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如何利用SQL進(jìn)行推理

更新時(shí)間：2015年07月20日 17:38:37 作者：Boss_he

本文介紹了如何利用SQL進(jìn)行推理，通過一個(gè)例子進(jìn)行實(shí)現(xiàn)推理的演示過程,需要了解的朋友可以參考下

數(shù)據(jù)庫環(huán)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(shí)現(xiàn)所有人住樓層的排列組合，然后把條件套進(jìn)去即求得。如何實(shí)現(xiàn)排列組合，

1.基礎(chǔ)數(shù)據(jù)準(zhǔn)備
--準(zhǔn)備基礎(chǔ)數(shù)據(jù)，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：