从内部联接的select中获取数据在同一个mysql表上

Question

这是我的案例表

id empid    reaction         date_t
1  emp109   interested       2017-09-12
2  emp109   not_interested   2017-09-13
3  emp109   maybe            2017-09-13

我想要实现的是特定员工ID

date_t  | number of interested  |number of not interested   | number of maybe

就像2017-09-12和2017-09-13的员工ID emp109一样，表格看起来像这样

 date_t     | number of interested  |number of not interested   | maybe
 2017-09-12    1                        0                        0
 2017-09-13    0                       1                         1

这是我到目前为止所尝试的

SELECT * FROM
(SELECT cases.date_t, cases.empid,COUNT(cases.reaction) as interested  FROM `cases` WHERE cases.reaction="interested" and cases.empid="EMP12654"  GROUP by cases.date_t)alpha
JOIN
(SELECT cases.date_t, cases.empid,COUNT(cases.reaction) as not_interested  FROM `cases` WHERE cases.reaction="not_interested" and cases.empid="EMP12654"  GROUP by cases.date_t)beta
on alpha.date_t=beta.date_t
JOIN
(SELECT cases.date_t, cases.empid,COUNT(cases.reaction) as maybe  FROM `cases` WHERE cases.reaction="maybe" and cases.empid="EMP12654"  GROUP by cases.date_t)zeta
on alpha.date_t=zeta.date_t

这不是给我想要的结果任何想法如何实现它？

Answer 1

这称为旋转。

您可以使用条件聚合在MySQL中执行此操作：

select date_t,
    sum(reaction = 'interested') as interested,
    sum(reaction = 'not_interested') as not_interested,
    sum(reaction = 'maybe') as maybe
from your_table
group by date_t;

以上使用的事实是MySQL将true视为1，将false视为0。