- 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...] ━━━