Open with in c# windows forms

Question

Last night I was created a windows from application it works is similar to notepad and I want to make it can open files from windows openwith dialog how to create it?
[In c#]

Answer 1

The Open With command just passes the path of the file as the first argument to the application so all you need to do is

public static void Main(string[] args)
{
if(args[0] != null)
{
//args[0] contans a path to the file do whatever you need to do to display it
}
else
{
//Start normally
}
}

To automaticly put your program in the open with list you will need to add some reg keys in HKEY_CLASSES_ROOT\YOUR_EXT\ or use advanced installer. This app helps you to add your program to open with menu

Answer 2

Find program.cs in solution explorer. You can find Main method. Add string[] args as parameter,
Use try{ if(args[0] != null)} catch{}

Answer 3

Opendialog is a standard component.

Have a look at these links.

https://docs.microsoft.com/en-us/dotnet/api/system.windows.forms.openfiledialog?view=netcore-3.1

https://www.c-sharpcorner.com/UploadFile/mahesh/openfiledialog-in-C-Sharp/

https://www.dotnetperls.com/openfiledialog

Answer 4

hossein B how to do it in windows forms?
I think we can do it in main.cs file
//I and you are from one
//country [Iran]

Answer 5

hossein B ok thanks

Answer 6

sneeze I know it so I want to do it on the windows explorer
[If you right click on a txt file and select [open with>] you will be able to open a file with multiple programs ] I want to insert my app into that list
[File assignment]