如果在SQL Server 2005中运行此脚本:
create table #xmltemp
(
id int,
data xml null
)
insert into #xmltemp
select 1, ''
update #xmltemp
set data.modify('insert <a id="1" /> into /')
update #xmltemp
set data.modify('insert <a id="2" /> into /')
update #xmltemp
set data.modify('insert <a id="3" /> into /')
update #xmltemp
set data.modify('insert <a id="4" /> into /')
select * from #xmltemp
update x
set data.modify('delete //a[@id=sql:column("t.xmlid")]')
from #xmltemp x
inner join (
select 1 as id, 1 as xmlid union
select 1 as id, 2 as xmlid union
select 1 as id, 3 as xmlid
) t on t.id = x.id
select * from #xmltemp
drop table #xmltemp
您将获得以下输出:
id data
1 <a id="1" /><a id="2" /><a id="3" /><a id="4" />
id data
1 <a id="2" /><a id="3" /><a id="4" />
我希望它删除所有三个节点而不是第一个节点,使第二个选择返回:
id data
1 <a id="4" />
有没有办法在单个查询中删除多个节点?具体来说,我想删除与另一个表中的列匹配条件的所有节点(在此示例中,t是动态创建的,但它可以很容易地成为实际的表)。
答案 0 :(得分:3)
您可以将加入的查询中的xmlid变为逗号分隔的字符串,并在谓词中使用该字符串。
create table #IdToDelete(id int, xmlid int)
insert into #IdToDelete values (1, 1)
insert into #IdToDelete values (1, 2)
insert into #IdToDelete values (1, 3)
insert into #IdToDelete values (2, 4)
update x
set data.modify('delete //a[contains(sql:column("t.xmlid"), @id)]')
from #xmltemp x
inner join (
select D1.id,
(select ','+cast(D2.xmlid as varchar(10))
from #IdToDelete as D2
where D1.id = D2.id
for xml path('')) as xmlid
from #IdToDelete as D1
group by D1.id
) t on t.id = x.id