如何利用SQL進行推理

數據庫環境：SQL SERVER 2008R2

有如下需求：
Baker, Cooper, Fletcher, Miller and Smith住在一座房子的不同樓層。
Baker 不住頂層。Cooper不住底層。
Fletcher 既不住頂層也不住底層。Miller住得比Cooper高。
Smith住的樓層和Fletcher不相鄰。
Fletcher住的樓層和Cooper不相鄰。
用SQL寫出來
 
解題思路:
先實現所有人住樓層的排列組合，然後把條件套進去即求得。如何實現排列組合，
 
1.基礎數據准備
--准備基礎數據，用A、B、C、D、E分別表示Baker, Cooper, Fletcher, Miller and Smith

CREATE TABLE ttb
 (
  subname VARCHAR(1) ,
  realname VARCHAR(10)
 )
INSERT INTO ttb
VALUES ( 'A', 'Baker' ),
  ( 'B', 'Cooper' ),
  ( 'C', 'Fletcher' ),
  ( 'D', 'Miller' ),
  ( 'E', 'Smith' )

2.生成所有可能情況的排列組合
--生成A、B、C、D、E所有的排列組合

WITH x0
   AS ( SELECT CONVERT(VARCHAR(10), 'A') AS hid
    UNION ALL
    SELECT CONVERT(VARCHAR(10), 'B') AS hid
    UNION ALL
    SELECT CONVERT(VARCHAR(10), 'C') AS hid
    UNION ALL
    SELECT CONVERT(VARCHAR(10), 'D') AS hid
    UNION ALL
    SELECT CONVERT(VARCHAR(10), 'E') AS hid
    ),
  x1
   AS ( SELECT hid
    FROM  x0
    WHERE LEN(hid) <= 5
    UNION ALL
    SELECT CONVERT(VARCHAR(10), a.hid + b.hid) AS hid
    FROM  x0 a
      INNER JOIN x1 b ON CHARINDEX(a.hid, b.hid, 1) = 0
    )
 SELECT hid AS name
 INTO #tt
 FROM x1
 WHERE LEN(hid) = 5
 ORDER BY hid

3.加入條件，找出滿足要求的樓層安排

WITH x2
   AS ( SELECT name
    FROM  #tt
    WHERE SUBSTRING(name, 5, 1) <> 'A'--Baker 不住頂層
      AND SUBSTRING(name, 1, 1) <> 'B'--Cooper不住底層
      AND ( SUBSTRING(name, 1, 1) <> 'C'
        AND SUBSTRING(name, 5, 1) <> 'C'--Fletcher 既不住頂層也不住底層
       )
      AND name LIKE '%B%D%'--Miller住得比Cooper高
      AND name NOT LIKE '%CE%' AND name NOT LIKE '%EC%' --Smith住的樓層和Fletcher不相鄰
      AND name NOT LIKE '%BC%' AND name NOT LIKE '%CB%' --Fletcher住的樓層和Cooper不相鄰
    ),
  x3--生成樓層號
   AS ( SELECT number AS id ,
      SUBSTRING(x2.name, number, 1) AS name
    FROM  master.dbo.spt_values
      INNER JOIN x2 ON 1 = 1
    WHERE type = 'P'
      AND number <= 5
      AND number >= 1
    )
 SELECT a.id AS 樓層,
   b.realname AS 姓名
 FROM x3 a
   INNER JOIN ttb b ON b.subname = a.name
 ORDER BY id

樓層安排如下：

通過以上的代碼的介紹，希望對大家的學習有所幫助。