporalType.TIMESTAMP)
private Date timestamp;
@JsonIgnore
@ManyToOne
@JoinColumn(name = "owner_id")
Employee owner;
@OneToMany(mappedBy = "ticket", cascade = CascadeType.ALL, orphanRemoval = true, fetch = FetchType.EAGER)
private Collection<Comment> comments = new ArrayList<>();
public Ticket() {
super();
this.title = "";
this.message = "";
timestamp = new Date();
this.status = TicketStatus.RAISED;
}
}
评论课
@Entity
public class Comment {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private int commentId;
private String message;
@OneToOne
@JoinColumn(name="comment_owner")
Employee employee;
@ManyToOne
@JoinColumn(name="ticket_id")
Ticket ticket;
}
我正在使用的查询是return getCurrentSession()。get(Ticket.class,id);
这是我得到的票证对象的toString
工单[ticketId = 5,title = WFH,消息=我明天需要在家工作,优先级=立即,状态= RAISED,类型= WFH_REQUEST,所有者=员工[empId = 1,employeeCode = 123,fullName = emp,emaillAddress = emp,密码= emp,角色= 2,票证=],评论= []]
您可以Employee
为同一表Employee创建两个不同的实体。
在其中一个中,您映射了列password
,在另一个实体中,您未映射password
。
因此,当您打算在没有密码的情况下检索实体时,请使用此新实体EmployeeWithoutPassword
。对于其余情况(插入,更新等),只需将常规实体与所有字段一起使用。
您也可以使用自定义的DTO来完成此操作,而无需创建新实体,只需返回所需的字段即可。
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