archangelmihail

ProcessShow

Jan 11th, 2014
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C# 0.22 KB | None | 0 0
  1. using System.Diagnostics;
  2.  
  3. Process openFile = new Process();
  4.         openFile.StartInfo.FileName = fileName;
  5.         openFile.Start();
  6.  
  7.         openFile.StartInfo.FileName = resultFileName;
  8.         openFile.Start();
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