SaveFileDialog.OpenFile Method ()


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Opens the file with read/write permission selected by the user.

Namespace:   System.Windows.Forms
Assembly:  System.Windows.Forms (in System.Windows.Forms.dll)

public Stream OpenFile()

Return Value

Type: System.IO.Stream

The read/write file selected by the user.


For security purposes, this method creates a new file with the selected name and opens it with read/write permissions. This can cause unintentional loss of data if you select an existing file to save to. To save data to an existing file while retaining existing data, use the File class to open the file using the file name returned in the FileName property.

The following code example illustrates creating a SaveFileDialog, setting members, calling the dialog box using the ShowDialog method, and opening the selected file. The example requires a form with a button placed on it.

private void button1_Click(object sender, System.EventArgs e)
     Stream myStream ;
     SaveFileDialog saveFileDialog1 = new SaveFileDialog();

     saveFileDialog1.Filter = "txt files (*.txt)|*.txt|All files (*.*)|*.*"  ;
     saveFileDialog1.FilterIndex = 2 ;
     saveFileDialog1.RestoreDirectory = true ;

     if(saveFileDialog1.ShowDialog() == DialogResult.OK)
         if((myStream = saveFileDialog1.OpenFile()) != null)
             // Code to write the stream goes here.


to save a file. Associated enumeration: FileDialogPermissionAccess.Save.

.NET Framework
Available since 1.1
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