sql - 根据postgresql中的条件左连接

Question

当我执行下面的查询时，它抛出错误提示:“as”处或附近的语法错误。

我们如何解决这个问题？

还有其他方法吗？

(select * from my_table1 where status='6') as a
left JOIN 
(select * from my_table1 where status='1') as b
ON (a.application_id = b.application_id)

Answer 1

使用子查询中的选择像下面这样尝试

select a.* from (select * from my_table1 where status='6') as a
left JOIN 
(select * from my_table1 where status='1') as b
ON a.application_id = b.application_id