Java JPA：加入 JPQL
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JPA: JOIN in JPQL
提问by Thang Pham
I thought I know how to use JOINin JPQLbut apparently not. Can anyone help me?
我以为我知道如何使用JOINinJPQL但显然不知道。谁能帮我？
select b.fname, b.lname from Users b JOIN Groups c where c.groupName = :groupName
This give me Exception
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
Usershave a OneToMany relationship with Groups.
Users有 OneToMany 关系Groups。
Users.java
Users.java
@Entity
public class Users implements Serializable{
@OneToMany(mappedBy="user", cascade=CascadeType.ALL)
List<Groups> groups = null;
}
Groups.java
Groups.java
@Entity
public class Groups implements Serializable {
@ManyToOne
@JoinColumn(name="USERID")
private Users user;
}
My second question is let say this query return a unique result, then if I do
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
String temp = (String) em.createNamedQuery("***")
.setParameter("groupName", groupName)
.getSingleResult();
***represent the query name above. So does fnameand lnameconcatenated together inside tempor I get a List<String>back?
***代表上面的查询名称。那么是fname和lname内部连接在一起temp还是我得到了List<String>回报？
采纳答案by axtavt
Join on one-to-many relation in JPQL looks as follows:
在 JPQL 中加入一对多关系如下所示：
select b.fname, b.lname from Users b JOIN b.groups c where c.groupName = :groupName
When several properties are specified in selectclause, result is returned as Object[]:
当select子句中指定了多个属性时，结果返回为Object[]：
Object[] temp = (Object[]) em.createNamedQuery("...")
.setParameter("groupName", groupName)
.getSingleResult();
String fname = (String) temp[0];
String lname = (String) temp[1];
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 @Tableto specify the table name for the entity explicitly, so it doesn't interfere with reserved words:
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显式指定实体的表名，这样它就不会干扰保留字：
@Entity @Table(name = "Users")
public class User implements Serializable { ... }
