Question

表mpkids_students AS A

id      BranchId    Email                   Mobile      StudentId
9497    25          mpsuraj2016@gmail.com   8700698773  25
9498    25          m016@gmail.com          8700698776  26

表mpkids_student_image_gallery AS B

id      like_count  student_id
1       25          27

表mpkids_visitors AS C

id      student_id
1       9497
2       9497
3       9497

表mpkids_visitors_count AS D

id  visitor_count   student_id
1   4               23

表mpkids_image_likes AS E

id  student_id
1   67

表mpkids_relatives_data AS F

id  student_id  rel_email               rel_mobile
1   9497        kushwahji@gmail.com     9009859691
2   9497        kushwah@gmail.com       7566403326
3   9497        kushwah@gmail.com       1236403326
4   9497        suraj@gmail.com         123640332

表mpkids_paidstatus AS G

id  student_id  Received
1   9497        7500
2   9497        3000
3   9497        3000

MYSQL查询

SELECT A.id as student_id,
COUNT(DISTINCT B.id) as images, 
COUNT(DISTINCT C.id)+ COUNT(DISTINCT D.visitor_count)  as visits, 
count(DISTINCT E.id) + SUM(B.like_count) as likes, 
COUNT(DISTINCT A.Email)+COUNT(DISTINCT F.rel_email)  as emails, 
COUNT(DISTINCT A.Mobile)+COUNT(DISTINCT F.rel_mobile)  as moibles, 
SUM(G.Received)  as Received 
FROM mpkids_students AS A 
LEFT JOIN mpkids_student_image_gallery AS B ON B.student_id = A.id 
LEFT JOIN mpkids_visitors AS C ON C.student_id = A.id 
LEFT JOIN mpkids_visitors_count AS D ON D.student_id = A.id 
LEFT JOIN mpkids_image_likes AS E ON E.student_id = A.id 
LEFT JOIN mpkids_relatives_data AS F ON F.student_id = A.id 
LEFT JOIN mpkids_paidstatus AS G ON G.student_id = A.id 
WHERE A.BranchId = 25 
GROUP BY A.id 
ORDER BY A.StudentId DESC

结果：

student_id  images  visits  likes   emails  moibles Received    
9497        0       3       NULL    4       5       202500  
9498        0       0       NULL    1       1       NULL

问题说明：

Received Field返回错误的值，我尝试了许多查询，但没有得到解决方案学生编号= 9497的接收字段正确值13500 请帮助我找到解决方法。

Answer 1

输出错误的原因是，当您基于Studentid加入时，您会从mpkids_paidstatus表中为每个学生获取多个记录，这将累加并返回错误的输出。

您还可以像使用子查询一样编写查询。

SELECT A.id as student_id,
COUNT(DISTINCT B.id) as images, 
COUNT(DISTINCT C.id)+ COUNT(DISTINCT D.visitor_count)  as visits, 
count(DISTINCT E.id) + SUM(B.like_count) as likes, 
COUNT(DISTINCT A.Email)+COUNT(DISTINCT F.rel_email)  as emails, 
COUNT(DISTINCT A.Mobile)+COUNT(DISTINCT F.rel_mobile)  as moibles, 
(select SUM(Received) from mpkids_paidstatus ps where ps.student_id=a.id)  as Received 
FROM mpkids_students AS A 
LEFT JOIN mpkids_student_image_gallery AS B ON B.student_id = A.id 
LEFT JOIN mpkids_visitors AS C ON C.student_id = A.id 
LEFT JOIN mpkids_visitors_count AS D ON D.student_id = A.id 
LEFT JOIN mpkids_image_likes AS E ON E.student_id = A.id 
LEFT JOIN mpkids_relatives_data AS F ON F.student_id = A.id 

WHERE A.BranchId = 25 
GROUP BY A.id 
ORDER BY A.StudentId DESC

多次计数和求和MySQL函数在MySQL结果中返回错误值并多次联接

1 个答案: