Open file with associated application

Question

i want to ask for help with opening a file from c# app with associated app. I tried this:

      ProcessStartInfo pi = new ProcessStartInfo(file);
      pi.Arguments = Path.GetFileName(file);
      pi.UseShellExecute = true;
      pi.WorkingDirectory = Path.GetDirectoryName(file);
      pi.FileName = file;
      pi.Verb = "OPEN";
      Process.Start(pi);

or this:

      Process.Start(file);

where string file in both examples represents full path to the file trying to open. Now, everything is working well, except the (jpg) images with ACDSee app. Irfanview associations works well, MS office documents too. After trying to open the jpg image associated with acdsee it just runs the acdsee in the notification area and does not open the file.

I discovered, that in the registry CLASSES_ROOT for *.jpg images, there is an ACDSee.JPG value as associated app, and under this key there is in the shell->Open->Command a path:

"C:\Program Files\ACD Systems\ACDSee\ACDSee.exe" /dde

and I thing that this weird /dde is the reason, why i cannot open the file. I realized that in the same reg key shell->Open there is some DDEExec key entry with value [open("%1")]

For Irfan view or other checked app there is not a ddeexec, just the normal command like

"C:\Program Files (x86)\IrfanView\i_view32.exe" "%1"

that can be run from command line after swaping the %1 for file name, but I could not run the command from acdsee entry in the command line :(

So my question is, how can I set up the ProcessStartInfo object to ensure that it will run all the files as it would be in the explorer by doubleclick, the standards and this DDEExec ones? Is there something other like DDEExec that I shoul be aware of? thanks and sorry for my EN

Answer 1

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.

Answer 2

This is an old thread but just in case anyone comes across it like I did. pi.FileName needs to be set to the file name (and possibly full path to file ) of the executable you want to use to open your file. The below code works for me to open a video file with VLC.

var path = files[currentIndex].fileName;
var pi = new ProcessStartInfo(path)
{
    Arguments = Path.GetFileName(path),
    UseShellExecute = true,
    WorkingDirectory = Path.GetDirectoryName(path),
    FileName = "C:\\Program Files (x86)\\VideoLAN\\VLC\\vlc.exe",
    Verb = "OPEN"
};
Process.Start(pi)

Tigran's answer works but will use windows' default application to open your file, so using ProcessStartInfo may be useful if you want to open the file with an application that is not the default.

Answer 3

In .Net Core (as of v2.2) it should be:

new Process
{
    StartInfo = new ProcessStartInfo(@"file path")
    {
        UseShellExecute = true
    }
}.Start();

Related github issue can be found here