## @SubQuery 注解@SubQuery 要求方法mapper方法返回LambdaQuery，不同于普通的LambdaQuery，此類實際上是其子類LambdaSubQuery，支持把子查詢作為”表名“，以源碼中的SubQueryCommonTest為例子說明 > 此想法和代碼來源于 https://gitee.com/xiandafu/beetlsql/issues/I1WRHZ ```java @SqlResource("lambda") public interface AnyMapper extends BaseMapper<User>{ /* 構造一個公共的子查詢Lambda,由lambda#allUserInDepartment構成 */ @SubQuery public LambdaQuery<User> allUserInDepartment(Integer deptId); } ``` 如上allUserInDepartment方法返回一個LambdaQuery<User>，lambda#allUserInDepartment提供了子查詢語句，其內容如下 ```sql allUserInDepartment === select * from sys_user where department_id=#{deptId} ``` 因此，BeetlSQL實際Lambda查詢的時候，構成的sql其實如下 ```java select * from (select * from sys_user where department_id=xxx) where ..... ``` 這樣的好處是可以把復雜的查詢轉成簡單的Lambda查詢 ```java AnyMapper anyMapper = sqlManager.getMapper(AnyMapper.class); LambdaQuery<User> lambdaQuery = anyMapper.allUserInDepartment(1); List<User> newList = lambdaQuery.andEq(User::getAge,42).select(); Assert.assertEquals(0,newList.size()); ``` 如上查詢，對應的sql實際上是 ```text ┏━━━━━ Debug [sql.SELECT * FROM ( select * from sys_user where d...] ━━━ ┣ SQL： SELECT * FROM ( select * from sys_user where department_id=? ) _t WHERE age = ? ┣ 參數： [1, 42] ┣ 位置： org.beetl.sql.core.mapper.SubQueryCommonTest.commonLambda(SubQueryCommonTest.java:32) ┣ 時間： 1ms ┣ 結果： [0] ┗━━━━━ Debug [sql.SELECT * FROM ( select * from sys_user where d...] ━━━ ```