Sql 每行计算第二列的数量

这是对请求的答复

问题是如何按每个选定的日期e.x进行计数：

2012-02-10:1

2012-02-15:0

2012-02-14:3

2012-02-11:0

如何提出这个要求

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

这可能会奏效：

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

那么，基本上是这样吗

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

---

COUNT（）

默认情况下忽略

NULL

值，因此它将只计算匹配项。

请检查此项以改进问题，您可以使用{}设置编码节的格式