Sql return results as xml - sql

Question

Given this data:
FName Lname ApartmentNumber
-----------------------------------
David Shumer 1
John Deer 1
Mark Ratz 2
Steven Styer 2
I would like to return it as xml, so I did this:
select * from Table1
for xml raw('person'), root('PeopleInApartment'), elements
And I got this result
<PeopleInApartment>
<Person>
<FName>David</FName>
<LName>Shumer</LName>
<ApartmentNumber>1</ApartmentNumber>
</Person>
<Person>
<FName>John</FName>
<LName>Deer</LName>
<ApartmentNumber>1</ApartmentNumber>
</Person>
<Person>
<FName>Mark</FName>
<LName>Ratz</LName>
<ApartmentNumber>2</ApartmentNumber>
</Person>
<Person>
<FName>Steven</FName>
<LName>Styer</LName>
<ApartmentNumber>2</ApartmentNumber>
</Person>
</PeopleInApartment>
Is it possible to group people by apartment so the result would look like this?:
<Apartments>
<PeopleInApartment>
<Person>
<FName>David</FName>
<LName>Shumer</LName>
<ApartmentNumber>1</ApartmentNumber>
</Person>
<Person>
<FName>John</FName>
<LName>Deer</LName>
<ApartmentNumber>1</ApartmentNumber>
</Person>
</PeopleInApartment>
<PeopleInApartment>
<Person>
<FName>Mark</FName>
<LName>Ratz</LName>
<ApartmentNumber>2</ApartmentNumber>
</Person>
<Person>
<FName>Steven</FName>
<LName>Styer</LName>
<ApartmentNumber>2</ApartmentNumber>
</Person>
</PeopleInApartment>
</Apartments>

Answer 1

Sql solution. Note subquery must depend only on outer columns in the GROUP BY, t1.ApartmentNumber specifically.
select t1.ApartmentNumber
, (select t2.FName, t2.Lname
from Table1 t2
where t2.ApartmentNumber = t1.ApartmentNumber
for xml path('Person'), type
) as PeopleInApartment
from Table1 t1
group by ApartmentNumber
for xml path('Apartment'), root('Apartments'), elements
Returns data apartment - wise
<Apartments>
<Apartment>
<ApartmentNumber>1</ApartmentNumber>
<PeopleInApartment>
<Person>
<FName>David </FName>
<Lname>Shumer</Lname>
</Person>
<Person>
<FName>John </FName>
<Lname>Deer </Lname>
</Person>
</PeopleInApartment>
</Apartment>
<Apartment>
<ApartmentNumber>2</ApartmentNumber>
<PeopleInApartment>
<Person>
<FName>Mark </FName>
<Lname>Ratz </Lname>
</Person>
<Person>
<FName>Steven</FName>
<Lname>Styer </Lname>
</Person>
</PeopleInApartment>
</Apartment>
</Apartments>

Answer 2

Use xml linq :
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Xml;
using System.Xml.Linq;
namespace ConsoleApplication3
{
class Program
{
const string FILENAME = #"c:\temp\test.xml";
static void Main(string[] args)
{
XDocument doc = XDocument.Load(FILENAME);
XElement apartments = new XElement("Apartments");
var groups = doc.Descendants("Person")
.GroupBy(x => (int)x.Element("ApartmentNumber"))
.ToList();
foreach (var group in groups)
{
XElement peopleInApartment = new XElement("PeopleInApartment", group);
apartments.Add(peopleInApartment);
}
}
}
}

Answer 3

Please check Below :
select
(select t2.FName, t2.Lname, t1.ApartmentNumber
from #TEMP t2
where t2.ApartmentNumber = t1.ApartmentNumber
for xml path('Person'), type
) as PeopleInApartment
from #TEMP t1
group by ApartmentNumber
for xml path(''), root ('Apartments')