Open File with Associated Application

Open file with associated application

Just write

System.Diagnostics.Process.Start(@"file path");

example

System.Diagnostics.Process.Start(@"C:\foo.jpg");
System.Diagnostics.Process.Start(@"C:\foo.doc");
System.Diagnostics.Process.Start(@"C:\foo.dxf");
...

And shell will run associated program reading it from the registry, like usual double click does.

c# open file with default application and parameters

If you want the file to be opened with the default application, I mean without specifying Acrobat or Reader, you can't open the file in the specified page.

On the other hand, if you are Ok with specifying Acrobat or Reader, keep reading:

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You can do it without telling the full Acrobat path, like this:

using Process myProcess = new Process();    
myProcess.StartInfo.FileName = "acroRd32.exe"; //not the full application path
myProcess.StartInfo.Arguments = "/A \"page=2=OpenActions\" C:\\example.pdf";
myProcess.Start();

If you don't want the pdf to open with Reader but with Acrobat, chage the second line like this:

myProcess.StartInfo.FileName = "Acrobat.exe";

You can query the registry to identify the default application to open pdf files and then define FileName on your process's StartInfo accordingly.

Follow this question for details on doing that: Finding the default application for opening a particular file type on Windows

How to open a file with the default associated program

You can use Desktop.getDesktop().open(File file). See the following question for other options: "[Java] How to open user system preffered editor for given file?"

C# Open File With Associated Application passing arguments

I believe this is expected. Behind the scenes, Windows is finding the default application in the registry and creating a new process and passing your file name to it. I get the same behavior if I go to a command prompt and type "filename.ext argument", that my arguments are not passed to the application.

What you probably need to do is find the default application yourself by looking in the registry. Then you can start that process with arguments, instead of trying to start by filetype association. There is an answer here on how to find the default application in the registry:

Finding the default application for opening a particular file type on Windows

How can you open a file with the program associated with its file extension?

You want to use the file to open as the file argument, not the parameter argument. No need to specify which program to use, ShellExecute will look it up for you.

ShellExecute(0, 0, L"c:\\outfile.txt", 0, 0 , SW_SHOW );

By leaving the verb as NULL (0) rather than L"open", you get the true default action for the file type - usually this is open but not always.

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