I am adding security to one of my application. I would like that on Enable, C:\Windows\System32\mstsc.exe c:\thinclient\pekts00.rdp with .rdp (remote desktop) configured is launched. It will continue to run to check if mstsc.exe is running and will logoff windows or start it if it is not.
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Try like below... It will help you...
System.Diagnostics.Process[] pname = System.Diagnostics.Process.GetProcessesByName("mstsc");
if (pname.Length == 0)
System.Diagnostics.Process.Start(Environment.GetEnvironmentVariable("windir") + @"\system32\mstsc.exe");
Pandian
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+0 so far - you are missing a way to specify RDP file and please show correct way (%windir%) to specify path to mstsc. – Alexei Levenkov Apr 16 '13 at 15:56
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@AlexeiLevenkov: i have edited my code... is this you except..? – Pandian Apr 16 '13 at 16:39
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yes it helps thank you... I don't want to kill the process if it is running though, I just want to make sure it is ALWAYS running – Drew Salesse Apr 16 '13 at 16:45
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@DrewSalesse: Thanks, If it helps you then mark it as answer, Then only it was easily identify to Others.. – Pandian Apr 16 '13 at 16:50
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I need to launch to c:\thinclient\pekts00.rdp – Drew Salesse Apr 16 '13 at 17:19
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@DrewSalesse: `System.Diagnostics.Process.Start(@"c:\thinclient\pekts00.rdp");` Try like this... it will launch the application – Pandian Apr 16 '13 at 17:22
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+1. Also actual sample still missing specifying parameters to the application, your last comment provides a way to solve it. – Alexei Levenkov Apr 17 '13 at 02:04