在每一行中计算第二列的数量

Question

以下是对请求的回答

问题是如何按每个selected_date e.x计算：

1. 2012-02-10：1
2. 2012-02-15：0
3. 2012-02-14：3
4. 2012-02-11：0

如何提出此请求

以下是获得上述答案的请求

 select selected_date, date1 from 
(select selected_date from 
       (select adddate('1970-01-01',t4.i*10000 + t3.i*1000 + t2.i*100 + t1.i*10 + t0.i) selected_date from
       (select 0 i union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t0,
        (select 0 i union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t1,
        (select 0 i union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t2,
         (select 0 i union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t3,
         (select 0 i union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t4) v
where selected_date between '2012-02-10' and '2012-02-15' ) vv left join clicker on clicker.date1=vv.selected_date

Answer 1

这可能有效：

SELECT selected_date, SUM(CASE WHEN date1 IS NULL THEN 0 ELSE 1 END) FROM table
GROUP BY selected_date

Answer 2

那么，基本上这个？

SELECT t.selected_date, COUNT(t.date1)
FROM ( Your Query Here )
GROUP BY t.selected_date

默认情况下，

COUNT()会忽略NULL个值，因此它只计算匹配项。