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im using this code that send only one image to 1 dialog i wan to send 2 image to 1 dialog :
C#
string fileName = @"" + Application.StartupPath + "\\Image\\" + "".ToString() + @"\" + @"\" + 2 + ".jpg";
var p = new Process();
p.StartInfo.FileName = fileName;
p.StartInfo.Verb = "Print";
p.Start();


What I have tried:

string fileName = @"" + Application.StartupPath + "\\Image\\" + "".ToString() + @"\" + @"\" + 2 + ".jpg";
var p = new Process();
p.StartInfo.FileName = fileName;
p.StartInfo.Verb = "Print";
p.Start();
Posted
Updated 11-May-16 13:36pm

1 solution

You're not sending anything to a "dialog". You're launching a separate process which takes command line arguments to tell it to print a file. Whether you can send multiple files to this process we have no idea. We don't know which application is registered to print .JPG files on your machine.
 
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