使用 Pivot 或 UnPivot 动态获取输出答案

【问题标题】：Using Pivot or UnPivot to get Output Dynamically使用 Pivot 或 UnPivot 动态获取输出
【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。

标签： sql sql-server sql-server-2008-r2

【解决方案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】：

【解决方案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】：

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