select rownumber in xquery and convert it to sql rows - sql

Question

i have a table which i save user's search result rows in xml format as follows :
<row id="5083" />
<row id="5085" />
<row id="5087" />
<row id="5090" />
<row id="5094" />
... (about 500,000 rows)
where each row element contains id of the result record which will be shown in result page.
now i need to select ids of one specific page for example page 2 ( 10th element to 20th element )
the first problem is how can i get that result in xquery ?
i tried to use position() function but it didn't work ...
select #results.query('for $x in (row)
where $x/position() > 10
return ($x)')
and the second prblem is how can get my result ids as sql rows instead of xmlnodelist ?

Answer 1

You can use the nodes() method to shred your XML and use the value() method to extract the value for attribute id.
SQL Fiddle
Query 1:
declare #results xml = '
<row id="5083" />
<row id="5085" />
<row id="5087" />
<row id="5090" />
<row id="5094" />'
select T.N.value('#id', 'int') as id
from #results.nodes('row[position() >= 2 and position() < 4]') as T(N)
Results:
| ID |
--------
| 5085 |
| 5087 |

Answer 2

In
for $x in (row) where $x/position() > 10 return ($x)
$x iterates over all the row children after the 10th but for each returns a sequence of length 1 with just one row element so $x/position() is always 1.
You could use
row[position() >=10 and position() < 20]
which will return elements in positions 10 to 19. If you want just the id's rather than the element nodes then
row[position() >=10 and position() < 20]/string(#id)