BeetlSQL 3.0.7 发布,新增 LambdaSubQuery

来源: 投稿
作者: 闲大赋
2020-10-27 10:35:00
  • 新增LambdaSubQuery,来源于 https://gitee.com/xiandafu/beetlsql/issues/I1WRHZ
  • 修复代码生成POJO类的错误
<dependency>
  <groupId>com.ibeetl</groupId>
  <artifactId>beetlsql</artifactId>
  <version>3.0.7-RELEASE</version>
</dependency

关于LambdaSubQuery使用说明如下,使用@SubQuery标注mapper方法

public  interface AnyMapper extends BaseMapper<User>{
/* 构造一个公共的子查询Lambda,由lambda#allUserInDepartment构成
*/
@SubQuery
public LambdaQuery<User> allUserInDepartment(Integer deptId);
}

如上allUserInDepartment方法返回一个LambdaQuery(是LambdaSubQuery),lambda#allUserInDepartment提供了子查询语句,其内容如下

allUserInDepartment
===
    
    select * from sys_user where department_id=#{deptId}

因此,BeetlSQL实际Lambda查询的时候,构成的sql其实如下

select * from (select * from sys_user where department_id=xxx) where .....

这样的好处是可以把复杂的查询转成简单的Lambda查询


LambdaQuery<User> lambdaQuery = anyMapper.allUserInDepartment(1);
List<User> newList = lambdaQuery.andEq(User::getAge,42).select();
Assert.assertEquals(0,newList.size());

如上查询,对应的sql实际上是

┏━━━━━ Debug [sql.SELECT * FROM ( select * from sys_user where d...] ━━━
┣ SQL:	 SELECT * FROM ( select * from sys_user where department_id=? ) _t WHERE age = ? 
┣ 参数:	 [1, 42]
┣ 位置:	 org.beetl.sql.core.mapper.SubQueryCommonTest.commonLambda(SubQueryCommonTest.java:32)
┣ 时间:	 1ms
┣ 结果:	 [0]
┗━━━━━ Debug [sql.SELECT * FROM ( select * from sys_user where d...] ━━━
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