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vb.net and launching an app from the cmd line

Posted on 2014-12-23
6
263 Views
Last Modified: 2015-01-15
attempting to load a program from command line and keep getting "cannot find file specified".

If i load the below from cmd line it works fine but not when i try in vb.
C:\Windows\SysWOW64\CCM\VAppLauncher.exe /launch "Avaya CMS Supervisor R16 -- English 16.1.0.0 R2"

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When i do the below, i get the error:

Dim process1 As New Process
        process1.StartInfo.FileName = "VAppLauncher.exe"
        process1.StartInfo.WorkingDirectory = "C:\\Windows\\SysWOW64\\CCM"
        process1.StartInfo.Arguments = (" /launch Avaya CMS Supervisor R16 -- English 16.1.0.0 R2")
        process1.Start()

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0
Comment
Question by:derek7467
6 Comments
 
LVL 13

Expert Comment

by:Jesus Rodriguez
ID: 40515032
Try to put everything on the Same line like
Process.Start("VAppLauncher.exe","/launch Avaya CMS Supervisor R16 -- English 16.1.0.0 R2")

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Just in case put the that this no work add the location of the program on the first part of the code like
Process.Start("C:\Windows\SysWOW64\CCM\VAppLauncher.exe","/launch Avaya CMS Supervisor R16 -- English 16.1.0.0 R2")

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0
 
LVL 52

Assisted Solution

by:Carl Tawn
Carl Tawn earned 250 total points
ID: 40515036
I imagine you need to quote the string after "launch" due to the spaces.

Try:
Dim process1 As New Process
        process1.StartInfo.FileName = "VAppLauncher.exe"
        process1.StartInfo.WorkingDirectory = "C:\\Windows\\SysWOW64\\CCM"
        process1.StartInfo.Arguments = (" /launch ""Avaya CMS Supervisor R16 -- English 16.1.0.0 R2""")
        process1.Start()

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0
 
LVL 44

Accepted Solution

by:
AndyAinscow earned 250 total points
ID: 40515058
You specify the working directory but not the path to where the app is (not the same thing).
Try this:
Dim process1 As New Process
        process1.StartInfo.FileName = "C:\\Windows\\SysWOW64\\CCM\\VAppLauncher.exe"
        process1.StartInfo.WorkingDirectory = "C:\\Windows\\SysWOW64\\CCM"
        process1.StartInfo.Arguments = (" /launch Avaya CMS Supervisor R16 -- English 16.1.0.0 R2")
        process1.Start()

or this as the next trial in case the arguments need quotes
Dim process1 As New Process
        process1.StartInfo.FileName = "C:\\Windows\\SysWOW64\\CCM\\VAppLauncher.exe"
        process1.StartInfo.WorkingDirectory = "C:\\Windows\\SysWOW64\\CCM"
        process1.StartInfo.Arguments = (" /launch 'Avaya CMS Supervisor R16 -- English 16.1.0.0 R2' ")
        process1.Start()
0
 

Author Comment

by:derek7467
ID: 40515104
doesnt work, now i get "cannot start application", but if i run that same code from a cmd line, it works.
0
 
LVL 44

Expert Comment

by:AndyAinscow
ID: 40515201
>>doesnt work, now i get "cannot start application"
Actually that sounds like one or more comments solved your original problem.  Now you have a rather different one.
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