6

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, Answer.QCode, Training.TTitle, Training.Tcomponent, Training.TImpliment_Partner, Training.Tvenue, Answer.Answer, Questions.Question, Training.TStartDate, Training.TEndDate, Evaluation.EDate, Questions.SL
ORDER BY Answer.QCode, Answer.Answer;

还有另一列 Training.TCode。我需要计算不同的 Training.TCode，有人可以帮助我吗？如果您需要更多信息，请告诉我

4

7 回答 7

4

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 服务器将如何执行此操作，第二个查询应该是访问兼容的......

declare @t table (eCode int, tcode int)
insert into @t values(1,1)
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)
于 2009-09-11T18:08:46.500 回答
2

大约一年前，我在 Google 群组中发布了一个类似的问题。我收到了一个很好的答案：

只要您计算基金或子基金，交叉表就可以（来自 Steve Dassin 的原始命题）：

  TRANSFORM COUNT(*) AS theCell
  SELECT ValDate,
      COUNT(*) AS StandardCount,
      COUNT(theCell) AS DistinctCount
  FROM tableName
  GROUP BY ValDate
  PIVOT fund IN(Null)

其中，对于每一天（组），将返回记录数和不同（不同）资金的数量。

改变

PIVOT fund IN(Null)

到

PIVOT subfund IN(Null)

得到同样的，对于子基金。

希望它可能会有所帮助，Vanderghast，Access MVP

我不知道这是否可行，但这是该帖子的链接。

于 2009-09-13T20:28:15.933 回答
2

萨达特，使用这样的子查询：

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,
(SELECT COUNT(*) FROM Training t2 WHERE t2.TCode = Evalution.ETCode) as TCodeCount
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, Answer.QCode, Training.TTitle, Training.Tcomponent, Training.TImpliment_Partner, Training.Tvenue, Answer.Answer, Questions.Question, Training.TStartDate, Training.TEndDate, Evaluation.EDate, Questions.SL
ORDER BY Answer.QCode, Answer.Answer;
于 2009-09-18T14:09:44.533 回答
2

通过执行以下操作，我设法在 Access 中计算不同的值：

select Job,sum(pp) as number_distinct_fruits

from

(select Job, Fruit, 1 as pp

from Jobtable group by Job, Fruit) t

group by Job

您必须小心，好像有一个空白/空字段（在我的代码水果字段中），分组依据将其计为记录。内部选择中的 Where 子句将忽略这些。我已经把它放在我的博客上，但我担心我太容易找到答案 - 这里的其他人似乎认为你需要两个子查询才能完成这项工作。我的解决方案可行吗？ Access 中的不同分组

于 2010-04-17T14:15:28.347 回答
0

看看这个博客条目，看来你可以用子查询来做到这一点......

http://blogs.msdn.com/access/archive/2007/09/19/writing-a-count-distinct-query-in-access.aspx

于 2009-09-12T23:17:02.240 回答
0

我会提议

select R_rep,sum(pp) as number_distinct_Billnos from (select R_rep, Billno, 1 as pp from `Vat_Sales` group by R_rep, Billno) t group by R_rep
于 2013-07-01T09:27:36.193 回答
-1

试试这个：

SELECT DISTINCT e.ETCode, t.TTitle, t.Tcomponent, 
      t.TImpliment_Partner, t.TVenue, t.TStartDate, 
      t.TEndDate, e.EDate, a.QCode, a.Answer, 
      q.SL, q.Question,
      Count(a.Answer) AnswerCount,
      Min(Select Count(*) 
       From (Select Distinct TCode From Training) As Z ) TCodeCount    
FROM Evaluation As e 
   JOIN Training AS t ON e.ETCode=t.TCode
   JOIN Answer AS a ON e.ECode=a.ECode
   JOIN Questions AS q ON a.QCode=q.QCode
GROUP BY e.ETCode, a.QCode, t.TTitle, t.Tcomponent, 
     t.TImpliment_Partner, t.Tvenue, a.Answer, q.Question, 
     t.TStartDate, t.TEndDate, Evaluation.EDate, q.SL
ORDER BY a.QCode, a.Answer;
于 2009-09-11T18:10:46.293 回答