选择查询和相同的查询具有结果

Question

我已经遇到了一些“复杂”的查询，如：

with q1 as (
select w.entity_id, p.actionable_group, p.error_type 
from issue_analysis p 
join log l on p.id = l.issue_analysis_id 
join website w on l.website_id = w.id 
where l.date >= '2016-06-01' 
group by 1, 2, 3 
), 
q2 as (
select w.entity_id, p.actionable_group, p.error_type 
from issue_analysis p 
join log l on p.id = l.issue_analysis_id 
join website w on l.website_id = w.id 
where l.date >= '2016-06-01' 
group by 1, 2, 3 
having count(1) = 1 
) 

并试图

SELECT q1.entity_id, count(q1.entity_id), count(q2.entity_id) 
from q1, q2 
group by 1 
order by 1 

但结果提供给我一个“错误”的数据，因为它不是真的包含两个计数...

请问您能描述一下最“清洁”的方式来解决这个问题，而没有大量的嵌套查询吗？

如果它可能会有所帮助 - q2是类似于q1但末尾having count(1) = 1。

P.S. 文档链接会很好，答案很简单。

Answer 1

毫无疑问，您得到一个笛卡尔积，这会影响聚合。相反，聚合之前做连接：

select q1.entity_id, q1_cnt, q2_cnt 
from (select q1.entity_id, count(*) as q1_cnt 
     from q1 
     group by q1.entity_id 
    ) q1 join 
    (select q2.entity_id, count(*) as q2_cnt 
     from q2 
     group by q2.entity_id 
    ) q2 
    on q1.entity_id = q2.entity_id;