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原帖：http://blog.csdn.net/htl258/archive/2009/04/13/4071123.aspx
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/*
SQLServer2005 XML在T－SQL查询中的典型应用
整理:fcuandy
时间:2008.11.7
前言:
??? 此文只讲xml数据类型及相应的一些操作方法在解决日常T－SQL编程中的一些应用，而避开xml modify,
xml schema,xml索引,命名空间等这些语法性或者生硬的一些问题(这些语法您可以查联机丛书),即此文主要
讲以xml的一些操作特性及xquery去解决编程问题.
Tags:
??? xquery,FLWOR迭带,sql:column,sql:variable,nodes,value,query,xpath,xquery function,if,聚合函数,xs:function等
典型应用举例:
*/
-- (1)
-- ====================================================================
-- 拆分
DECLARE @s VARCHAR ( 100 )
SET @s = ' a,b,c,dd,ee,f,aa,a,f '
-- 常规做法(sql2000常用),以一split函数拆分串为表类型结构,如
-- SELECT * FROM dbo.split(@s,',') a
-- 当然，也可能是循环去拆分,或者以一输助表的identity列利用charindex等函数拿identity列值与','的位置匹配实现拆分
-- 这些做法，roy_88及本人以前都整理过，不再累赘，可见推荐贴。即便 是xml法，也贴过多次，下面一笔带过
-- XML做法:
SELECT b.v FROM
??? ( SELECT CAST ( ' <r> ' + REPLACE ( @s , ' , ' , ' </r><r> ' ) + ' </r> ' AS XML) x) a?? -- 将字串","换换为"</r><r>"并前后拼上<r>,</r>以用来构造xml串
CROSS APPLY
??? ( SELECT v = t.x.value( ' . ' , ' VARCHAR(10) ' ) FROM a.x.nodes( ' //r ' ) AS t(x) ) b? -- 使用 xml.nodes函数将xml串拆分为行
/*
a
b
c
dd
ee
f
aa
a
aa
f
*/
-- (2)
-- ====================================================================
-- 去重,@s中出现的元素，重复的只要一个,希望结果为 'a,f'
-- 常规做法，循环或函数,或临时表拆后distinct
-- XML做法:
-- a.在(1)的基础上进行
; WITH fc AS ?? -- 定义cte命名,将@s转换为一个表结构
(
??? SELECT DISTINCT b.v v
??????????? FROM
??????????????? ( SELECT CAST ( ' <r> ' + REPLACE ( @s , ' </r><r> ' ) + ' </r> ' AS XML) x) a
??????????? CROSS APPLY
??????????????? ( SELECT v = t.x.value( ' . ' , ' VARCHAR(10) ' ) FROM a.x.nodes( ' //r ' ) AS t(x) ) b
)
-- 对这个表利用xml方法进行行值拼接
SELECT STUFF (b.v.value( ' /r[1] ' , ' varchar(100) ' ), 1 , '' )
??? FROM
??? ( SELECT v = ( SELECT ' , ' + v FROM fc FOR XML PATH( '' ),ROOT( ' r ' ),TYPE)) b
/*
a,f
*/
-- b FLWOR语句 + T-SQL组合:
SELECT STUFF (v, '' ) FROM
??? ( SELECT CAST ( ' <r> ' + REPLACE ( @s , ' </r><r> ' ) + ' </r> ' AS XML) x) a
CROSS APPLY
??? ( SELECT x = ( SELECT t.x.value( ' . ' , ' varchar(10) ' ) v,idx = ROW_NUMBER() OVER ( ORDER BY GETDATE ()) FROM a.x.nodes( ' //r ' ) AS t(x) FOR XML PATH( ' r ' ),TYPE)) b -- 利用row_number得到唯一idx
CROSS APPLY
