29.09.2010, 20:38
(
Последний раз редактировалось JaTochNietDan; 30.09.2010 в 15:30.
)
Quote:
I tryed, but:
This works Process.Start("samp.exe") This doens't works: Process.Start("samp.exe -(ip) -(port)") = Says this file couldnt found bla bla |
Код:
Dim SA-MP As New ProcessStartInfo("samp.exe", "127.0.0.1:7777") Process.Start(SA-MP)