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我从第三方MS Access 2000文件格式获取数据在下面表格形式选择记录,其中多个项目匹配

  • 论文和
  • PaperTags
  • 条件

下面是来自这些表格的示例数据。

Papers表和样本数据

+----+----------+ 
| ID | PaperID | 
+----+----------+ 
| 1 | 658  | 
| 2 | 659  | 
| 3 | 660  | 
| 4 | 661  | 
| 5 | 662  | 
| 6 | 663  | 
| 7 | 664  | 
+----+----------+ 

PaperTags表和样本数据

+----+----------+----------------------------------------+ 
| ID | PaperID |     TagID    | 
+----+----------+----------------------------------------+ 
| 1 | 663  | 3         | 
| 2 | 663  | 15 --Y        | 
| 3 | 663  | 17         | 
| 4 | 663  | 18 --Y        | 
| 5 | 664  | 14         | 
| 62 | 658  | 9         | 
| 63 | 658  | 14         | 
| 64 | 658  | 17         | 
| 65 | 659  | 15 --Y        | 
| 66 | 659  | 17         | 
| 67 | 659  | 18 --Y        | 
| 68 | 660  | 17         | 
| 69 | 660  | 18 --N as it has only 18 and not 15 | 
| 70 | 661  | 10         | 
| 71 | 661  | 17         | 
| 72 | 661  | 18 --N as it has only 18 and not 15 | 
| 73 | 662  | 18 --N as it has only 18 and not 15 | 
| 74 | 662  | 14         | 
| 75 | 662  | 17         | 
| 76 | 662  | 18 --N as it has only 18 and not 15 | 
+----+----------+----------------------------------------+ 

现在我的最终用户将通过例如15和18我的目标的一个或多个TagIDs是找到所有的这些TagID具有全部的PaperID。在这些例子中,我需要返回663和659

我试过下面的查询,但如果数据有任何故障,那么它不起作用。例如,PaperID 662在表格中出现两次,具有相同的TagID,因此计数(PaperID)= 2的结果是正确的,并且最终会显示在我的结果中。

select Count(PaperID), PaperID from PaperTags 
group by TagID, PaperID 
having TagID = 15 or TagID = 18 
and count(PaperID) = 2 

,我试过其他查询是

select * from Papers 
where Papers.PaperID 
in 
(
select PaperTags.PaperID from PaperTags 
where (PaperTags.Tagid = 15 or PaperTags.Tagid = 18) 
and PaperTags.PaperID = Papers.PaperID 
) 

我已经直通下面的文章,但因为我使用MSACCESS我不能使用这种方法。

Select records from a table where all other records with same foreign key have a certain value

我认为必须有更好的方式来进行筛选。任何帮助深表感谢。

回答

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事情是这样的:

SELECT G.ContentID 
FROM (
SELECT PT.ContentID, PT.TagID 
FROM PaperTags AS PT 
WHERE PT.TagID IN (15, 18) 
GROUP BY PT.ContentID, PT.TagID 
) AS G 
GROUP BY G.ContentId 
HAVING Count(*) = 2 
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@T基翁 - 在访问我收到错误“语法错误(缺少操作员)在查询表达式‘计数(DISTINCT内容识别)’”我测试了一个简单的查询与伯爵( DISTINCT ContentID),仍然得到了同样的错误,我认为MS Access不支持Count(DISTINCT ColumnName) – ndd

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呃..嗯...我想... –

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我已经更新了我的答案。 –