【发布时间】：2016-08-13 00:39:11
【问题描述】：
我在 MySQL 上有 User 和 Employee 表，并且在 User 表中有 employeeId 作为外键。
现在我需要获取没有用户的员工。
我在 MySQL Workbench 中编写了这个 SQL，这正是我想要的：
SELECT * FROM HUMANRESOURCE.EMPLOYEE E LEFT JOIN AUTHORIZE.USER U
ON U.EMPLOYEEOBJID = E.OBJID
WHERE U.EMPLOYEEOBJID is NULL;
但是当我尝试将此 SQL 实现为 JPA 查询时，它什么也不返回。这是 JPA 查询：
Query query = em.createQuery("SELECT e FROM Employee e LEFT JOIN User u
WHERE u.employee.objid = e.objid
AND u.employee IS NULL");
这是真正有效的 JPA 查询，我用它来获取拥有用户的员工：
Query query = em.createQuery("SELECT e FROM Employee e INNER JOIN User u
WHERE u.employee.objid = e.objid");
我在这里做错了什么？
实体类更新：
Base.java
package com.kadir.entity;
import java.math.BigInteger;
import java.sql.Timestamp;
import java.util.Date;
import javax.persistence.Cacheable;
import javax.persistence.Column;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.MappedSuperclass;
import javax.persistence.Version;
@Cacheable
@MappedSuperclass
public abstract class Base {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Column(name = "OBJID")
private BigInteger objid;
@Column(name = "CREATEDBY")
private String createdby;
@Column(name = "CREATEDDATE")
private Timestamp createddate;
@Version
@Column(name = "ROWVERSION")
private Integer rowversion;
@Column(name = "UPDATEDBY")
private String updatedby;
@Column(name = "UPDATEDDATE")
private Timestamp updateddate;
@Column(name = "ARCHIVED", columnDefinition = "int default 0")
private int archived;
public BigInteger getObjid() {
return this.objid;
}
public void setObjid(BigInteger objid) {
this.objid = objid;
}
public String getCreatedby() {
return this.createdby;
}
public void setCreatedby(String createdby) {
this.createdby = createdby;
}
public Date getCreateddate() {
return this.createddate;
}
public void setCreateddate(Timestamp createddate) {
this.createddate = createddate;
}
public Integer getRowversion() {
return this.rowversion;
}
public void setRowversion(Integer rowversion) {
this.rowversion = rowversion;
}
public String getUpdatedby() {
return this.updatedby;
}
public void setUpdatedby(String updatedby) {
this.updatedby = updatedby;
}
public Timestamp getUpdateddate() {
return this.updateddate;
}
public void setUpdateddate(Timestamp updateddate) {
this.updateddate = updateddate;
}
public int getArchived() {
return archived;
}
public void setArchived(int archived) {
this.archived = archived;
}
}
Employee.java
package com.kadir.entity.humanresource;
import com.kadir.entity.corporation.Company;
import com.kadir.entity.Base;
import java.io.Serializable;
import javax.persistence.*;
/**
* The persistent class for the EMPLOYEE database table.
*
*/
@Cacheable
@Entity
@Table(name = "EMPLOYEE", schema = "HUMANRESOURCE")
@NamedQuery(name = "Employee.findAll", query = "SELECT e FROM Employee e")
public class Employee extends Base implements Serializable {
private static final long serialVersionUID = 1L;
@ManyToOne
@JoinColumn(name = "COMPANYOBJID")
private Company company;
@Column(name = "FIRSTNAME")
private String firstname;
@Column(name = "GENDER")
private int gender;
@Column(name = "EMAIL")
private String email;
@Column(name = "PHONE")
private String phone;
@Column(name = "LASTNAME")
private String lastname;
public Employee() {
}
public Company getCompany() {
return this.company;
}
public void setCompany(Company company) {
this.company = company;
}
public String getFirstname() {
return this.firstname;
}
public void setFirstname(String firstname) {
this.firstname = firstname;
}
public int getGender() {
return this.gender;
}
public void setGender(int gender) {
this.gender = gender;
}
public String getEmail() {
return this.email;
}
public void setEmail(String email) {
this.email = email;
}
public String getPhone() {
return this.phone;
}
public void setPhone(String phone) {
this.phone = phone;
}
public String getLastname() {
return this.lastname;
}
public void setLastname(String lastname) {
this.lastname = lastname;
}
}
User.java
package com.kadir.entity.authorize;
import com.kadir.entity.Employee;
import com.kadir.entity.Base;
import java.io.Serializable;
import javax.persistence.*;
import java.util.List;
/**
* The persistent class for the USER database table.
*
*/
@Cacheable
@Entity
@Table(name="USER", schema="AUTHORIZE")
@NamedQuery(name="User.findAll", query="SELECT u FROM User u")
public class User extends Base implements Serializable {
private static final long serialVersionUID = 1L;
@OneToOne
@JoinColumn(name="EMPLOYEEOBJID")
private Employee employee;
@Column(name="NAME")
private String name;
@Column(name="PASSWORD")
private String password;
public User() {
}
public Employee getEmployee() {
return this.employee;
}
public void setEmployee(Employee employee) {
this.employee = employee;
}
public String getName() {
return this.name;
}
public void setName(String name) {
this.name = name;
}
public String getPassword() {
return this.password;
}
public void setPassword(String password) {
this.password = password;
}
}
【问题讨论】：
-
基本上：您犯了将 SQL 转换为 JPQL 的常见错误，这是行不通的，因为它们绝对不一样。解决问题的唯一方法是尝试从您拥有的模型和您当前未显示的模型（JPA 实体）的角度来回答问题。如果 Employee 有一个到 User 的反向映射，那么这个问题就会变得非常简单。
-
Employee 没有反向映射，User 类中只有 Employee 对象。这种情况有什么解决方法还是我必须添加反向映射？ @金比
-
实体类可以同时看到吗？
-
我用实体类@Zia编辑了问题
-
尝试在查询中使用“IS EMPTY”而不是 IS NULL
标签： java jpa join eclipselink jpql