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[数据库]利用SQL进行推理


数据库环境:SQL SERVER 2008R2

有如下需求:

Baker, Cooper, Fletcher, Miller and Smith住在一座房子的不同楼层。
Baker 不住顶层。Cooper不住底层。
Fletcher 既不住顶层也不住底层。Miller住得比Cooper高。
Smith住的楼层和Fletcher不相邻。
Fletcher住的楼层和Cooper不相邻。
用SQL写出来
 
解题思路:
先实现所有人住楼层的排列组合,然后把条件套进去即求得。如何实现排列组合,
具体可以参考我前面的文章 http://www.cnblogs.com/boss-he/p/4534017.html
 
1.基础数据准备
--准备基础数据,用A、B、C、D、E分别表示Baker, Cooper, Fletcher, Miller and SmithCREATE TABLE ttb  (   subname VARCHAR(1) ,   realname VARCHAR(10)  )INSERT INTO ttbVALUES ( 'A', 'Baker' ),    ( 'B', 'Cooper' ),    ( 'C', 'Fletcher' ),    ( 'D', 'Miller' ),    ( 'E', 'Smith' )

2.生成所有可能情况的排列组合

--生成A、B、C、D、E所有的排列组合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.加入条件,找出满足要求的楼层安排

WITH  x2     AS ( SELECT  name        FROM   #tt        WHERE  SUBSTRING(name, 5, 1) <> 'A'--Baker 不住顶层            AND SUBSTRING(name, 1, 1) <> 'B'--Cooper不住底层            AND ( SUBSTRING(name, 1, 1) <> 'C'               AND SUBSTRING(name, 5, 1) <> 'C'--Fletcher 既不住顶层也不住底层              )            AND name LIKE '%B%D%'--Miller住得比Cooper高            AND name NOT LIKE '%CE%' AND name NOT LIKE '%EC%' --Smith住的楼层和Fletcher不相邻            AND name NOT LIKE '%BC%' AND name NOT LIKE '%CB%' --Fletcher住的楼层和Cooper不相邻       ),    x3--生成楼层号     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 楼层,      b.realname AS 姓名  FROM  x3 a      INNER JOIN ttb b ON b.subname = a.name  ORDER BY id

楼层安排如下:

(本文完)