FileSystem.FileOpen Method


Opens a file for input or output. The My feature gives you better productivity and performance in file I/O operations than FileOpen. For more information, see FileSystem.

public static void FileOpen (int FileNumber, string FileName, Microsoft.VisualBasic.OpenMode Mode, Microsoft.VisualBasic.OpenAccess Access = Microsoft.VisualBasic.OpenAccess.Default, Microsoft.VisualBasic.OpenShare Share = Microsoft.VisualBasic.OpenShare.Default, int RecordLength = -1);
static member FileOpen : int * string * Microsoft.VisualBasic.OpenMode * Microsoft.VisualBasic.OpenAccess * Microsoft.VisualBasic.OpenShare * int -> unit
Public Sub FileOpen (FileNumber As Integer, FileName As String, Mode As OpenMode, Optional Access As OpenAccess = Microsoft.VisualBasic.OpenAccess.Default, Optional Share As OpenShare = Microsoft.VisualBasic.OpenShare.Default, Optional RecordLength As Integer = -1)



Required. Any valid file number. Use the FreeFile function to obtain the next available file number.


Required. A string expression that specifies a file name - may include directory or folder, and drive.


Required. Enumeration specifying the file mode: Append, Binary, Input, Output, or Random. For more information, see OpenMode .


Optional. Enumeration specifying the operations permitted on the open file: Read, Write, or ReadWrite. Defaults to ReadWrite. For more information, see OpenAccess .


Optional. Enumeration specifying the operations not permitted on the open file by other processes: Shared, Lock Read, Lock Write, and Lock Read Write. Defaults to Lock Read Write. For more information, see OpenShare .


Optional. Number less than or equal to 32,767 (bytes). For files opened for random access, this value is the record length. For sequential files, this value is the number of characters buffered.


Record length is negative (and not equal to -1).

FileName is already open, or FileName is invalid.


This example illustrates various uses of the FileOpen function to enable input and output to a file.

The following code opens the file TestFile in Input mode.

FileOpen(1, "TESTFILE", OpenMode.Input)
' Close before reopening in another mode.

This example opens the file in Binary mode for writing operations only.

FileOpen(1, "TESTFILE", OpenMode.Binary, OpenAccess.Write)
' Close before reopening in another mode.

The following example opens the file in Random mode. The file contains records of the structure Person.

Structure Person
    <VBFixedString(30)> Dim Name As String
    Dim ID As Integer
End Structure
Public Sub ExampleMethod()
    ' Count 30 for the string, plus 4 for the integer.
    FileOpen(1, "TESTFILE", OpenMode.Random, , , 34)
    ' Close before reopening in another mode.
End Sub

This code example opens the file in Output mode; any process can read or write to file.

FileOpen(1, "TESTFILE", OpenMode.Output, OpenAccess.Default, OpenShare.Shared)
' Close before reopening in another mode.

This code example opens the file in Binary mode for reading; other processes cannot read file.

FileOpen(1, "TESTFILE", OpenMode.Binary, OpenAccess.Read,


The FileOpen function is provided for backward compatibility and may affect performance. For non-legacy applications, the My.Computer.FileSystem object provides better performance. For more information, see File Access with Visual Basic.

You must open a file before any I/O operation can be performed on it. FileOpen allocates a buffer for I/O to the file and determines the mode of access to use with the buffer.


When writing to a file, an application may have to create a file, if the file to which it is trying to write does not exist. To do so, it needs permission for the directory in which the file is to be created. However, if the file specified by FileName does exist, the application needs Write permission only to the file itself. Wherever possible, to help improve security, create the file during deployment and grant Write permission to that file only, instead of to the whole directory. To help improve security, write data to user directories instead of to the root directory or the Program Files directory.

The channel to open can be found by using the FreeFile() function.


The FileOpen function requires Read access from the FileIOPermissionAccess enumeration, which may affect its execution in partial trust situations. For more information, see FileIOPermissionAccess enumeration.

Applies to

See also