通过 oneToMany 关系加入两个 JPA 实体答案

【问题标题】：Joining two JPA Entities with oneToMany relationship通过 oneToMany 关系加入两个 JPA 实体
【发布时间】：2017-04-03 05:57:47
【问题描述】：

我有两个实体 A,B。在这个 A 中是父级并且具有下表结构。

表-A

A1-->栏目

表-B

B1-->A1 B2-->A1

我需要根据表 A 和 B 上的 where 条件获取表 B(B1,B2) 中的 A1 行及其关联行。

当我使用左外连接获取时，JPA 将返回 Table-b（B1-A1 或 B2-A1）中具有相同值的两行。

即使我尝试过使用普通连接。请让我知道我缺少什么。

我正在使用@Query 来指定查询。 SELECT a From TableA a LEFT OUTER JOIN FETCH a.TablesBCollection p where a.TableAColum =?1 and p.TableBColumnp.TableBColumn.

@Entity
@Table(name = "USER", schema = "USER_PROFILE")
public class User implements java.io.Serializable {

private BigDecimal userId;
private String firstNm;
private String lastNm;
private String userIndex;
private Set<UserPhone> userPhones = new HashSet<UserPhone>(0);

public User(){
}

public User(BigDecimal userId,String firstNm,String lastNm){
this.userId=userId;
this.firstNm=firstNm;
this.lastNm=lastNm;
}

public User(BigDecimal userId,String firstNm,String lastNm,Set<UserPhone> userPhones){
this.userId=userId;
this.firstNm=firstNm;
this.lastNm=lastNm;
this.userPhones=userPhones;
}

@Id
@Column(name = "USER_ID", unique = true, nullable = false, precision = 22, scale = 0)
public BigDecimal userId() {
 return this.userId;
    }

public void setuserId(BigDecimal userId) {
        this.userId = userId;
    }


@Column(name = "FIRST_NM", nullable = false, length=50)
public String firstNm() {
 return this.firstNm;
    }

public void setfirstNm(String firstNm) {
        this.firstNm = firstNm;
    }

@Column(name = "LAST_NM", nullable = false, length=50 )
public String lastNm() {
 return this.lastNm;
    }

public void setlastNm(String lastNm) {
        this.lastNm = lastNm;
    }

@Column(name = "USER_INDEX", nullable = false, length=50 )
public String userIndex() {
 return this.userIndex;
    }

public void setUserIndex(String userIndex) {
        this.userIndex = userIndex;
    }

@OneToMany(fetch = FetchType.LAZY, mappedBy = "user")
public Set<UserPhone> getuserPhones() {
    return this.userPhones;
    }

public void setUserPhones(Set<UserPhone> userPhones) {
    this.userPhones = userPhones;
    }

}



@Entity
@Table(name = "USER_PHONE", schema = "USER_PROFILE")
public class UserPhone implements java.io.Serializable {

private BigDecimal userPhoneId;
private String city;
private String State;
private User user;

public UserPhone(){
}

public UserPhone(BigDecimal userPhoneId){
this.userPhoneId = userPhoneId;
}

public UserPhone(BigDecimal userPhoneId,String city,String State,User user){
this.userPhoneId = userPhoneId;
this.city = city;
this.State = State;
this.user = user;
}

@Id
@Column(name = "USER_PHONE_ID", unique = true, nullable = false, precision = 22, scale = 0)
public BigDecimal userPhoneId() {
 return this.userPhoneId;
    }

public void setUserPhoneId(BigDecimal userPhoneId) {
        this.userPhoneId = userPhoneId;
    }

@Column(name = "CITY", nullable = false, length=50 )
public String city() {
 return this.city;
    }

public void setCity(String city) {
        this.city = city;
    }

@Column(name = "STATE", nullable = false, length=50 )
public String state() {
 return this.state;
    }

public void setState(String state) {
        this.state = state;
    }

@ManyToOne(fetch = FetchType.LAZY)
@JoinColumn(name = "USER_ID")
public User getUser() {
        return this.user;
    }

public void setUser(User user) {
        this.user = user;
    }

}

@QUERY("Select a from User a LEFT OUTER JOIN FETCH a.userPhones p where a.userIndex =?1 and p.city<>p.state")

【问题讨论】：

JPA 围绕类而不是表进行操作。发布实际实体

标签： hibernate jpa spring-data-jpa

【解决方案1】：

为什么不通过将 IDE 连接到数据库来将数据库表生成为 POJO。您可以在 IDE 中 create a JPA project 并连接到您的数据库，it will ask for the relationship between columns, you can mention it there 它将为您创建所需的 POJO。

【讨论】：