Access Query; Creating Master/Slave from list - sql

Question

Asked a question a few days ago, but have decided to go down a differnt route so am re-doing the question as the edit's were getting a bit messy.
I have a list of data containing two columns:
pID sKey
100 8611
100 2318
101 3516
101 5413
102 6546
102 5646
102 8411
103 8795
103 5845
The first sKey to appear would become the Master sKey for that pID and each sKey after would be a slave. The data would look like this.
pID sKey sKey_1
100 8611 2318
101 3516 5413
102 6546 5646
102 6546 8411
103 8795 5845
This query gets me close
SELECT MyTable.pID, MyTable.sKey, MyTable_1.sKey
FROM MyTable
INNER JOIN MyTable AS MyTable_1
ON MyTable.pID = MyTable_1.pID
WHERE (((IIf([MyTable.sKey]=[MyTable_1.sKey],"Y","N"))="N"))
pID sKey sKey
100 2318 8611
100 8611 2318
101 3516 5413
101 5413 3516
102 5646 6546
102 5646 8411
102 6546 5646
102 6546 8411
102 8411 5646
102 8411 6546
103 5845 8795
103 8795 5845
But as you can see it reverses the order and doubles each one up, and when it hits an instance where there is 3 or more sKey's it goes a bit crazy :\
Anyone have any ideas, or can point me in the right direction?

Answer 1

If you're trying to use the MIN(skey) as your Master, then something like this should work:
select
p.pId,
p.skey,
p3.skey skey1
from mytable p
join (select pID, min(skey) minskey
from mytable
group by pID
) p2 on p.pid = p2.pid and p.skey = p2.minskey
join mytable p3 on p.pid = p3.pid and p.skey != p3.skey
SQL Fiddle Demo
This produces slightly different results than yours above.
If your desired results are to use the first skey that shows up, then I'd recommend adding an Identity/AutoNumber column to your table just to seed from. You can't guarantee the order of the results without that column. So assuming you were to add such a column, then something like this should work:
select
p.pId,
p.skey,
p3.skey skey1
from mytable p
join (select pID, min(id) minId
from mytable
group by pID
) p2 on p.id = p2.minId
join mytable p3 on p.pid = p3.pid and p.id <> p3.id
order by p.pid, p3.id
SQL Fiddle Demo With AutoNumber