Question

我有学生记录，他们每天都进行驾驶考试=，我想计算今天通过并且未通过驾驶考试的学生总人数，例如

StudentName  Status     date          pass     fail
test1        something    2014-05-14     1        0
test2        something    2014-05-14     0        1
test1        something    2014-04-14     0        1
test1        something    2014-03-14     0        1

目前我有这个查询 - select count(case pass where 1 then 1 else null end) As TotalPassed, count(case pass where 1 then 1 else null end) as Totalfailed from DatabaseName where date>='2014-05-14'问题是此查询仅搜索2014-05-14的总记录，但我想要的是特定学生的记录总数。对于此查询，我得到结果为;

StudentName Status       date          pass      fail
test1       Something    2014-05-14     1          0
test2       something    2014-05-14     0          1

但我需要的是查询计算每个人通过或未通过考试的总时间，因此预期结果应为：

    StudentName  Status      date          pass      fail
    test1        something   2014-05-14     1          2
    test2        something   2014-05-14     0          1

Answer 1

您应该使用SUM代替COUNT，最后添加GROUP BY。

SELECT StudentName, 
MAX(date) as date,
SUM(pass) As TotalPassed,
SUM(fail) As Totalfailed,
SUM(pass) + SUM(fail) as TotalTriesVersion1,
COUNT(*) as TotalTriesVersion2
FROM Table1 
GROUP BY StudentName

here是sqlfiddle（下次请自己提供:)）

奖励：我给了你两种方法来计算每个学生参加考试的总时间，正如我从你想要的那样的评论中所理解的那样。

Answer 2

你需要像这样使用GROUP BY子句：

SELECT COUNT(CASE pass WHERE 1 THEN 1 ELSE NULL END) AS TotalPassed,
COUNT(CASE pass WHERE 1 THEN 1 ELSE NULL END) AS Totalfailed FROM DatabaseName
WHERE date>='2014-05-14' GROUP BY StudentName

Answer 3

select studentname , date , sum(pass) as Totalpass, sum(fail) as TotalFail from DatabaseName where date ='2014-05-14'
group by studentname,date

Answer 4

通过

小组尝试

从表中选择StudentName，date，sumpass，fail，count（*）按学生姓名分组，日期，通过，失败

计算总记录数

4 个答案: