MS Access中的重复计数(不同[f]) [英] Duplicate Count(Distinct [f]) in MS Access

问题描述

有没有一种快速的方法可以在MS Access中复制Count(Distinct [f])的效果?

Is there a quick way to duplicate the effect of the Count(Distinct [f]) in MS Access?

例如:

单个引用的数据表(实际数据中有数千个):

Data Table for a single referral (there are a few thousand of these in the real data):

| Referral ID | Assessment Date | Assessment Team | Service Provided | Service Team
| 1           | 02/01/2008      | AAA             | BBB              | AAA
| 1           | 02/01/2008      | AAA             | CCC              | AAA
| 1           | 02/01/2008      | AAA             | DDD              | BBB
| 1           | 03/01/2008      | BBB             | EEE              | CCC

我想要一个查询，给出:

I want a query that gives:

| Referral ID | Number of Assessments | Teams Assessing | Services Provided | No Teams Providing
| 1           | 2                     | 2               | 4                 | 3

任何帮助表示赞赏！

推荐答案

这有点复杂，但可能适合.小鬼是你的桌子.

This is somewhat complicated, but might suit. Imp is your table.

SELECT DISTINCT a.[Referral ID], b.CountOfADate, c.CountOfATeam, d.CountOfService, e.CountOfSTeam

FROM (((imp AS a 

INNER JOIN 
    (SELECT b1.[Referral ID], Count(b1.ADate) AS CountOfADate
     FROM (SELECT DISTINCT t.[Referral ID], t.[Assessment Date] As ADate FROM imp As t) AS b1
     GROUP BY b1.[Referral ID]) AS b 
ON a.[Referral ID] = b.[Referral ID]) 

INNER JOIN 
    (SELECT c1.[Referral ID], Count(c1.ATeam) AS CountOfATeam
    FROM (SELECT DISTINCT t.[Referral ID], t.[Assessment Team] As ATeam FROM imp As t) AS c1
    GROUP BY c1.[Referral ID]) AS c 
ON a.[Referral ID] = c.[Referral ID]) 

INNER JOIN 
    (SELECT d1.[Referral ID], Count(d1.Service) AS CountOfService
    FROM (SELECT DISTINCT t.[Referral ID], t.[Service Provided] As Service FROM imp As t) AS d1
    GROUP BY d1.[Referral ID]) AS d 
ON a.[Referral ID] = d.[Referral ID]) 

INNER JOIN 
    (SELECT e1.[Referral ID], Count(e1.STeam) AS CountOfSTeam
    FROM (SELECT DISTINCT t.[Referral ID], t.[Service Team] As STeam FROM imp As t) AS e1
    GROUP BY e1.[Referral ID]) AS e 
ON a.[Referral ID] = e.[Referral ID];

这篇关于MS Access中的重复计数(不同[f])的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！