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Deleting multiple nodes in SQL Server XML field

开发者 https://www.devze.com 2023-04-11 12:17 出处:网络
If you run this script in SQL Server 2005: create table #xmltemp ( id int, data xml null ) insert into #xmltemp

If you run this script in SQL Server 2005:

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

You will get the following output:

id  data
1   <a id="1" /><a id="2" /><a id="3" /><a id="4" />

id  data
1   <a id="2" /><a id="3" />开发者_C百科;<a id="4" />

I would expect it to delete all three nodes rather than just the first, making the second select return:

id  data
1   <a id="4" />

Is there a way to delete multiple nodes in a single query? Specifically, I want to delete all nodes that match criteria from a column in another table (in this example t is created on the fly but it could just as easily have been an actual table).


You can make xmlid from the joined query into a comma separated string and use that string in the predicate.

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
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