How to get Physical Drives associated with SQL Instance

I need to get a list of physical drives (ie: drive
letters) of all the drives associated with an instance of
SQL 2000 running on a cluster.
The ideal solution would be a stored procedure that given
the instance name you would get back the drives.
Can anyone make a useful suggestion? Is this possible with
a stored procedure?
Thanks,
kenKen,
You can use master..xp_fixeddrives to get the fixed drives name as well as
free space in MB.Since its a undocumented call, please refrain from using it
in production code.An alternative would be to use master..xp_cmdshell
--
Dinesh.
SQL Server FAQ at
http://www.tkdinesh.com
"Ken McClain" <ken@.softbreeze.net> wrote in message
news:0b8e01c35062$4e3466e0$a501280a@.phx.gbl...
> I need to get a list of physical drives (ie: drive
> letters) of all the drives associated with an instance of
> SQL 2000 running on a cluster.
> The ideal solution would be a stored procedure that given
> the instance name you would get back the drives.
> Can anyone make a useful suggestion? Is this possible with
> a stored procedure?
> Thanks,
> ken|||In addition to Dinesh's suggestion which I can't recall if it works on a
cluster there is also the system function fn_servershareddrives() that will
give you the shared cluster drives, possibly combining them will give you
the answer, I haven't got a cluster to test on at the moment. See BOL for
details of this function
SELECT *
FROM ::fn_servershareddrives()
--
HTH
Jasper Smith (SQL Server MVP)
I support PASS - the definitive, global
community for SQL Server professionals -
http://www.sqlpass.org
"Ken McClain" <ken@.softbreeze.net> wrote in message
news:0b8e01c35062$4e3466e0$a501280a@.phx.gbl...
I need to get a list of physical drives (ie: drive
letters) of all the drives associated with an instance of
SQL 2000 running on a cluster.
The ideal solution would be a stored procedure that given
the instance name you would get back the drives.
Can anyone make a useful suggestion? Is this possible with
a stored procedure?
Thanks,
ken