Question

我有一张表（苹果），其中包含：

cid  date_am date_pm 
----------------------
1      1       1
2      2       1
3      1       3  
1      1       2

我之前（严重地）问了一个问题，即如何根据客户的数量（1）对客户进行排名。解决方案是（基于一栏）：

SELECT cid, sum( date_pm ) AS No_of_ones
FROM apples
WHERE date_am =1
GROUP BY cid
ORDER BY no_of_ones DESC

这适用于一列，但我如何对两列的总和做同样的事情。即。

SELECT cid, sum( date_pm ) AS No_of_ones
FROM apples
WHERE date_am =1
add to
SELECT cid, sum( date_am ) AS No_of_ones
FROM apples
WHERE date_pm =1
GROUP by cid
ORDER by no_of_ones(added)

希望我已经设法清楚地表达了你的帮助-thanks

Answer 1

select cid, sum(case when date_pm = 1 then 1 else 0 end) + sum(case when date_am = 1 then 1 else 0 end)
from apples
group by cid

Answer 2

select cid, sum(addone) as total from 
  (select cid, 1 as addone
   from apples
    where date_pm = 1 group by cid
    union
    select cid, 1 as addone
    from apples
    where date_am = 1 group by cid) 
group by cid order by total DESC

OR

select cid, sum(case when date_am=1 then 1 else 0 end) 
            + sum(case when date_pm=1 then 1 else 0 end) as total 
from apples 
group by CID 
order by total DESC

将两列组合在一起

2 个答案: