我有一张桌子
StudentID StudentName Subject Marks
1 Savita EC1 50
1 Savita EC2 55
1 Savita EC3 45
1 Savita EC4 34
1 Savita EC5 23
2 Rajesh EC1 34
2 Rajesh EC2 56
2 Rajesh EC3 12
2 Rajesh EC4 45
2 Rajesh EC5 23
3 Smita EC1 76
3 Smita EC2 45
3 Smita EC3 67
3 Smita EC4 56
3 Smita EC5 76
4 Rahul EC1 66
4 Rahul EC2 34
4 Rahul EC3 22
4 Rahul EC4 18
4 Rahul EC5 33
我写了一个像
这样的查询SELECT StudentName, EC1,EC2,EC3,EC4,EC5,TotalMarks, case
when EC1<30 and ec2<30 then 'fail'
when EC1<30 and EC3<30 then 'fail'
when EC1<30 and EC4<30 then 'fail'
when EC1<30 and EC5<30 then 'fail'
when EC2<30 and EC3<30 then 'fail'
when EC2<30 and EC4<30 then 'fail'
when EC2<30 and EC5<30 then 'fail'
when EC3<30 and EC4<30 then 'fail'
when EC3<30 and EC5<30 then 'fail'
when EC4<30 and EC5<30 then 'fail'
else 'pass'
end as Result
FROM (SELECT StudentName, EC1, EC2, EC3, EC4, EC5, TotalMarks=EC1+EC2+EC3+EC4+EC5
FROM Student
PIVOT(sum(Marks) for subject in([EC1],[EC2],[EC3],[EC4],[EC5],[TotalMarks]))as pivotTable) A
其中2个科目中少于30分的学生输出失败或传递
Rahul 66 34 22 18 33 173 fail
Rajesh 34 56 12 45 23 170 fail
Savita 50 55 45 34 23 207 pass
Smita 76 45 67 56 76 320 pass
我想为每个小于30的主题添加7个标记，并在添加7个标记后查看通过失败的学生。例如 - 添加7个标记后，rajesh记录应该像
Rajesh 34 56 19 45 30 170 pass
答案 0 :(得分：4)
也许这是你正在寻找的东西：
SELECT A.StudentName, EC1,EC2,EC3,EC4,EC5,Total,
case when fail2 >= 2 then 'Failure'
when fail >= 2 then 'Near Pass'
else 'Pass' end as Result
FROM
( SELECT StudentName, EC1, EC2, EC3, EC4, EC5
FROM Student
PIVOT(sum(Marks) for subject in([EC1],[EC2],[EC3],[EC4],[EC5]))as pt) A,
( select
studentName,
sum(case when Marks < 30 then 1 else 0 end) as fail,
sum(case when Marks < 23 then 1 else 0 end) as fail2,
sum(case when Marks >= 30 then 1 else 0 end) as pass,
sum(marks) as total
from
student
group by
studentname
) B
where
A.StudentName = B.StudentName
我删除了你所有失败组合的比较逻辑，并将其替换为来自原始表的sum + group by + case，以便您可以确定每个失败，近通道和通过的计数学生无需单独列出所有案例。
您可以在SQL Fiddle
中对此进行测试答案 1 :(得分：2)
您可以使用SUM，COUNT聚合函数的窗口版本来简化查询：
SELECT StudentID, StudentName,
[EC1],[EC2],[EC3],[EC4],[EC5],
Total,
CASE WHEN Below30 >= 2 THEN 'Fail'
ELSE 'Pass'
END AS WithoutBonus,
CASE WHEN Below23 >= 2 THEN 'Fail'
ELSE 'Pass'
END AS WithBonus
FROM (SELECT StudentID, StudentName, Subject, Marks,
SUM(Marks) OVER (PARTITION BY StudentID) AS Total,
