如何在使用连接时区分sum（）

Question

我有2张桌子。 一个保存OK部件的数据，另一个保存拒绝的数据。

我喜欢以下列格式接收数据：

| YYYY-MM | SUM(ok.qty) | SUM(reject.qty) | reject % |

1）是否可以使用JOIN然后某种DISTINCT sum(ok.qty)？

SELECT 
    DATE_FORMAT(date, '%Y-%m'), 
    SUM(ok.qty), 
    SUM(reject.qty), 
    SUM(reject.qty) / SUM(reject.qty + ok.qty) as reject_percent
FROM ok
JOIN reject ON ok.id = reject.ok_id
GROUP BY DATE_FORMAT(date, '%Y-%m')

2）如果不是如何消除重复的子查询？

SELECT 
    DATE_FORMAT(date, '%Y-%m'), 
    SUM(ok.qty), 
    (SELECT qty FROM reject WHERE ok_id = ok.id), 
    (SELECT qty FROM reject WHERE ok_id = ok.id) / 
        ( (SELECT qty FROM reject WHERE ok_id = ok.id) + SUM(ok.qty) ) 
        as reject_percent
FROM ok
GROUP BY DATE_FORMAT(date, '%Y-%m')

现在必须写3次查询。

数据库设计：

ok.id可能有多次拒绝。

Answer 1

你可以尝试这样..

SELECT 
    DATE_FORMAT(date, '%Y-%m'), 
    SUM(ok.qty), 
    SUM(reject.qty), 
    ((SUM(reject.qty) / SUM(reject.qty + ok.qty))*100)as reject_percent
FROM ok
inner JOIN (Select Ok_Id,Sum(qty) as qty from Reject group by OK_id) as reject ON ok.id = reject.ok_id
GROUP BY DATE_FORMAT(date, '%Y-%m')