use the button click event then type the following codes

opendialog dp= new;

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Use control flow to do your jobs. Like

If (dp.DialogResult=Windows.Forms.DialogResult.OK)
''''Do Your Jobs''''''

You can use it like so

if(openFileDialog1.ShowDialog() == DialogResult.OK)
    //use the OpenFileDialog as you like, you can get the
    //file name with "openFileDialog1.FileName;"    

That above assumes your OpenFileDialog is named "openFileDialog1". Just place that piece of code within the button's click event (that's usually how I do it)

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