G
Guest
Guest
Archived from groups: microsoft.public.windows.networking.wireless (More info?)
I know how to use batch files and "net stop" etc.
I just need to know if "net stop" has a switch in command line (available in
a batch file) that will allow it to be run under a user level account.
My situation is as follows:
I have a WiFi 802.11g card (linksys) and it keeps dropping every 10-15 min.
I know what the problem is, WCZ enabled looking for a good network. So
everytime I boot my computer I have to go in and stop the WCZ service. This
is far from convienient, esp if I am not here to turn it off.
As I am the only one in the house that knows anything about computers, I
need a batch file/script/anything that will allow me to stop WCZ while logged
in under a user account, _AFTER_ it has connectted to the WiFi net.
how can I go about this?
You help is appreciated, thanks.
p.s. - do not just say "Use the software from the manufacturer". I do not
want to use it as windows SP2 touts having great flexability when dealing
with networks/NICs of all types (which is true).
I know how to use batch files and "net stop" etc.
I just need to know if "net stop" has a switch in command line (available in
a batch file) that will allow it to be run under a user level account.
My situation is as follows:
I have a WiFi 802.11g card (linksys) and it keeps dropping every 10-15 min.
I know what the problem is, WCZ enabled looking for a good network. So
everytime I boot my computer I have to go in and stop the WCZ service. This
is far from convienient, esp if I am not here to turn it off.
As I am the only one in the house that knows anything about computers, I
need a batch file/script/anything that will allow me to stop WCZ while logged
in under a user account, _AFTER_ it has connectted to the WiFi net.
how can I go about this?
You help is appreciated, thanks.
p.s. - do not just say "Use the software from the manufacturer". I do not
want to use it as windows SP2 touts having great flexability when dealing
with networks/NICs of all types (which is true).