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Open File button Code help plz

SepticZaps

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Joined: Aug, 2018

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Posted

Hey can anyone help me with the open file button code please.

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Grahicaf

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Joined: Apr, 2018

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            // Displays an OpenFileDialog so the user can select a LUA. Â 
            OpenFileDialog open = new OpenFileDialog();
            open.Filter = "Text Files|*.txt|Lua Files|*.lua|Lua-C Files|*.luac";
            open.Title = "Select a script";

            // Show the Dialog. Â 
            // If the user clicked OK in the dialog and   
            if (open.ShowDialog() == DialogResult.OK)
            {
REPLACETHISWITHYOURTEXTBOXNAME.Text = File.ReadAllText(open.FileName)
    }

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