Excel If Value Is Less Than Or Equal To Formula Opening and Reading a Text File Using the OpenFileDialog Form in C#.NET

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Opening and Reading a Text File Using the OpenFileDialog Form in C#.NET

The OpenFileDialog object communicates with the computer’s Application Programming Interface (API) to present the user with available files and return the user’s file selection to the program. This object is part of the System.Windows.Forms library, so no additional usage reference is needed.

This dialog can be customized to display the file type, starting directory, and the title displayed in the dialog itself.

If you limit the file type to a .txt extension, as we did in this sample code, only those specific file types are visible to the user, although they have the option to select All files

as well.

Let’s look at the function below (the button click event in this example) that is used to interact with this OpenFileDialog.

private void btnOpenTextFile_Click ( object sender , EventArgs e )

First, declare a variable that holds the user’s file selection.

String input = string.Empty;

Since OpenFileDialog is an object, we create a new instance by declaring a variable with data type OpenFileDialog and setting it equal to the new instance.

OpenFileDialog dialog = new OpenFileDialog();

Now we specify the file type we want to make available to the user. In this case, text files.

dialog.Filter =

“txt files (*.txt)|*.txt|All files (*.*)|*.*”;

Next, the starting directory of the dialog and the title of the dialog box are set.

dialog.InitialDirectory = “C:”;

dialog.Title = “Select text file”;

Once the dialog properties are set, it is ready to be presented to the user.

if (dialog.ShowDialog() == DialogResult.OK)

strFileName = dialog.FileName;

if (strFileName == String.Empty)

return;//user did not select a file to open

If the user selected a file, the DialogResult property is set to “OK” and we also have the file name and path of that file.

Now we can use StreamWriter and StreamReader to read/write text file.

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