COUNT(CASE WHEN Marks < 30 THEN 1 END)
OVER (PARTITION BY StudentID) AS Below30,
COUNT(CASE WHEN Marks < 23 THEN 1 END)
OVER (PARTITION BY StudentID) AS Below23
FROM mytable ) AS src
PIVOT (
MAX (Marks)
FOR Subject IN ([EC1],[EC2],[EC3],[EC4],[EC5]) ) pvt
ORDER BY StudentName
Below30计算具有<1的Mark的受试者的数量。每个StudentID 30 。 Below23计算具有<1的Mark的受试者的数量。 23 per StudentID 。
因此，WithoutBonus是'Fail' ，如果特定学生有2个或更多主题Mark<30，而WithBonus如果特定学生有2个或更多主题'Fail'，则为Mark<23 。
如果您希望输出Marks值增加7，如果它们是< 30，并且您不关心WithoutBonus估算值，那么您可以使用以下内容版本：
SELECT StudentID, StudentName,
[EC1],[EC2],[EC3],[EC4],[EC5],
Total,
CASE WHEN Below23 >= 2 THEN 'Fail'
ELSE 'Pass'
END AS WithBonus
FROM (SELECT StudentID, StudentName, Subject,
CASE WHEN Marks < 30 THEN Marks+7 ELSE Marks END AS Marks,
SUM(Marks) OVER (PARTITION BY StudentID) AS Total,
COUNT(CASE WHEN Marks < 23 THEN 1 END) OVER (PARTITION BY StudentID) AS Below23
FROM mytable ) AS src
PIVOT (
MAX (Marks)
FOR Subject IN ([EC1],[EC2],[EC3],[EC4],[EC5]) ) pvt
ORDER BY StudentName
<强>输出：
StudentID StudentName EC1 EC2 EC3 EC4 EC5 Total WithBonus
--------------------------------------------------------------
4 Rahul 66 34 29 25 33 173 Fail
2 Rajesh 34 56 19 45 30 170 Pass
1 Savita 50 55 45 34 30 207 Pass
3 Smita 76 45 67 56 76 320 Pass
答案 2 :(得分：1)
你可以试试这个
SELECT
StudentName,
CASE WHEN EC1 < 30 THEN EC1 + 7 ELSE EC1 END AS EC1,
CASE WHEN EC2 < 30 THEN EC2 + 7 ELSE EC2 END AS EC2,
CASE WHEN EC3 < 30 THEN EC3 + 7 ELSE EC3 END AS EC3,
CASE WHEN EC4 < 30 THEN EC4 + 7 ELSE EC4 END AS EC4,
CASE WHEN EC5 < 30 THEN EC5 + 7 ELSE EC5 END AS EC5,
Total = (EC1 + EC2 + EC3 + EC4 + EC5),
CASE
WHEN EC1 < 23 AND EC2 < 23 THEN 'FAIL'
WHEN EC1 < 23 AND EC3 < 23 THEN 'FAIL'
WHEN EC1 < 23 AND EC4 < 23 THEN 'FAIL'
WHEN EC1 < 23 AND EC5 < 23 THEN 'FAIL'
WHEN EC2 < 23 AND EC3 < 23 THEN 'FAIL'
WHEN EC2 < 23 AND EC4 < 23 THEN 'FAIL'
WHEN EC2 < 23 AND EC5 < 23 THEN 'FAIL'
WHEN EC3 < 23 AND EC4 < 23 THEN 'FAIL'
WHEN EC3 < 23 AND EC5 < 23 THEN 'FAIL'
WHEN EC4 < 23 AND EC5 < 23 THEN 'FAIL'
ELSE 'PASS'
END AS Result
FROM
(
SELECT * FROM Student
) AS ST
PIVOT
(
SUM(Marks) For [Subject] IN (EC1, EC2, EC3, EC4, EC5)
) AS PV
<强>输出
Rahul 66 34 29 25 33 173 FAIL
Rajesh 34 56 19 45 30 170 PASS
Savita 50 55 45 34 30 207 PASS
Smita 76 45 67 56 76 320 PASS