对于具有多个字段的XML路径的SQL东西
问题描述:
我正在使用SQL Server 2008.我试图编写一个使用STUFF和FOR XML PATH的select语句在一个字段中创建逗号分隔的字符串。只有一个记录,但有31个字段。对于具有多个字段的XML路径的SQL东西
我从1个表中获取数据,每个月的每一天都有一个smallint字段--0表示当天不是,1表示是。 (我没有设计这个表,不幸的是,我不能改变它。)像这样:
z1st z2nd z3rd z4th z5th z6th z7th z8th z9th z10th
------ ------ ------ ------ ------ ------ ------ ------ ------ ------
0 1 0 0 0 0 0 0 0 0
0 0 0 0 1 0 0 0 0 1
0 0 1 0 0 0 0 0 0 0
0 0 0 1 0 0 0 0 0 0
0 1 0 0 0 0 0 0 0 1
等等,通过z31st一路。
我的语句传递整数变量@order_snbr和@admin_nbr,它总是返回一行。
DECLARE @strSelectedDayList AS VARCHAR(MAX)
SELECT @strSelectedDayList = STUFF((SELECT ', ' + CASE WHEN td1.z1st = 1 THEN '1st' END
, CASE WHEN td1.z2nd = 1 THEN '2nd' END
, CASE WHEN td1.z3rd = 1 THEN '3rd' END
, CASE WHEN td1.z4th = 1 THEN '4th' END
, CASE WHEN td1.z5th = 1 THEN '5th' END
, CASE WHEN td1.z6th = 1 THEN '6th' END
, CASE WHEN td1.z7th = 1 THEN '7th' END
, CASE WHEN td1.z8th = 1 THEN '8th' END
, CASE WHEN td1.z9th = 1 THEN '9th' END
, CASE WHEN td1.z10th = 1 THEN '10th' END
, CASE WHEN td1.z11th = 1 THEN '11th' END
, CASE WHEN td1.z12th = 1 THEN '12th' END
, CASE WHEN td1.z13th = 1 THEN '13th' END
, CASE WHEN td1.z14th = 1 THEN '14th' END
, CASE WHEN td1.z15th = 1 THEN '15th' END
, CASE WHEN td1.z16th = 1 THEN '16th' END
, CASE WHEN td1.z17th = 1 THEN '17th' END
, CASE WHEN td1.z18th = 1 THEN '18th' END
, CASE WHEN td1.z19th = 1 THEN '19th' END
, CASE WHEN td1.z20th = 1 THEN '20th' END
, CASE WHEN td1.z21st = 1 THEN '21st' END
, CASE WHEN td1.z22nd = 1 THEN '22nd' END
, CASE WHEN td1.z23rd = 1 THEN '23rd' END
, CASE WHEN td1.z24th = 1 THEN '24th' END
, CASE WHEN td1.z25th = 1 THEN '25th' END
, CASE WHEN td1.z26th = 1 THEN '26th' END
, CASE WHEN td1.z27th = 1 THEN '27th' END
, CASE WHEN td1.z28th = 1 THEN '28th' END
, CASE WHEN td1.z29th = 1 THEN '29th' END
, CASE WHEN td1.z30th = 1 THEN '30th' END
, CASE WHEN td1.z31st = 1 THEN '31st' END
FROM TableDays td1 WITH(NOLOCK)
WHERE td1.order_snbr = td2.order_snbr
AND td1.admin_nbr = td2.admin_nbr
FOR XML PATH('')), 1, 1, '')
FROM TableDays td2 WITH(NOLOCK)
WHERE td2.order_snbr = @order_snbr
AND td2.admin_nbr = @admin_nbr
PRINT @strSelectedDayList
这将产生一个结果,像这样:
1st14th28th
我想是这样的:
1st, 14th, 28th
任何帮助,将不胜感激!如果有更好的方法来做到这一点(没有STUFF或FOR XML PATH),我很乐意切换。
答
也许是另一种方法
例
Declare @YourTable Table ([z1st] varchar(50),[z2nd] varchar(50),[z3rd] varchar(50),[z4th] varchar(50),[z5th] varchar(50),[z6th] varchar(50),[z7th] varchar(50),[z8th] varchar(50),[z9th] varchar(50))
Insert Into @YourTable Values
(0,1,0,0,0,1,0,0,0)
,(0,0,0,0,1,0,0,1,0)
,(0,0,1,0,0,0,0,0,0)
,(1,1,1,1,0,0,0,0,0)
,(0,1,0,0,0,0,1,0,1)
Select A.*
,C.*
From @YourTable A
Cross Apply (Select XMLData = cast((select A.* for XML Raw) as xml)) B
Cross Apply (
Select S = Stuff((Select ', ' +replace(Item,'z','')
From (
Select Item = attr.value('local-name(.)','varchar(100)')
From B.XMLData.nodes('/row') as A(r)
Cross Apply A.r.nodes('./@*') AS B(attr)
Where attr.value('local-name(.)','varchar(100)') like 'z[0-9]%'
and attr.value('.','varchar(max)') =1
) C1
For XML Path ('')),1,2,'')
) C
返回
这真是聪明! – n8wrl
@ n8wrl嘘......别告诉我的妻子。她会要我做东西。 –
太棒了!像梦一样工作.... –