Spawn legacy app in CMD window in C#

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2.00/5 (1 vote)

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I need to run a legacy app that is run from a cmd window using the Process class.

C#

ProcessStartInfo startInfo = new ProcessStartInfo(); 
startInfo.FileName = "cmd.exe"; 
startInfo.Arguments = "/C \"C:\\UTDSys\\UTD2TCM.exe –r " + Parameters.Candidate + "\"";
startInfo.CreateNoWindow = true; 
startInfo.UseShellExecute = false;  

try 
{ // Start the process with the info we specified.     
  // Call WaitForExit and then the using statement will close.     
  using (Process exeProcess = Process.Start(startInfo))     
  {         
    exeProcess.WaitForExit();     
  }  
} 
catch (Exception e) 
{     
  string sMsg = "Error copying the files to " + Parameters.FullPath + ".";
  HandleErrorMsg(e, sMsg);
  return; 
}

The process My2Com.exe should run in the background, however, I consistantly get the message that a file, used when run from the cmd line with different flags, is missing. If I run the command as indicated in a cmd window, C:\MySys\My2Com.exe –r FullyQualPath, it works as expected. I have tried several different ways to set up the Process class without success.

Any suggestions would be appreciated.

Thank you,

Tom R.

Posted 15-Aug-12 8:39am

CliffRat

Updated 15-Aug-12 9:11am

v2

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Comments

[no name] 15-Aug-12 14:44pm

You have a missing set of double quotes in your string. \"C:\\MySys\\My2Com.exe\", I think

[no name] 15-Aug-12 14:45pm

Or \"C:\\MySys\\My2Com.exe –r " + Parameters.FullPath + "\""

CliffRat 15-Aug-12 15:12pm

Thank you, Wes, for spotting that. It is actually correct in the program and I have corrected it here.

[no name] 15-Aug-12 15:53pm

Back again. Any particular reason that you are running cmd.exe and not your application directly?

CliffRat 15-Aug-12 16:41pm

This was just one of the ways I attempted to get this to work. I started out with...
startInfo.FileName = "c:\MySys\My2Com.exe";
startInfo.Arguments = "–r " + Parameters.FilePath;
and got the same results.

[no name] 15-Aug-12 16:58pm

What, exactly, is the error message that you are getting?

CliffRat 16-Aug-12 8:28am

The error message is specific to the app. It comes up when you attempt to run it without parameter so it appears that the arguments are not there when the process gets spawned.

2 solutions

Solution 2

The fix involves how the arguments string is constructed:

C#

startInfo.FileName = "C:\\UTDSys\\UTD2TCM.exe";
startInfo.Arguments = "-r " + Parameters.FilePath.ToString();

or

C#

startInfo.FileName = "C:\\UTDSys\\UTD2TCM.exe";
startInfo.Arguments = string.Format("-r {0}", Parameters.FilePath);

Thanks to everyone for your help.

Posted 16-Aug-12 3:00am

CliffRat

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Sergey Alexandrovich Kryukov · Accepted Answer · 2012-08-15T10:26:00

You don't need "CMD" to start your application. "CMD" is another application, nothing more, it does not add anything to your functionality. Simply use a static method System.Diagnostics.Process.Start:
http://msdn.microsoft.com/en-us/library/h6ak8zt5.aspx[^].

In your case:

C#

System.Diagnostics.Process.Start("C:\\UTDSys\\UTD2TCM.exe", string.Format("-r {0}", Parameters.Candidate);

Now, please understand that there is no cases where the hard-coded path names like "C:\UTDSys\UTD2TCM.exe" can be useful, unless you want to develop an application which can work only on your development computer. This file may not be placed in this directory, the directory might not exist, and even the presence of the volume "C:" is not a must (on one of my computers, there is no such drive). The file name is always calculated based on user input, some configuration files, assembly location, "special directories" (related to users' accounts, those per user or for all users), etc.

—SA