让我们假设一个带有theese字段的用户类:
@Entity
public class User extends Model {
public String email;
public String password;
@ElementCollection
public List<String> stringList;
}
我正在寻找一种方法来执行数据库请求,以查找在stringList中具有给定字符串的所有用户,例如
List<User> usersHelloWorld = User.find("byStringList", "HelloWorld").fetch();
但当然,这不起作用。有没有办法让它发挥作用?
修改:
以下是我的实际课程字段:
@Entity
public class User extends Model {
public String email;
public String password;
public String firstname;
public String lastname;
public String gender;
public String fbId;
public String googleId;
@ElementCollection
public List<String> eventTopicIds;
@Transient
UserEventBuffer eventBuffer;
@Transient
public String fbAccessToken;
public String googleAccessToken;
}
有了这个请求:
List<User> u = User.find("SELECT u from User u where ? IN u.eventTopicIds", "internalns_rootTopic1").fetch();
我收到了这个错误:
发生JPAQueryException:执行查询SELECT时出错 来自用户你在哪里? IN u.eventTopicIds:SQL语句中的语法错误 “SELECT USER0_.ID AS ID3_,USER0_.EMAIL AS EMAIL3_,USER0_.FBID AS FBID3_,USER0_.FIRSTNAME AS FIRSTNAME3_,USER0_.GENDER为GENDER3_, USER0_.GOOGLEACCESSTOKEN作为GOOGLEAC6_3_,USER0_.GOOGLEID AS GOOGLEID3_,USER0_.LASTNAME AS LASTNAME3_,USER0_.PASSWORD AS PASSWORD3_ FROM USER USER0_ CROSS JOIN USER_EVENTTOPICIDS EVENTTOPIC1_ WHERE USER0_.ID = EVENTTOPIC1_.USER_ID AND(?IN(。[*]))“;预期 “NOT,EXISTS,SELECT,FROM”; SQL语句:选择user0_.id为id3_, user0_.email as email3_,user0_.fbId as fbId3_,user0_.firstname as firstname3_,user0_.gender为gender3_,user0_.googleAccessToken为 googleAc6_3_,user0_.googleId为googleId3_,user0_.lastname为 lastname3_,user0_.password as password3_ from User user0_ cross join User_eventTopicIds eventtopic1_其中user0_.id = eventtopic1_.User_id 和(?in(。))[42001-149]
答案 0 :(得分:2)
通常,您应该能够运行像
这样的JPA查询User.find("Select u from User as u inner join u.stringList as strings where ? in strings", "HelloWorld").fetch();
做了一个小测试,我的对象有电子邮件而不是字符串,但它应该是相同的,除非你遇到一些保留的话。问题不在于关键字,我必须内部加入stringList才能使用in关键字。稍早一点标记:)
希望这有帮助。