OpenFileDialog.OpenFile Method

Definition

Opens the file selected by the user, with read-only permission. The file is specified by the FileName property.

public System.IO.Stream OpenFile ();

Returns

A Stream that specifies the read-only file selected by the user.

Exceptions

The file name is null.

An I/O error occurred while opening the file.

Examples

The following code example demonstrates how to use the OpenFile method.

var fileContent = string.Empty;
var filePath = string.Empty;

using (OpenFileDialog openFileDialog = new OpenFileDialog())
{
    openFileDialog.InitialDirectory = "c:\\";
    openFileDialog.Filter = "txt files (*.txt)|*.txt|All files (*.*)|*.*";
    openFileDialog.FilterIndex = 2;
    openFileDialog.RestoreDirectory = true;

    if (openFileDialog.ShowDialog() == DialogResult.OK)
    {
        //Get the path of specified file
        filePath = openFileDialog.FileName;

        //Read the contents of the file into a stream
        var fileStream = openFileDialog.OpenFile();

        using (StreamReader reader = new StreamReader(fileStream))
        {
            fileContent = reader.ReadToEnd();
        }
    }
}

MessageBox.Show(fileContent, "File Content at path: " + filePath, MessageBoxButtons.OK);

Remarks

The OpenFile method is used to provide a facility to quickly open a file from the dialog box. The file is opened in read-only mode for security purposes. To open a file in read/write mode, you must use another method, such as FileStream.

Applies to

Produk Versi
.NET Framework 1.1, 2.0, 3.0, 3.5, 4.0, 4.5, 4.5.1, 4.5.2, 4.6, 4.6.1, 4.6.2, 4.7, 4.7.1, 4.7.2, 4.8, 4.8.1
Windows Desktop 3.0, 3.1, 5, 6, 7, 8, 9

See also