Open a File through Open File Dialog in any Event using C#

Hi All,

Here i am given a code snippet for to Open a file in any event with based on that file format.

// Here i am explain in Button Click Event

  Private void button_Click(object sender, EventArgs e)
  {
     OpenFileDialog opObject=new OpenFileDialog();
     opObject.Filter="All files (*.*)| *.* ";
     if(opObject.ShowDialog() ==DialogResult.OK)
     {
       System.Diagnostics.Process processObj = new System.Diagnostics.Process();
       processObj.StartInfo.UseShellExecute = true;
       processObj.StartInfo.FileName = opObject.FileName;  \\ Ex: C:\\Test\\Sample.xls
       processObj.Start();  
     } 
  }

If you have a file name in text file means (For Example in Text File have C:\Test\Sample.xls)

Private void button_Click(object sender, EventArgs e)
  {
     OpenFileDialog opObject=new OpenFileDialog();
     opObject.Filter="(*.txt)| *.txt ";
     if(opObject.ShowDialog() ==DialogResult.OK)
     {
       string sFilePath = System.IO.File.ReadAllText(opObject.FileName);  \\ Its working only single line in Text File else use Split function in String
       System.Diagnostics.Process processObj = new System.Diagnostics.Process();
       processObj.StartInfo.UseShellExecute = true;
       processObj.StartInfo.FileName = sFilePath;
       processObj.Start();  
     } 

}

Cheers :)