问题
I am generating a XML file from a SQL Server query.
I have some problem about sorting elements issues.
For example, there is simple code as below.
<test>
<tree abc="123"/>
<tree abc="789"/>
<tree-order abc="456"/>
</test>
As you can see tree and tree-order are different elements and I want to sort by attributes as
<test>
<tree abc="123"/>
<tree-order abc="456"/>
<tree abc="789"/>
</test>
like this.
Anyone has good idea to solve this problem?
Thank you.
Sorry I have detail question as below
AS is
<Main>
<test>
<tree abc="123"/>
<tree abc="789"/>
<tree-order abc="456"/>
</test>
<test>
<tree abc="123"/>
<tree abc="789"/>
<tree-order abc="456"/>
</test>
</Main>
To be :
<Main>
<test>
<tree abc="123"/>
<tree abc="456"/>
<tree-order abc="789/>
</test>
<test>
<tree abc="123"/>
<tree abc="456"/>
<tree-order abc="789/>
</test>
</Main>
Thank you.
回答1:
If you want to reorder existing xml, you can use XQuery:
declare @data xml = '
<test>
<tree abc="123"/>
<tree abc="789"/>
<tree-order abc="456"/>
</test>
'
select @data.query('<test>{for $i in test/* order by $i/@abc return $i}</test>')
Result:
<test>
<tree abc="123" />
<tree-order abc="456" />
<tree abc="789" />
</test>
sql fiddle demo
update:
To reorder multiple nodes, you can use XQuery like this:
select @data.query('
element Main {
for $j in Main/test
return element test {
for $i in $j/* order by $i/@abc return $i
}
}
')
sql fiddle demo
update 2
To order by integer values of attributes, use cast as <type>:
select @data.query('
element Main {
for $j in Main/test
return element test {
for $i in $j/* order by $i/@abc cast as xs:integer?
return $i
}
}
')
sql fiddle demo
来源：https://stackoverflow.com/questions/20613146/sql-server-for-xml-sorting-control-by-attribute