Splitting columns using xml path

Question

Please see my script below,
This is one of the msdn user request but as he is newbie unable to put ddl and dmls.. but all he wants is doing pivot on values.. but I'm trying the same using dynamic pivot which I'm not able to achieve. Help is appreciated.
Table Script:
create table #temp(
ProjectNO int,projectname varchar(10),client varchar(10),programmers varchar(10));
declare @cols AS nvarchar(MAX)
,@QUERY AS NVARCHAR(MAX);
Insert into #temp
Select 01 , 'ave' , 'zica' , 'dee'
UNION
Select 01, 'ave', 'zica ', 'law'
UNION
Select 01 ,'ave', 'zica', 'amy'
UNION
Select 01 , 'ave' , 'rowan' , 'dee'
UNION
Select 01 ,'ave' ,'rowan' ,'law'
UNION
Select 01 ,'ave' ,'rowan' ,'amy'
Select * from #temp
Question 1: Desired output:
ProjectNO. Prjtname Clt1 Clt2 Prog1 Prog2 Prog3
01 ave zica rowan dee law amy
Question 2: Query:
Select distinct Replace((select distinct convert(char(8),client)
from #temp
FOR XML path('')) ,'";;#;;"',' ') as clients
FROM #temp
GROUP BY ProjectNO
In the above query I'm doing xml path and I'm getting result in single row like
clients 
rowan    zica
Instead can I get result like below using xml path?
client1 client2
rowan  zica
- please mark correct answers

Answer 1

For this particular problem we don't want to use XML path. I already answered in the other thread with a link to this article
T-SQL:
Dynamic Pivot on Multiple Columns
Let me know if using that article code you will be able to achieve the goal. It's rather simple, actually, I can write up the mock up idea.
declare @MaxClients int, @MaxProgrammers int
select @MaxClients = max(ClientsCount) from (select ProjectNo, count(distinc(Client)) as ClientsCount from #Temp GROUP BY ProjectNo) X
select @MaxProgrammers = max(ProgrammersCount) from (select ProjectNo, count(distinc(Programmer)) as ProgrammersCount from #Temp GROUP BY ProjectNo) X
declare @SQL nvarchar(max), @Loop int
set @Loop = 1
Actually, that't the beginning of the idea and you should be able to figure this out till completion based on the article I pointed out. We would need to use 2 dense_rank functions.
For every expert, there is an equal and opposite expert. - Becker's Law
My blog
My TechNet articles