Sql MS ACCESS：如何使用ACCESS查询计算不同的值？

下面是当前的复杂查询

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;

还有一个列Training.TCode。我需要数一数不同的Training.t代码，有人能帮我吗？ 如果您需要更多信息，请告诉我

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

编辑-请现在查看此…

以下面的SQL代码为例。第一个选择是SQL server将如何执行此操作，第二个查询应与access兼容

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)

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

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;

---

看看这篇博文，似乎你可以通过子查询来实现这一点

---

大约一年前，我在谷歌集团发布了一个类似的问题。我得到了一个极好的回答：

---

交叉表可以做到（根据Steve Dassin的原始提议）只要 当您计算基金或子基金时：

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

对于每天（组），将返回记录数和 不同（不同）基金的数量

改变

PIVOT fund IN(Null)

到

同样，对于子基金

希望能有所帮助， Vanderghast，访问MVP

---

---

我不知道这是否有效，但是。

Sadat，使用如下子查询：

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;

---

通过执行以下操作，我成功地在Access中计算了不同的值：

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)

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，因为如果有一个空白/空字段（在我的代码结果字段中），则group by会将其作为记录计算。但是，内部select中的Where子句将忽略这些内容。 我已经在我的博客上发表了这篇文章，但我担心我太容易找到答案了——这里的其他人似乎认为需要两个子查询来实现这一点。我的解决方案可行吗？ 我建议

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

---

我想你们并没有试过，或者根本不关心access数据库。在ms中，不允许使用访问计数（distinct col），它会抛出语法错误为“缺少运算符”@Sadat，你是对的，我忘记了访问SQL有多么不同。。。然后您将需要一个子查询来代替。。。我已经编辑过了，但仍然不起作用。首先，必须显式地将“as”用作别名。我已经放置了它，但仍然存在错误。@Sadat，我编辑添加了“As”，并在子查询周围添加了聚合函数，但您得到了什么错误？Jet/ACE SQL中的表别名不需要As，因此我认为这可能不是问题的原因。抱歉，忽略我，我错过了访问标签抱歉，我正在复习我的访问技能。。。。看这里，这表明您只能通过使用两个子查询进行计数（不同）。。。希望这有帮助。萨达特，看看我编辑的版本。这将使您达到您需要的位置。请不要尝试使用计数（distinct col），因为MS Access不支持计数（distinct）。感谢您的合作。我打算建议您根据需要查看培训表，并填写您的计数（*）。然后在视图版本上而不是直接在表上进行连接。这个建议也会起到同样的作用。