微信公众号搜"智元新知"关注
微信扫一扫可直接关注哦!

sql – MS ACCESS:如何使用访问查询计算不同的值?

这是下面给出的当前复杂查询.
SELECT disTINCT Evaluation.ETCode,Training.TTitle,Training.Tcomponent,Training.TImpliment_Partner,Training.TVenue,Training.TStartDate,Training.TEndDate,Evaluation.EDate,Answer.QCode,Answer.Answer,Count(Answer.Answer) AS [Count],Questions.SL,Questions.Question
FROM ((Evaluation INNER JOIN Training ON Evaluation.ETCode=Training.TCode) INNER JOIN Answer ON Evaluation.ECode=Answer.ECode) INNER JOIN Questions ON Answer.QCode=Questions.QCode
GROUP BY Evaluation.ETCode,Training.Tvenue,Questions.Question,Questions.SL
ORDER BY Answer.QCode,Answer.Answer;

另一栏是Training.TCode.我需要计算明显的Training.TCode,有人可以帮帮我吗?
如果您需要更多信息,请告诉我

解决方法

尝试
select ...,count(distinct Training.Tcode) as ...,...

编辑 – 请现在看看这个……

采用以下sql代码.第一个选择是sql服务器如何执行此操作,第二个查询应该是访问兼容…

declare @t table (eCode int,tcode int)
insert into @t values(1,1)
insert into @t values(1,2)
insert into @t values(1,3)
insert into @t values(2,2)
insert into @t values(2,3)
insert into @t values(3,1)    

select 
    ecode,count(distinct tCode) countof
from
    @t
group by
    ecode

select ecode,count(*)
from
    (select distinct tcode,ecode
    from  @t group by tcode,ecode) t
group by ecode

它返回以下内容

ecode tcode
1       3 (there are 3 distinct tcode for ecode of 1)
2       2 (there are 2 distinct tcode for ecode of 2)
3       1 (there is 1 distinct tcode for ecode of 3)

原文地址:https://www.jb51.cc/mssql/83992.html

版权声明:本文内容由互联网用户自发贡献,该文观点与技术仅代表作者本人。本站仅提供信息存储空间服务,不拥有所有权,不承担相关法律责任。如发现本站有涉嫌侵权/违法违规的内容, 请发送邮件至 dio@foxmail.com 举报,一经查实,本站将立刻删除。

相关推荐