MYSQL中的列分组，复杂查询

我有一个MYSQL查询，我需要将这些列彼此相邻地分组，但我似乎无法做到这一点

SELECT Supervisor,
   (CASE WHEN DAYOFWEEK(workdate) = 2 THEN (COUNT(`ID Number`)) END) AS `Monday`,
   (CASE WHEN DAYOFWEEK(workdate) = 3 THEN (COUNT(`ID Number`)) END) AS `Tuesday`,
   (CASE WHEN DAYOFWEEK(workdate) = 4 THEN (COUNT(`ID Number`)) END) AS `Wednesday`,
   (CASE WHEN DAYOFWEEK(workdate) = 5 THEN (COUNT(`ID Number`)) END) AS `Thursday`,
   (CASE WHEN DAYOFWEEK(workdate) = 6 THEN (COUNT(`ID Number`)) END) AS `Friday`,
   (CASE WHEN DAYOFWEEK(workdate) = 7 THEN (COUNT(`ID Number`)) END) AS `Saturday`
FROM payroll.employeedatanew_copy
INNER JOIN payroll.employeehours ON employeedatanew_copy.`ID Number` = employeehours.employeeid
WHERE workdate BETWEEN "2013-10-28" AND "2013-11-02"
GROUP BY workdate, supervisor

以下是我的结果：

这就是我想要的：

---

我想你在找这个：

SELECT Supervisor,
   COUNT(CASE WHEN DAYOFWEEK(workdate) = 2 THEN 1 END) `Monday`,
   COUNT(CASE WHEN DAYOFWEEK(workdate) = 3 THEN 1 END) `Tuesday`,
   COUNT(CASE WHEN DAYOFWEEK(workdate) = 4 THEN 1 END) `Wednesday`,
   COUNT(CASE WHEN DAYOFWEEK(workdate) = 5 THEN 1 END) `Thursday`,
   COUNT(CASE WHEN DAYOFWEEK(workdate) = 6 THEN 1 END) `Friday`,
   COUNT(CASE WHEN DAYOFWEEK(workdate) = 7 THEN 1 END) `Saturday`
FROM payroll.employeedatanew_copy
JOIN payroll.employeehours ON employeedatanew_copy.`ID Number` = employeehours.employeeid
WHERE workdate BETWEEN "2013-10-28" AND "2013-11-02"
GROUP BY supervisor

或者，

计数

s可以替换为：

SUM(DAYOFWEEK(workdate) = 2) `Monday`

---

这是基于MySQL中

true

等于

1

这一事实而起作用的，但不一定在其他DBMS中起作用。

我想您正在寻找这个：

SELECT Supervisor,
   COUNT(CASE WHEN DAYOFWEEK(workdate) = 2 THEN 1 END) `Monday`,
   COUNT(CASE WHEN DAYOFWEEK(workdate) = 3 THEN 1 END) `Tuesday`,
   COUNT(CASE WHEN DAYOFWEEK(workdate) = 4 THEN 1 END) `Wednesday`,
   COUNT(CASE WHEN DAYOFWEEK(workdate) = 5 THEN 1 END) `Thursday`,
   COUNT(CASE WHEN DAYOFWEEK(workdate) = 6 THEN 1 END) `Friday`,
   COUNT(CASE WHEN DAYOFWEEK(workdate) = 7 THEN 1 END) `Saturday`
FROM payroll.employeedatanew_copy
JOIN payroll.employeehours ON employeedatanew_copy.`ID Number` = employeehours.employeeid
WHERE workdate BETWEEN "2013-10-28" AND "2013-11-02"
GROUP BY supervisor

或者，

计数

s可以替换为：

SUM(DAYOFWEEK(workdate) = 2) `Monday`

---

这是基于这样一个事实：MySQL中的

true

等于

1

，但不一定在其他DBMS中工作。

在“我想要什么”图像中，每一行的值对于每个工作日都是相同的，这是预期的吗？请从

group by

@MikeBrant中删除

组中的
workdate
字段，是的，我刚刚为每个日期使用了相同的数据集，所以值是相同的：）。谢谢Barranka，但这还不够，下面是完美的答案！在“我想要什么”图像中，每一行的值对于每个工作日都是相同的，这是预期的吗？从
分组中删除
workdate
字段by
@MikeBrant，是的，我刚刚对每个日期使用了相同的数据集，所以值是相同的：）。谢谢Barranka，但这还不够，下面是完美的答案！需要注意的是，按
workdate
分组基本上是先将所有workdate=2行放在前面，然后再将workdate=3行，以此类推。您根本不希望进行分组。这就是为什么Darrent Bent不止一次出现在不同的一排。非常感谢你们，非常感谢你们的帮助！需要注意的是，按
workdate
分组基本上是先将所有workdate=2行放在前面，然后再将workdate=3行，以此类推。您根本不希望进行分组。这就是为什么Darrent Bent不止一次出现在不同的一排。非常感谢你们，非常感谢你们的帮助！