How to start .exe file with arguments in C#?

Question

I need to start an exe file from my app in C#.

The query string should look like this:

Extractor.exe "base\sql" "develop" "..\Data\System" "Grids"

This is my code:

public static void StartExtractor()
{
    System.Diagnostics.Process process = new System.Diagnostics.Process();
    process.StartInfo.FileName = OurFields.keskullPass + @"\Extractor.Deploy\Extractor.exe";
    process.StartInfo.Arguments = "\"base\\sql\" \"develop\" \"..\\Data\\System\" \"Grids\"";
    process.Start();
    process.WaitForExit();
}

Answer 1

A simple way to start a process with arguments in C#:

Process.Start("yourapp.exe", "your arguments");

If you really need to wait for process to exit, then it becomes:

var process = Process.Start("yourapp.exe", "your arguments");

process.WaitForExit();

Usually you do not create Process object manually.

Process is defined in System.Diagnostics, so you also need to add it to your using clause.

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And your problem is probably related to specific expectations of external application to arguments' format, or a mistake in concatenation of full path to exe.

To quote arguments you can do it manually as you did, or use a routine like that: How to correctly escape command line arguments in c#

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More examples on how to start a process:

• How do I start a process from C#?
• Wait till a process ends
• Passing quotes in Process.Start arguments
• MSDN: Process.Start Method
• Microsoft: How to wait for a shelled application to finish by using Visual C# (also has an example on how to wait for process to complete with timeout)