如何在 Querydsl 中创建子查询

Posted 2023-03-29

【中文标题】如何在 Querydsl 中创建子查询

【英文标题】：How to create subquery in Querydsl

【发布时间】：2018-08-30 06:26:37

【问题描述】：

我是 Querydsl 的新手。我必须将以下查询转换为 Querydsl。我尝试如下，但我没有得到结果。

谁能告诉我我在查询中遗漏了什么或做错了什么？

select * from room  as room 
      where room.nroom_id not in(
                          select rdm.nroom_id from roomdepartmentmapping as rdm)

我试过这样

    JPAQuery<Tuple> query = new JPAQuery<Tuple>(em);  

    query.from(room) 
         .where(room.nRoomId.notIn
                         (query.select(roomDepartmentMapping.nRoomId)
                               .from(roomDepartmentMapping)
                         )
               );

控制台

   at com.querydsl.core.support.SerializerBase.visit(SerializerBase.java:31) ~[querydsl-core-4.1.4.jar:na]
        at com.querydsl.core.types.OperationImpl.accept(OperationImpl.java:83) ~[querydsl-core-4.1.4.jar:na]
        at com.querydsl.core.support.SerializerBase.handle(SerializerBase.java:92) ~[querydsl-core-4.1.4.jar:na]
        at com.querydsl.jpa.JPQLSerializer.serialize(JPQLSerializer.java:220) ~[querydsl-jpa-4.1.4.jar:na]
        at com.querydsl.jpa.JPQLSerializer.visit(JPQLSerializer.java:358) ~[querydsl-jpa-4.1.4.jar:na]
        at com.querydsl.jpa.JPQLSerializer.visit(JPQLSerializer.java:39) ~[querydsl-jpa-4.1.4.jar:na]
        at com.querydsl.core.types.SubQueryExpressionImpl.accept(SubQueryExpressionImpl.java:57) ~[querydsl-core-4.1.4.jar:na]
        at com.querydsl.core.support.SerializerBase.handle(SerializerBase.java:92) ~[querydsl-core-4.1.4.jar:na]            
        at com.querydsl.jpa.JPQLSerializer.visitOperation(JPQLSerializer.java:403) ~[querydsl-jpa-4.1.4.jar:na]
        at com.querydsl.core.support.SerializerBase.visit(SerializerBase.java:231) ~[querydsl-core-4.1.4.jar:na]
        at com.querydsl.core.support.SerializerBase.visit(SerializerBase.java:31) ~[querydsl-core-4.1.4.jar:na]
        at com.querydsl.core.types.OperationImpl.accept(OperationImpl.java:83) ~[querydsl-core-4.1.4.jar:na]
        at com.querydsl.core.support.SerializerBase.handle(SerializerBase.java:92) ~[querydsl-core-4.1.4.jar:na]
        at com.querydsl.jpa.JPQLSerializer.serialize(JPQLSerializer.java:220) ~[querydsl-jpa-4.1.4.jar:na]
        at com.querydsl.jpa.JPQLSerializer.visit(JPQLSerializer.java:358) ~[querydsl-jpa-4.1.4.jar:na]
        at com.querydsl.jpa.JPQLSerializer.visit(JPQLSerializer.java:39) ~[querydsl-jpa-4.1.4.jar:na]
        at com.querydsl.core.types.SubQueryExpressionImpl.accept(SubQueryExpressionImpl.java:57) ~[querydsl-core-4.1.4.jar:na]
        at com.querydsl.core.support.SerializerBase.handle(SerializerBase.java:92) ~[querydsl-core-4.1.4.jar:na]
        at com.querydsl.core.support.SerializerBase.visitOperation(SerializerBase.java:270) ~[querydsl-core-4.1.4.jar:na]
        at com.querydsl.jpa.JPQLSerializer.visitOperation(JPQLSerializer.java:403) ~[querydsl-jpa-4.1.4.jar:na]
        at com.querydsl.core.support.SerializerBase.visit(SerializerBase.java:231) ~[querydsl-core-4.1.4.jar:na]           
        at com.querydsl.jpa.JPQLSerializer.visit(JPQLSerializer.java:39) ~[querydsl-jpa-4.1.4.jar:na]
        at com.querydsl.core.types.SubQueryExpressionImpl.accept(SubQueryExpressionImpl.java:57) ~[querydsl-core-4.1.4.jar:na]

【参考方案1】：

需要JPASubQuery 的实例，而不是使用封闭查询的实例。您可以使用new JPASubQuery() 或便捷方法JPAExpressions.select。您的查询应如下所示：

JPAQuery<Tuple> query = new JPAQuery<Tuple>(em);  

query.from(room) 
     .where(room.nRoomId.notIn
                     (JPAExpressions.select(roomDepartmentMapping.nRoomId)
                           .from(roomDepartmentMapping)
                     )
           );

【讨论】：

坦克它的工作。你能检查一下这个https://***.com/questions/52111072/spring-with-querydsl-null-pointer-exception