每周工作日出勤的Mysql数据透视表答案

【问题标题】：Mysql Pivot table for attendance per weekdays of week每周工作日出勤的Mysql数据透视表
【发布时间】：2016-12-07 18:04:11
【问题描述】：

对于一个日期间隔，我想显示学生每个工作日的出勤记录，记录在三个表 [Attendance, Students, Person] 中。表的架构（相关字段）：

Attendance Table
--------------------- 
Attendance_Identifier 
Student_Identifier 
Classroom_Identifier 
Attendance_Datetime 
Attendance_Value 
...

Student
------------------
Student_Identifier
Person_Identifier
Classroom_Identifier
...

Person
-----------------
Person_Identifier
Frist_Name
Last_Name
Gender
...

预期报告输出：

Gender    Student     Mon     Tue     Wed     Thu     Fri     WeeklyTotal
------------------------------------------------------------------------
Male     Ab Stain     2/10    3/12    1/9     1/10    0/10    7/51
Male     Pre Senter   10/10   12/12   9/9     9/10    10/10   50/51
...
Female    Al Ways     10/10   12/12   9/9     10/10   10/10   51/51
Female    Not Often   5/10    5/12    4/9     4/10    5/10    23/51
...

我还有一个函数可以从日期间隔中获取特定星期几的计数，例如日期间隔 @s、@e 中的总星期一，只需执行以下操作：select get_weekday(@s,@e,0).

所以我对存储过程的查询是：

set @s = '2016-01-01';
set @e = '2016-12-07';
SELECT 
concat(p.Frist_Name, ' ',p.Last_Name) as Student, p.Gender
    ,GROUP_CONCAT(CASE WHEN DATE_FORMAT(Attendance_Datetime, '%a') = 'Mon' 
        THEN CONCAT(COUNT(Attendance_Value),'/',get_weekday(@s,@e,0)) ELSE NULL END) AS Mon
    ,GROUP_CONCAT(CASE WHEN DATE_FORMAT(Attendance_Datetime, '%a') = 'Tue' 
        THEN CONCAT(COUNT(Attendance_Value),'/',get_weekday(@s,@e,1)) ELSE NULL END) AS Tue
    ,GROUP_CONCAT(CASE WHEN DATE_FORMAT(Attendance_Datetime, '%a') = 'Wed' 
        THEN CONCAT(COUNT(Attendance_Value),'/',get_weekday(@s,@e,2)) ELSE NULL END) AS Wed
    ,GROUP_CONCAT(CASE WHEN DATE_FORMAT(Attendance_Datetime, '%a') = 'Thu' 
        THEN CONCAT(COUNT(Attendance_Value),'/',get_weekday(@s,@e,3)) ELSE NULL END) AS Thu
    ,GROUP_CONCAT(CASE WHEN DATE_FORMAT(Attendance_Datetime, '%a') = 'Fri' 
        THEN CONCAT(COUNT(Attendance_Value),'/',get_weekday(@s,@e,4)) ELSE NULL END) AS Fri
    , SUM(COUNT(Attendance_Value)) as WeeklyTotal
FROM Attendance a JOIN Student s ON s.Student_Identifier=a.Student_Identifier 
JOIN Person p ON p.Person_Identifier=s.Person_Identifier 
WHERE date(Attendance_Datetime) BETWEEN @s AND @e AND a.Classroom_Identifier = '363'
AND (Attendance_Value = 'Present' OR Attendance_Value = 'Late') AND (p.Gender = 'Male' OR p.Gender = 'Female') 
AND DATE_FORMAT(Attendance_Datetime, '%a') !='Sat' AND DATE_FORMAT(Attendance_Datetime, '%a')!='Sun' 
GROUP BY p.Gender, Student, WeeklyTotal ORDER BY p.Gender, Student;

