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如何利用SQL进行推理

2019-01-03 MsSQL 前端之家

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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,Miller and Smith

<div class="jb51code">
<pre class="brush:sql;">
CREATE TABLE ttb
(
subname VARCHAR(1),realname VARCHAR(10)
)
INSERT INTO ttb
VALUES ( 'A','Baker' ),( 'B','Cooper' ),( 'C','Fletcher' ),( 'D','Miller' ),( 'E','Smith' )

2.生成所有可能情况的排列组合
--生成A、B、C、D、E所有的排列组合

<div class="jb51code">
<pre class="brush:sql;">
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

<div class="jb51code">
<pre class="brush:sql;">
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) <> 'C'
AND SUBSTRING(name,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