基本上我需要列中的不同值通过Query
作为列名示例如下
id status userId date
------------------------
1 p u12 01/01/2014
2 p u13 01/01/2014
3 a u14 01/01/2014
4 hd u15 01/01/2014
5 a u16 01/01/2014
6 p u12 01/02/2014
7 a u13 01/02/2014
8 p u14 01/02/2014
date p a hd
------------------------
01/01/2014 2 2 1
02/01/2014 2 1 0
状态'p','a','hd'用作列名,按日期分组
答案 0 :(得分:2)
试试这个:
SELECT a.date, SUM(a.status='p') AS p, SUM(a.status='a') AS a,
SUM(a.status='hd') AS hd
FROM tableA a
GROUP BY a.date
<强>输出强>
| DATE | P | A | HD |
|--------------------------------|---|---|----|
| January, 01 2014 00:00:00+0000 | 2 | 2 | 1 |
| January, 02 2014 00:00:00+0000 | 2 | 1 | 0 |
答案 1 :(得分:0)
尝试此查询,但它会强制您了解所有可能的状态(结果字段名称) -
SELECT
date,
COUNT(IF(status = 'p', status, NULL)) AS p,
COUNT(IF(status = 'a', status, NULL)) AS a,
COUNT(IF(status = 'hd', status, NULL)) AS hd
FROM
table
GROUP BY
date;
你也可以尝试自动化这个逻辑,看一下这个链接 - Dynamic pivot tables (transform rows to columns)。
答案 2 :(得分:0)
此SQL已经过测试
select p.date1,P.p,A.a,case when HD.hd is null then 0 else HD.hd end hd from (select COUNT(status1) p,date1 from tbl1 where status1 ='p'
group by date1) as P LEFT JOIN (select COUNT(status1) a,date1 from tbl1 where status1 ='a'
group by date1) as A on P.date1 = A.date1 LEFT JOIN (select COUNT(status1) hd,date1 from tbl1 where status1 ='hd'
group by date1) as HD on A.date1 = HD.date1
只需根据需要更改表名或列名。