即使对代码进行了多次调整，我每次都会收到以下组函数错误。

Error Code: 1111. Invalid use of group function

编辑

根据@Solarflare 的第一个建议（我不知道如何尝试第二个），我将查询改进为：

SELECT 
concat(p.Frist_Name, ' ',p.Last_Name) as Student, p.Gender
    ,CONCAT(CASE WHEN DATE_FORMAT(Attendance_Datetime, '%a') = 'Mon' 
        THEN COUNT(Attendance_Value) ELSE 0 END,'/',get_weekday(@s,@e,0) ) AS Mon
    ,CONCAT(CASE WHEN DATE_FORMAT(Attendance_Datetime, '%a') = 'Tue' 
        THEN COUNT(Attendance_Value) ELSE 0 END,'/',get_weekday(@s,@e,1) ) AS Tue
    ,CONCAT(CASE WHEN DATE_FORMAT(Attendance_Datetime, '%a') = 'Wed' 
        THEN COUNT(Attendance_Value) ELSE 0 END,'/',get_weekday(@s,@e,2) ) AS Wed
    ,CONCAT(CASE WHEN DATE_FORMAT(Attendance_Datetime, '%a') = 'Thu' 
        THEN COUNT(Attendance_Value) ELSE 0 END,'/',get_weekday(@s,@e,3) ) AS Thu
    ,CONCAT(CASE WHEN DATE_FORMAT(Attendance_Datetime, '%a') = 'Fri' 
        THEN COUNT(Attendance_Value) ELSE 0 END,'/',get_weekday(@s,@e,4) ) AS Fri
    , COUNT(Attendance_Value) as WeeklyTotal
FROM Attendance a JOIN Student s ON s.Student_Identifier=a.Student_Identifier 
JOIN Person p ON p.Person_Identifier=s.Person_Identifier 
WHERE date(Attendance_Datetime) BETWEEN @s AND @e AND a.Classroom_Identifier = '363'
AND (Attendance_Value = 'Present' OR Attendance_Value = 'Late') AND (p.Gender = 'Male' OR p.Gender = 'Female') 
AND DATE_FORMAT(Attendance_Datetime, '%a') !='Sat' AND DATE_FORMAT(Attendance_Datetime, '%a')!='Sun' 
GROUP BY Student,Gender,DATE_FORMAT(Attendance_Datetime, '%a') ORDER BY p.Gender, Student,Mon DESC;

结果如下：

【问题讨论】：

您不能使用group_concat(count())，这将是一个双组。我认为您根本不需要 group_concat ；你可能可以把箱子放在count() 里面，比如concat(count(case when ... = 'Wed' ... else null end), ' /', ...) 等等。如果没有，请先为枢轴创建子查询，然后是格式化内容。
感谢@Solarflare 的建议，第一个帮助使输出几乎达到预期，请参阅主要问题编辑，看看您是否提供进一步帮助:)
您按工作日分组。所以你每天得到一排。你不想要那个。将整个 case...end 放在计数中。如果您提供表格创建和示例数据（例如，作为 sql fiddle），我可以对其进行测试，但我认为我在第一条评论中发布的代码应该可以工作。
嗨@Solarflare，您的建议有所帮助，但仍不如预期。请看一下sql fiddle。 sqlfiddle.com/#!2/fd93f/1

标签： mysql sql pivot-table mysql-workbench

【解决方案1】：

只剩下一些小事让它工作，试试这个：

SELECT 
  concat(p.Frist_Name, ' ',p.Last_Name) as Student, p.Gender
  ,concat(cast(COUNT(CASE WHEN weekday(Attendance_Datetime) = 0
     then 1 END) as char), '/', get_weekday(@s,@e,0)) as Mon
  ,concat(cast(COUNT(CASE WHEN weekday(Attendance_Datetime) = 1
     then 1 END) as char), '/', get_weekday(@s,@e,1)) as Tue
  ,concat(cast(COUNT(CASE WHEN weekday(Attendance_Datetime) = 2
     then 1 END) as char), '/', get_weekday(@s,@e,2)) as Wed
  ,concat(cast(COUNT(CASE WHEN weekday(Attendance_Datetime) = 3
     then 1 END) as char), '/', get_weekday(@s,@e,3)) as Thu
  ,concat(cast(COUNT(CASE WHEN weekday(Attendance_Datetime) = 4
     then 1 END) as char), '/', get_weekday(@s,@e,4)) as Fri
  ,COUNT(Attendance_Value) as WeeklyTotal
FROM Attendance a 
JOIN Student s ON s.Student_Identifier=a.Student_Identifier 
JOIN Person p ON p.Person_Identifier=s.Person_Identifier 
WHERE date(Attendance_Datetime) BETWEEN @s AND @e 
  AND a.Classroom_Identifier = '363'
  AND (Attendance_Value = 'Present' OR Attendance_Value = 'Late') 
  AND (p.Gender = 'Male' OR p.Gender = 'Female') 
  AND weekday(Attendance_Datetime) not in (5,6)
GROUP BY Student, Gender ORDER BY p.Gender, Student, Mon DESC;

枢轴是用case 完成的，它在日期具有正确的工作日时计数。否则值为null（不是0），才不算数（else null不必写）。

【讨论】：