System.Diagnostics.Process.Start(“net”,“use”\ server \ z)
如何在进程启动时隐藏弹出窗口?谢谢
答案 0 :(得分:0)
使用Startinfo属性。看看这样的东西是否有效:
Dim NewProcess As New Process()
NewProcess.StartInfo.UseShellExecute = False
NewProcess.StartInfo.FileName = "net"
NewProcess.StartInfo.Arguments = "use \server\z"
NewProcess.StartInfo.CreateNoWindow = True
NewProcess.Start()