使用 Pivot 或 UnPivot 动态获取输出
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【中文标题】使用 Pivot 或 UnPivot 动态获取输出【英文标题】：Using Pivot or UnPivot to get Output Dynamically 【发布时间】：2021-11-28 10:06:48 【问题描述】：我有一个具有以下结构的表，我想从该表中动态接收输出。但我不知道如何使用 Pivot 和 UnPivot。
我使用的字段列表如下。
SELECT [RoomID]
,[RoomNumber]
,[RoomType]
,[RoomTypeDescription]
,[RoomBed]
,[PriceOfPerNight]
,[RoomStatuse]
,[RoomStatuseDesc]
,[RoomFloorID]
,[RoomFloorTitle]
FROM [HotelOnline].[dbo].[XtblRooms]
RoomID RoomNumber RType RDesc Beds Price RoomStatuse RDesc FloorID RoomFloorTitle
1 RM100001 2 Degree 2 6 9000000.00 1 Free 1 Floor 001
2 RM100002 1 Degree 1 4 6000000.00 1 Free 1 Floor 001
3 RM100003 2 Degree 2 3 4500000.00 1 Free 1 Floor 001
4 RM100004 3 Degree 3 5 4800000.00 1 Free 1 Floor 001
5 RM100005 1 Degree 1 3 4700000.00 1 Free 1 Floor 001
6 RM100006 1 Degree 1 6 7500000.00 1 Free 1 Floor 001
7 RM100007 1 Degree 1 5 7000000.00 1 Free 1 Floor 001
8 RM100008 1 Degree 1 2 2500000.00 1 Free 1 Floor 001
9 RM100009 3 Degree 3 3 3500000.00 1 Free 1 Floor 001
10 RM100010 3 Degree 3 8 8000000.00 1 Free 1 Floor 001
11 RM100011 2 Degree 2 5 6500000.00 1 Free 2 Floor 002
12 RM100012 3 Degree 3 2 3800000.00 1 Free 2 Floor 002
13 RM100013 2 Degree 2 5 9650000.00 1 Free 2 Floor 002
14 RM100014 3 Degree 3 2 2500000.00 1 Free 2 Floor 002
15 RM100015 2 Degree 2 2 4500000.00 1 Free 2 Floor 002
16 RM100016 3 Degree 3 4 4000000.00 1 Free 2 Floor 002
17 RM100017 1 Degree 1 2 2500000.00 1 Free 2 Floor 002
18 RM100018 3 Degree 3 3 4500000.00 1 Free 2 Floor 002
19 RM100019 2 Degree 2 5 5000000.00 1 Free 2 Floor 002
20 RM100020 2 Degree 2 4 4500000.00 1 Free 2 Floor 002
21 RM100021 1 Degree 1 6 7500000.00 1 Free 3 Floor 003
22 RM100022 2 Degree 2 3 3000000.00 1 Free 3 Floor 003
23 RM100023 3 Degree 3 3 2500000.00 1 Free 3 Floor 003
24 RM100024 1 Degree 1 3 2500000.00 1 Free 3 Floor 003
25 RM100025 2 Degree 2 5 4800000.00 1 Free 3 Floor 003
26 RM100026 3 Degree 3 4 4000000.00 1 Free 3 Floor 003
27 RM100027 2 Degree 2 2 1800000.00 1 Free 3 Floor 003
28 RM100028 3 Degree 3 5 4700000.00 1 Free 3 Floor 003
29 RM100029 1 Degree 1 3 3500000.00 1 Free 3 Floor 003
30 RM100030 2 Degree 2 6 4600000.00 1 Free 3 Floor 003
31 RM100031 2 Degree 2 5 4500000.00 1 Free 4 Floor 004
32 RM100032 1 Degree 1 2 3500000.00 1 Free 4 Floor 004
33 RM100033 3 Degree 3 4 3700000.00 1 Free 4 Floor 004
34 RM100034 2 Degree 2 3 2800000.00 1 Free 4 Floor 004
35 RM100035 3 Degree 3 6 5500000.00 1 Free 4 Floor 004
36 RM100036 2 Degree 2 4 3700000.00 1 Free 4 Floor 004
37 RM100037 3 Degree 3 6 5800000.00 1 Free 4 Floor 004
38 RM100038 1 Degree 1 3 4000000.00 1 Free 4 Floor 004
39 RM100039 1 Degree 1 5 5500000.00 1 Free 4 Floor 004
40 RM100040 1 Degree 1 6 6500000.00 1 Free 4 Floor 004
41 RM100041 1 Degree 1 4 4500000.00 1 Free 5 Floor 005
42 RM100042 2 Degree 2 6 5500000.00 1 Free 5 Floor 005
43 RM100043 2 Degree 2 4 4000000.00 1 Free 5 Floor 005
44 RM100044 2 Degree 2 3 3500000.00 1 Free 5 Floor 005
45 RM100045 3 Degree 3 3 3000000.00 1 Free 5 Floor 005
46 RM100046 3 Degree 3 5 4000000.00 1 Free 5 Floor 005
47 RM100047 3 Degree 3 4 3900000.00 1 Free 5 Floor 005
48 RM100048 2 Degree 2 5 4700000.00 1 Free 5 Floor 005
49 RM100049 2 Degree 2 3 3800000.00 1 Free 5 Floor 005
50 RM100050 3 Degree 3 5 4700000.00 1 Free 5 Floor 005
这是我需要的输出。
Floor 001 Floor 002 Floor 003 Floor 004 Floor 005
==========================================================
RM100001 RM100012 RM100028 RM100033 RM100049
RM100002 RM100013 RM100029 RM100033 RM100050
.. . . . . .
