JPA:加入 JPQL [英] JPA: JOIN in JPQL

问题描述

我以为我知道如何在 JPQL 中使用 JOIN 但显然不知道.有人可以帮我吗?

I thought I know how to use JOIN in JPQL but apparently not. Can anyone help me?

select b.fname, b.lname from Users b JOIN Groups c where c.groupName = :groupName

这给了我例外

org.eclipse.persistence.exceptions.JPQLException
Exception Description: Syntax error parsing the query
Internal Exception: org.eclipse.persistence.internal.libraries.antlr.runtime.EarlyExitException

Users 与 Groups 有一对多的关系.

Users have a OneToMany relationship with Groups.

Users.java

@Entity
public class Users implements Serializable{

    @OneToMany(mappedBy="user", cascade=CascadeType.ALL)
    List<Groups> groups = null;
}

Groups.java

@Entity
public class Groups implements Serializable {
    @ManyToOne
    @JoinColumn(name="USERID")
    private Users user;
}

我的第二个问题是说这个查询返回一个唯一的结果，然后如果我这样做

My second question is let say this query return a unique result, then if I do

String temp = (String) em.createNamedQuery("***")
    .setParameter("groupName", groupName)
    .getSingleResult();

*** 代表上面的查询名称.那么 fname 和 lname 是在 temp 内部连接在一起还是我得到一个 List 回来?

*** represent the query name above. So does fname and lname concatenated together inside temp or I get a List<String> back?

推荐答案

在JPQL中加入一对多关系如下:

Join on one-to-many relation in JPQL looks as follows:

select b.fname, b.lname from Users b JOIN b.groups c where c.groupName = :groupName 

在select子句中指定多个属性时，返回结果为Object[]:

When several properties are specified in select clause, result is returned as Object[]:

Object[] temp = (Object[]) em.createNamedQuery("...")
    .setParameter("groupName", groupName)
    .getSingleResult(); 
String fname = (String) temp[0];
String lname = (String) temp[1];

顺便说一下，为什么您的实体以复数形式命名，这令人困惑.如果你想有复数的表名，你可以使用 @Table 来明确指定实体的表名，这样它就不会干扰保留字:

By the way, why your entities are named in plural form, it's confusing. If you want to have table names in plural, you may use @Table to specify the table name for the entity explicitly, so it doesn't interfere with reserved words:

@Entity @Table(name = "Users")     
public class User implements Serializable { ... } 

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