Xquery Get Position() - sql-server

Question

This is a part of our xml file.
<point distanceTotal="162" seqNo="189">
<lineSection id="395" track="1" direction="1">
<outInfos>
<comment commentTypeId="4" priority="1"oneLiner="BOT">
<layerVPK seasonValue="S0"/>
<vectors>
<vector dateFrom="2016-12-11"/>
</vectors>
<frenchText>1x3 MH</frenchText>
</comment>
<comment commentTypeId="4" priority="1" oneLiner="bot">
<layerVPK seasonValue="S0"/>
<frenchText>Réception voie occupée</frenchText>
<dutchText>Test</dutchText>
</comment>
</outInfos>
</point>
We are uploading this to a SqlServer column and with XQuery we are fetching the values.
But, I can't find a way to get the position() coded, and basically T-SQL ROW_NUMBER or dense rank can't be used as not always all data exists.
As example the dutchText only exists on the second comment and there is no field that identifies the 2 comments....
This is the SQL Code
SELECT fi.file_uid,
fi.file_date,
T1.ref.value('#id', 'varchar(100)') AS gTV_id,
T2.ref.value('#id', 'varchar(100)') AS gTrn_id,
T4.ref.value('#seqNo', 'varchar(100)') AS gTrnTPp_seqNo,
T7.ref.value('text()[1]', 'varchar(1000)') AS gTrnTPpOiCDT_Text,
T6.ref.query('/globalTrainVariant/trains/globalTrainVariant/train/timetablePoints/point/outInfos/comment[position()]') AS Test
FROM ods.filesin fi
CROSS APPLY fi.file_xml.nodes('declare namespace cern="http://...";
(/cern:trains/globalTrainVariant)') T1(ref)
CROSS APPLY T1.ref.nodes('declare namespace cern="http://...";
(train)') T2(ref)
CROSS APPLY T2.ref.nodes('declare namespace cern="http://...";
(timetablePoints)') T3(ref)
CROSS APPLY T3.ref.nodes('declare namespace cern="http://...";
(point)') T4(ref)
CROSS APPLY T4.ref.nodes('declare namespace cern="http://...";
(outInfos)') T5(ref)
CROSS APPLY T5.ref.nodes('declare namespace cern="http://...";
(comment)') T6(ref)
CROSS APPLY T6.ref.nodes('declare namespace cern="http://...";
(dutchText)') T7(ref)
WHERE fi.file_type = 'trains'
The code gives no errors, but the Test field is always blank.
Any suggestions ?

Answer 1

If you would look up the documentation, you would see that, as of now, you can't return the result of the position() function directly:
In SQL Server, fn:position() can only be used in the context of a
context-dependent predicate. Specifically, it can only be used inside
brackets ([ ]).
However, there is a neat trick you can employ to get it. Namely, you can compare the position of the element with a known sequence and then return the matched value from that sequence. An example below illustrates that.
declare #x xml = N'<point distanceTotal="162" seqNo="189">
<outInfos>
<comment commentTypeId="4" priority="1" oneLiner="BOT">
<layerVPK seasonValue="S0" />
<vectors>
<vector dateFrom="2016-12-11" />
</vectors>
<frenchText>1x3 MH</frenchText>
</comment>
<comment commentTypeId="4" priority="1" oneLiner="bot">
<layerVPK seasonValue="S0" />
<frenchText>Réception voie occupée</frenchText>
<dutchText>Test</dutchText>
</comment>
</outInfos>
</point>';
with cte as (
select top (1000) row_number() over(order by ac.object_id) as [RN]
from sys.all_columns ac
)
select t.c.query('.') as [OutInfos], sq.RN as [TextPosition], x.c.query('.') as [DutchComment]
from #x.nodes('/point/outInfos') t(c)
cross join cte sq
cross apply t.c.nodes('./comment[position() = sql:column("sq.RN")]/dutchText') x(c);
In it, the CTE produces an ordered set of integers (I usually keep a special table around, but you can always construct one as you go), and match condition is specified in the XQuery expression that defines the x(c) output.

Answer 2

I agree with Roger that the position() fuction cannot be called directly and should be inside []. However, there is a solution which doesn't require any additional tables and supports any number of rows by using recursion:
declare #Xml xml = N'<?xml version="1.0" encoding="utf-16"?>
<root>
<n>1</n>
<n>10</n>
<n>5</n>
<n>3</n>
<n>11</n>
</root>';
with cte as
(
select t.c.value(N'n[1]', N'int') n, 1 RowNum
from #Xml.nodes(N'root[1]') t(c)
where t.c.exist(N'n[1]') = 1
union all
select t.c.value(N'n[position() = sql:column("cte.RowNum") + 1][1]', N'int') n, cte.RowNum + 1
from #Xml.nodes(N'root[1]') t(c)
cross join cte
where t.c.exist(N'n[position() = sql:column("cte.RowNum") + 1]') = 1
)
select *
from cte;

Answer 3

It might be simpler and better performing to use Node Order Comparison Operators in order to count the preceding //comment nodes in the XML tree.
I didn't test on huge XML documents, but it's definitely less I/O intensive, and was less CPU intensive on my contrived tests.
declare #x xml = N'<point distanceTotal="162" seqNo="189">
<outInfos>
<comment commentTypeId="4" priority="1" oneLiner="BOT">
<layerVPK seasonValue="S0" />
<vectors>
<vector dateFrom="2016-12-11" />
</vectors>
<frenchText>1x3 MH</frenchText>
</comment>
<comment commentTypeId="4" priority="1" oneLiner="bot">
<layerVPK seasonValue="S0" />
<frenchText>Réception voie occupée</frenchText>
<dutchText>Test</dutchText>
</comment>
</outInfos>
</point>';
select
[OutInfos] = t.c.query('../..'),
[TextPosition] = t.c.value('let $dutchText := . return count(../../comment[. << $dutchText])', 'int'),
[DutchComment] = t.c.query('.')
from #x.nodes('/point/outInfos/comment/dutchText') t(c)