選擇查詢和相同的查詢具有結果

Question

我已經遇到了一些「複雜」的查詢，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 
) 

並試圖

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

但結果提供給我一個「錯誤」的數據，因爲它不是真的包含兩個計數...

請問您能描述一下最「清潔」的方式來解決這個問題，而沒有大量的嵌套查詢嗎？

如果它可能會有所幫助 - q2是類似於q1但末尾having count(1) = 1。

P.S. 文檔鏈接會很好，答案很簡單。

Answer 1

毫無疑問，您得到一個笛卡爾積，這會影響聚合。相反，聚合之前做連接：

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;