如何利用SQL进行推理

数据库  /  管理员 发布于 5年前   229

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，
 
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 Smith

 

CREATE 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：

通过以上的代码的介绍，希望对大家的学习有所帮助。