我尝试了几种方法，但没有得到答案。如果可能，请提供帮助。我尝试了几种方法，但没有得到答案。如果可能，请提供帮助。如果可能，请使用动态方法来获得答案。不适合你也没关系。我的问题会像往常一样得到解决。
【问题讨论】：
请看“这是我需要的输出”。在写别的之前。谢谢。 如果您想帮助我，请按照我发送的查询中提到的方法进行。谢谢。 @Larnu 选择 [RoomID] ,[RoomNumber] ,[RoomType] ,[RoomTypeDescription] ,[RoomBed] ,[PriceOfPerNight] ,[RoomStatuse] ,[RoomStatuseDesc] ,[RoomFloorID] ,[RoomFloorTitle] FROM [ HotelOnline].[dbo].[XtblRooms] @Larnu 你说“不适合你也没关系。我的问题会像往常一样解决。”这是什么意思“和往常一样”？您要求动态查询....这是因为您事先不知道楼层数吗？可能的最大楼层数是多少？ 你问了关于你的工作的问题，@AliNajafZadeh，你要求我们干预......如果你不想要答案（链接的重复项算作）不要发布问题。 【参考方案1】：因为我是“文盲”，所以无法回答这个问题：
--If you want a full working example for your data, provide full DDL and DML statements.
WITH YourTable AS(
SELECT *
FROM (VALUES
('1','RM100001','2','Degree 2','6','9000000.00','1','Free','1','Floor 001'),
('2','RM100002','1','Degree 1','4','6000000.00','1','Free','1','Floor 001'),
('3','RM100003','2','Degree 2','3','4500000.00','1','Free','1','Floor 001'),
('4','RM100004','3','Degree 3','5','4800000.00','1','Free','1','Floor 001'),
('5','RM100005','1','Degree 1','3','4700000.00','1','Free','1','Floor 001'),
('6','RM100006','1','Degree 1','6','7500000.00','1','Free','1','Floor 001'),
('7','RM100007','1','Degree 1','5','7000000.00','1','Free','1','Floor 001'),
('8','RM100008','1','Degree 1','2','2500000.00','1','Free','1','Floor 001'),
('9','RM100009','3','Degree 3','3','3500000.00','1','Free','1','Floor 001'),
('10','RM100010','3','Degree 3','8','8000000.00','1','Free','1','Floor 001'),
('11','RM100011','2','Degree 2','5','6500000.00','1','Free','2','Floor 002'))V(RoomID,RoomNumber,RType,RDesc,Beds,Price,RoomStatuse,RDesc2,FloorID,RoomFloorTitle)),
/*
Having multiple columns with the same name is a design flaw (not allowed).
I have named the second Rdesc as RDesc2, but this is also flawed (it breaks normalisation rules).
Fix your design, normalise your data.
*/
RNs AS(
SELECT RoomNumber,
RoomFloorTitle,
ROW_NUMBER() OVER (PARTITION BY RoomFloorTitle ORDER BY RoomNumber) AS RN --ORDER BY should be your ID/always ascending column
FROM YourTable)
SELECT MAX(CASE RoomFloorTitle WHEN 'Floor 001' THEN RoomNumber END) AS Floor001,
MAX(CASE RoomFloorTitle WHEN 'Floor 002' THEN RoomNumber END) AS Floor002
FROM RNs
GROUP BY RN;
db<>fiddle
【讨论】：
【参考方案2】：SELECT *
FROM (
select RoomStatuseDesc,
RoomNumber,
count(*) over (partition by RoomStatuseDesc order by
RoomNumber) rm2 from XtblRooms)as XtblRooms
PIVOT (max(RoomNumber) --as R_count, min(RoomNumber) as r_start
for RoomStatuseDesc in('Floor1','Floor2', 'Floor3', 'Floor4', 'Floor5'))pvt
【讨论】：
您的答案可以通过额外的支持信息得到改进。请edit 添加更多详细信息，例如引用或文档，以便其他人可以确认您的答案是正确的。你可以找到更多关于如何写好答案的信息in the help center。以上是关于使用 Pivot 或 UnPivot 动态获取输出的主要内容，如果未能解决你的问题，请参考以下文章
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