Question

Given the two sample tables here:

Tickets Table

ID  User    Description

0   James   This is a support ticket
1   Fred    This is a ticket too

Properties Table

ID  TicketID    Label           Value

0   0           Engineer        Scott
1   1           Engineer        Dale
2   0           Manu            Dell
3   1           Manu            HP
4   0           OS              Windows
5   1           OS              Linux

How can I arrive at a view like this:

ID  User    Description                 Engineer    Manu    OS

1   James   This is a support ticket    Scott       Dell    Windows
2   Fred    This is a ticket too        Dale        HP      Linux

It is important to note that the properties table would not always be the same. Some “Tickets” may have properties that others do not.

Is this even possible?

Answer 1

You can perform this with a PIVOT. When doing the PIVOT you can do it one of two ways, with a Static Pivot that you will code the rows to transform or a Dynamic Pivot which will create the list of columns at run-time:

Static Pivot (See SQL Fiddle for Demo):

select id, [user], [engineer], [manu], [OS]
from 
(
    select t.id
        , t.[user]
        , p.ticketid
        , p.label
        , p.value
    from tickets t
    inner join properties p
        on t.id = p.ticketid
) x
pivot
(
    min(value)
    for label in ([engineer], [manu], [OS])
) p

Or you can use a Dynamic Pivot (See SQL Fiddle for Demo):

DECLARE @cols AS NVARCHAR(MAX),
    @query  AS NVARCHAR(MAX);

select @cols = STUFF((SELECT distinct ',' + QUOTENAME(p.label) 
                    from tickets t
                    inner join properties p
                        on t.id = p.ticketid
            FOR XML PATH(''), TYPE
            ).value('.', 'NVARCHAR(MAX)') 
        ,1,1,'')

set @query = 'SELECT id, [user], ' + @cols + ' from 
             (
                 select t.id
                        , t.[user]
                        , p.ticketid
                        , p.label
                        , p.value
                    from tickets t
                    inner join properties p
                        on t.id = p.ticketid
            ) x
            pivot 
            (
                min(value)
                for label in (' + @cols + ')
            ) p '

execute(@query)

Both query will return the same results.