??? ( SELECT v = CAST (b.x.query( ' for $r in //r where count(//r[v=$r/v and idx<$r/idx])=0 return concat(",",xs:string($r/v[1])) ' ) AS VARCHAR ( MAX ))) c? -- 类似count计数法，取得v相同的节点集idx值最小的节点,原型为:
-- SELECT * FROM tb a WHERE 1>(SELECT COUNT(*) FROM tb WHERE v=a.v AND id<a.id)
/*
a,f
*/
-- c distinct-values
SELECT REPLACE (v, ' ' , ' ) FROM
??? ( SELECT CAST ( ' <r> ' + REPLACE ( @s , ' </r><r> ' ) + ' </r> ' AS XML) x) a
CROSS APPLY
??? ( SELECT CAST (a.x.query( ' distinct-values(//r) ' ) AS VARCHAR ( MAX )) v) b? -- 直接调用distinct-values函数来操作
/*
a,aa
*/
-- 导入去重,last(),position()
DECLARE ?? @doc ? xml
SET ?? @doc ?? = ' <?xml version="1.0" encoding="gb2312" ?>
<employees>
??? <employee>
??????? <empid>e0001</empid>
??????? <name>萧峰</name>
??? </employee>
??? <employee>
??????? <empid>e0002</empid>
??????? <name>段誉</name>
??? </employee>
??? <employee>
??????? <empid>e0003</empid>
??????? <name>王语嫣</name>
??? </employee>
??? <employee>
??????? <empid>e0003</empid>
??????? <name>张无忌</name>
??? </employee>
</employees>
'
create table people2
(
??? personid varchar ( 10 )? primary key ,
??? name varchar ( 20 )
)
INSERT people2
SELECT DISTINCT b. * FROM
??? ( SELECT x = @doc .query( ' for $e in //employee? return? //employee[empid = $e/empid][last()] ' )) a? -- FLWOR时,用当前节点去//emploee节点集中找节点集中empid等于当前节点的empid,在找到的集合中取最后一个利用last()函数
CROSS APPLY
??? ( SELECT id = t.x.value( ' empid[1] ' ,name = t.x.value( ' name[1] ' , ' varchar(100) ' ) FROM a.x.nodes( ' //employee ' ) AS t(x)) b
SELECT * FROM people2
/*
e0001??? 萧峰
e0002??? 段誉
e0003??? 张无忌
*/
GO
drop table people2
GO
-- 同组一选多，也可应用此方法，不过没有必要，就不再累赘了。
-- (3)
-- ====================================================================
-- 列名,列值相关
-- a,按行聚合
declare @t table (Sname nvarchar ( 5 ),? V1 float ,??? V2 float ,??? V3 float ,????? V4 float ,??? V5 float ,????? V6 float )
insert @t select N ' 张三 ' ,??? 0.11 , 0.21 , 0.29 ,? 0.32 ,?? 0.11 ,??? 0.08
insert @t select N ' 李四 ' ,??? 0.01 , 0.61 ,? 0.73 ,?? 0.21 ,??? 0.12
insert @t select N ' 张五 ' ,??? 0.31 , 0.23 ,? 0.33 ,?? 0.91 ,??? 0.65
insert @t select N ' 张六 ' ,??? 0.59 , 0.11 ,? 0.26 ,? 0.13 ,??? 0.15
select b. * from
??? ( select x = cast (( select * from @t for xml path( ' r ' )) as xml)) a
cross apply
??? (
??????? select name = x.query( ' ./Sname/text() ' ),v = x.query( ' max(./*[local-name(.)!="Sname"]) ' ) from a.x.nodes( ' //r ' ) as t(x)?
