How do I setup Vista so it only connects to ONE manually added Wireless LAN
network?

There's a lot of pages about using the setting "Automatically connect to
non-preferred networks" to disable Vista's ability to roam to others but I
don't want them in the list to start with, so they can never be selected. I
don't want to ever aquire any new networks unless manually added. I'm looking
for a one-off solution that requires no policing. If the network isn't
present then there's no connection.

Incidently the setting "Automatically connect to non-preferred networks"
doesn't seem to be present in Windows Vista home premium, is it hidden or
does it have an equivalent? Its setting of "Connect to a more preferred
network if available" just controls automatic switching and not their
availability in the first place.

This thread (
http://windowshelp.microsoft.com/com...=en-us&m=1&p=1 )
is the most similar I've found on this group but the solution for me isn't
valid.

I should have phrased that slightly differently and said

How do I setup Vista so it only connects to ONE manually added Wireless LAN,
and not aquire or use any other networks.

"Pete" wrote:

How do I setup Vista so it only connects to ONE manually added Wireless LAN
network?

There's a lot of pages about using the setting "Automatically connect to
non-preferred networks" to disable Vista's ability to roam to others but I
don't want them in the list to start with, so they can never be selected. I
don't want to ever aquire any new networks unless manually added. I'm looking
for a one-off solution that requires no policing. If the network isn't
present then there's no connection.

Incidently the setting "Automatically connect to non-preferred networks"
doesn't seem to be present in Windows Vista home premium, is it hidden or
does it have an equivalent? Its setting of "Connect to a more preferred
network if available" just controls automatic switching and not their
availability in the first place.

This thread (
http://windowshelp.microsoft.com/com...=en-us&m=1&p=1 )
is the most similar I've found on this group but the solution for me isn't
valid.