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用于MS-Access中数据透视的SQL查询

[英]SQL Query for Pivoting data in MS-Access

I have a table with three fields: ticketNumber, attendee and tableNumber. 我有一个包含三个字段的表:ticketNumber,与会者和tableNumber。

eg 例如

ticketNumber | attendee | tableNumber
------------ | -------- | -----------
     A1      | alex     |     3  
     A2      | bret     |     2  
     A3      | chip     |     1  
     A4      | dale     |     2  
     A5      | eric     |     2  
     A6      | finn     |     3

I'd like to generate a table with each tableNumber as a field and the list of names sitting at that tableNumber. 我想生成一个表,其中每个tableNumber作为一个字段,而该tableNumber的名称列表。

1  |  2   | 3
-----|------|-----
chip | bret | alex 
     | dale | finn
     | eric |

The query I've been working on is: 我一直在处理的查询是:

transform attendee
select attendee
from registration
group by tableNumber, name
pivot tableNumber

This gives me: 这给了我:

attendee |  1   |  2   |  3 
---------|------|------|----- 
chip     | chip |      | 
bret     |      | bret |
dale     |      | dale |
eric     |      | eric |
alex     |      |      | alex
finn     |      |      | finn

I know how to get the table I require using PivotTableView but I'd like to know how to do it with a query so that I can use it in a code I'm working on. 我知道如何使用PivotTableView获取所需的表,但我想知道如何通过查询进行操作,以便可以在正在使用的代码中使用它。 I'm unsure of how I can write this query without having to select a field (in my case, select attendee). 我不确定如何在不选择字段的情况下编写此查询(在我的情况下,选择与会者)。 Also is it possible to generate the table without the empty cells? 还有可能生成没有空单元格的表吗?

Thank you :) 谢谢 :)

The table needs a unique identifier field that will properly sort and I don't think the ticketNumber can be relied on for that. 该表需要一个唯一的标识符字段,该字段将正确排序,我认为不能依赖ticketNumber。 An autonumber should serve. 应使用自动编号。 Then try: 然后尝试:

TRANSFORM First(Table1.attendee) AS FirstOfattendee SELECT DCount("*","Table1","tableNumber=" & [tableNumber] & " AND ID<" & [ID])+1 AS RowSeq FROM Table1 GROUP BY DCount("*","Table1","tableNumber=" & [tableNumber] & " AND ID<" & [ID])+1 PIVOT Table1.tableNumber;

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