??????? -- r为二级节点(因为文档本身无根节点，即为每项的顶级节点)即为一个r节点表示一条记录. r下级节点，每个表示一个列，因为列名未知，所以用/*匹配所有节点，因为name为区别列，不参与聚合运算，故用local-name取得来过滤
??? ) b
/*
张三??? 0.32
李四??? 0.73
张五??? 0.91
张六??? 0.59
*/
-- b,由值引到取列
if not object_id ( ' T1 ' ) is null
??? drop table T1
GO
Create table T1( [ tId ] int , [ tName ] nvarchar ( 4 ))
Insert T1
select 1 ,N ' zhao ' union all
select 2 ,N ' qian ' union all
select 3 ,N ' sun '
Go
-- > --> 借且(Roy)生成測試數據
if not object_id ( ' T2 ' ) is null
??? drop table T2
Go
Create table T2( [ tId ] int , [ zhao ] nvarchar ( 1 ), [ qian ] nvarchar ( 1 ), [ sun ] nvarchar ( 1 ))
Insert T2
select 1 ,N ' a ' ,N ' b ' ,N ' c ' union all
select 2 ,N ' d ' ,N ' e ' ,N ' f ' union all
select 3 ,N ' g ' ,N ' h ' ,N ' i '
Go
SELECT c.tid,c.tName,v FROM t1 c
CROSS APPLY
??? ( SELECT x = ( SELECT * FROM t2 WHERE tid = c.tid FOR XML PATH( ' r ' ),TYPE)) a
CROSS APPLY
??? ( SELECT v = t.x.query( ' ./*[local-name(.)=xs:string(sql:column("c.tName")) ]/text() ' )
??????? FROM a.x.nodes( ' //r ' ) AS t(x)
??? ) b
/*
1??? zhao??? a
2??? qian??? e
3??? sun??? i
*/
-- c,列名，列值，与系统表
CREATE TABLE tb(f1 INT ,f2 INT ,x INT ,z INT ,d INT ,ex INT ,dd INT ,vv INT )
INSERT tb SELECT 1 , 2 , 3 , 5 , 11 , 2423 , 33
GO
SELECT * FROM tb
GO
SELECT name,v FROM
? ( SELECT name FROM sys.columns WHERE object_id = object_id ( ' tb ' , ' u ' ) ) a
CROSS JOIN
? ( SELECT x = ( SELECT * FROM tb FOR XML PATH( ' r ' ),TYPE)) b
CROSS APPLY
( SELECT v = t.x.query( ' ./*[local-name(.)=xs:string(sql:column("a.name")) ]/text() ' ) FROM b.x.nodes( ' //r ' ) AS t(x) ) c
/*
f1??? 1
f2??? 2
x??? 3
z??? 5
d??? 11
ex??? 3
dd??? 2423
vv??? 33
*/
GO
DROP TABLE tb
GO
-- (4)
-- 一些综合计算
-- 以下表 ta.a值 yyyymmdd-yyyymmdd表连续时间段，","表单个日期
If object_id ( ' ta ' , ' u ' ) is not null
??? Drop table ta
Go
Create table ta(a varchar ( 100 ))
Go
Insert into ta
select ' 1 | |20080101-20080911 '
union all
select ' 2 | |20080101，20080201，20080301，20080515，20080808 '
union all
select ' 3 | |20080101，20080201，20080301，20080515，20081108 '
Go
declare @s varchar ( 8 )
select @s = convert ( varchar ( 8 ), getdate (), 112 )
select stuff ( replace ( replace ( cast (x as varchar ( 1000 )), ' </item><item> ' , case when type = ' 1 ' then ' - ' else ' ， ' end ), ' </item> ' , '' ), 6 ,type + ' | | ' ) a
??? from
??? (
??????? select left (a, 1 ) type,
??????????? cast (
??????????????????? ' <item> '
??????????????????? +
??????????????????? replace (
??????????????????????? stuff (a,
??????????????????????? case when left (a, 1 ) = 1 then ' - ' else ' ， ' end ,
??????????????????????? ' </item><item> '
??????????????????????? )
??????????????????? +
??????????????????? ' </item> '
??????????????? AS XML
??????????????? ) x
??????? from ta
??? ) base
??? where x.value( '
??????????? if (sql:column("base.type")="1") then
??????????????? if(
??????????????????? (/item/text())[1]<sql:variable("@s")
??????????????????? and
??????????????????? (/item/text())[2]>sql:variable("@s")
??????????????? )
??????????????? then 1
??????????????? else 0
??????????? else
??????????????? count(//item[text()>sql:variable("@s")])
??????????? '
???????????,
??????????? ' int '
??????????? ) > 0
go
（编辑：李大同）
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