open url in c#
var uri = "https://www.google.com";
var psi = new System.Diagnostics.ProcessStartInfo();
psi.UseShellExecute = true;
psi.FileName = uri;
System.Diagnostics.Process.Start(psi);open url in c#
var uri = "https://www.google.com";
var psi = new System.Diagnostics.ProcessStartInfo();
psi.UseShellExecute = true;
psi.FileName = uri;
System.Diagnostics.Process.Start(psi);Copyright © 2021 Codeinu
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