SQL Server : creating XML with child nodes for non unique values

Question

I am having a hard time figuring out how to create an XML file from a SQL select that has sub nodes to store the not unique values.

I am using Microsoft SQL Server 2012

e.g.

select company, founder
from companies

The result of the query is:

 microsoft corp       Bill Gates
 microsoft corp       Paul Allen
 apple inc            Steve Jobs
 apple inc            Ronald Wayne
 apple inc            Steve Wozniak

I would like to generate an XML like

<values>
       <company>microsoft corp</company>
       <founders>
           <founder id="1">Bill Gates</founder>
           <founder id="2">Paul Allen</founder>
       </founders>
</values>
<values>
       <company>apple inc</company>
       <founders>
           <founder id="1">Steve Jobs</founder>
           <founder id="2">Ronald Wayne</founder>
           <founder id="3">Steve Wozniak</founder>
        </founders>
       </company>
</values>

I am not sure the node <founders> is required, I think it would also work fine for me to have the founders directly under the <values> node, since they keep in the right company, getting an id to have a list.

What I get right now, using FOR XML, then I tried different options is the following:

<values>
       <company>apple inc</company>
       <founder>Steve Jobs</founder>
</values>
<values>
       <company>apple inc</company>
       <founder>Ronald Wayne</founder>
</values>
<values>
       <company>apple inc</company>
       <founder>Steve Wozniak</founder>
</values>

which is not compatible with what I need to achieve then.

Any help to get all the founders under the same <values> node is highly appreciated.

Thanks!

just updated sorry about that

user3700389
– user3700389

2014-06-02 18:18:05 +00:00
Commented Jun 2, 2014 at 18:18 — user3700389
– user3700389, Commented Jun 2, 2014 at 18:18

Darka · Accepted Answer · 2014-06-02 19:08:09Z

Searched a bit internet and found solution:

DECLARE @table TABLE (company VARCHAR(100), founder VARCHAR(100))

 insert into @table SELECT 'microsoft corp', 'Bill Gates'
 insert into @table SELECT 'microsoft corp', 'Paul Allen'
 insert into @table SELECT 'apple inc', 'Steve Jobs'
 insert into @table SELECT 'apple inc', 'Ronald Wayne'
 insert into @table SELECT 'apple inc', 'Steve Wozniak'


 SELECT company, 
    (
        SELECT ROW_NUMBER()OVER( ORDER BY founder ) as 'founder/@id',
            founder as 'founder'
        FROM @table as F
        WHERE F.company = Comp.company
        FOR XML PATH(''), TYPE
    ) AS founders
 FROM @table as Comp
 GROUP BY company
 FOR XML PATH('VALUES')

Result:

<VALUES>
  <company>apple inc</company>
  <founders>
    <founder id="1">Ronald Wayne</founder>
    <founder id="2">Steve Jobs</founder>
    <founder id="3">Steve Wozniak</founder>
  </founders>
</VALUES>
<VALUES>
  <company>microsoft corp</company>
  <founders>
    <founder id="1">Bill Gates</founder>
    <founder id="2">Paul Allen</founder>
  </founders>
</VALUES>

M.Ali · Accepted Answer · 2014-06-02 19:12:23Z

Test Data

DECLARE @Companies TABLE (company VARCHAR(100), founder VARCHAR(100))
INSERT INTO @Companies VALUES
 ('microsoft corp',       'Bill Gates'),
 ('microsoft corp',       'Paul Allen'),
 ('apple inc',            'Steve Jobs'),
 ('apple inc',            'Ronald Wayne'),
 ('apple inc',            'Steve Wozniak')

Query

SELECT company
       ,(SELECT ROW_NUMBER() OVER (PARTITION BY Company ORDER BY Founder ASC)  AS [@ID] 
               ,B.founder [text()]
         FROM @companies B
        WHERE B.company = A.company
        FOR XML PATH('Founder'),TYPE) AS Founders
 FROM  @companies A
GROUP BY A.Company       
FOR XML PATH('values'),ROOT('Doc'), ELEMENTS

OUTPUT

<Doc>
  <values>
    <company>apple inc</company>
    <Founders>
      <Founder ID="1">Ronald Wayne</Founder>
      <Founder ID="2">Steve Jobs</Founder>
      <Founder ID="3">Steve Wozniak</Founder>
    </Founders>
  </values>
  <values>
    <company>microsoft corp</company>
    <Founders>
      <Founder ID="1">Bill Gates</Founder>
      <Founder ID="2">Paul Allen</Founder>
    </Founders>
  </values>
</Doc>

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