MYSQL每周和每天的学生出勤率汇总报告

我面临一个挑战，即报告给定日期范围内学生每周每天的出勤情况。因此，我目前可以提取每天的每周出勤率，但它会在一周的几天内进行汇总，因此，如果有4周和4个星期一（即每周1个星期一），我会得到4个星期一，但每周不显示

我目前的表格涉及：

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
...

我当前的sql（给定日期范围为“2017-06-25”和“2017-07-20”，给定类别为“365”）：

---

考虑处理应用程序代码中的数据显示问题谢谢@草莓的建议，但我将此作为存储过程编写，只显示结果，而不使用任何其他应用程序呈现结果（以csv或excel导出的结果除外）

SELECT 
concat(p.Frist_Name, ' ',p.Last_Name) AS Student, p.Gender
,cast(COUNT(CASE WHEN weekday(date(Attendance_Datetime)) = 0
then 1 END) AS CHAR) AS Mon
,cast(COUNT(CASE WHEN weekday(date(Attendance_Datetime)) = 1
then 1 END) AS CHAR) AS Tue
,cast(COUNT(CASE WHEN weekday(date(Attendance_Datetime)) = 2
then 1 END) AS CHAR) AS Wed
,cast(COUNT(CASE WHEN weekday(date(Attendance_Datetime)) = 3
then 1 END) AS CHAR) AS Thu
,cast(COUNT(CASE WHEN weekday(date(Attendance_Datetime)) = 4
then 1 END) AS  CHAR) AS Fri
,cast(COUNT(distinct date(Attendance_Datetime)) AS  CHAR) AS WeeklyTotal
FROM (
SELECT distinct date(Attendance_Datetime) as Attendance_Date,a.* 
FROM Attendance a WHERE date(Attendance_Datetime) BETWEEN '2017-06-25' AND '2017-07-20'
AND a.Classroom_Identifier = 365 AND a.Course_Identifier IS NULL
AND (Attendance_Value = 'Present' OR Attendance_Value = 'Late')
AND weekday(date(Attendance_Datetime)) not in (5,6) GROUP BY Student_Identifier,Attendance_Date
) a
JOIN Student s ON s.Student_Identifier=a.Student_Identifier
JOIN Person p ON p.Person_Identifier=s.Person_Identifier
WHERE (Attendance_Value = 'Present' OR Attendance_Value = 'Late')
AND (p.Gender = 'Male' OR p.Gender = 'Female')
GROUP BY Student, Gender ORDER BY p.Gender, Student, Mon DESC;