Question

当我执行下面的查询时，它抛出错误：语法错误位于“ as”或附近。

我们该如何解决？

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

通过使用子查询中的选择，像下面一样尝试

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

Answer 2

您缺少初始的select * from子句。此外，不需要子查询；您只需要注意条件的放置。

select *  
from my_table1 a 
left join my_table1 b on b.status='1' and
                         b.application_id = a.application_id
where a.status='6'

Answer 3

您已经从子查询中选择了所有值并将它们联接：

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

Answer 4

select * from my_table1 as A 
LEFT JOIN  my_table1 as B
ON (A.application_id = A.application_id)
WHERE A.status='6' AND B.status='1'

根据PostgreSQL中的条件左联接

4 个答案: