OpenFile(String,FileAccess) Method (FileManager)

Opens an existing file with the specified access mode.

Syntax

public static System.IO.FileStream OpenFile(
   System.String filePath, 
   System.IO.FileAccess fileAccess
)

public: static System.IO.FileStream* OpenFile(
   System.String filePath, 
   System.IO.FileAccess fileAccess
)

public:
static System.IO.FileStream^ OpenFile(
   System.String filePath, 
   System.IO.FileAccess fileAccess
)

'Declaration

Public Shared Function OpenFile( _
   ByVal filePathAn absolute path to a file.
 As System.String, _
   ByVal fileAccessA constant that defines how the file can be accessed.
 As System.IO.FileAccess _
) As System.IO.FileStream

Parameters

filePath: An absolute path to a file.
fileAccess: A constant that defines how the file can be accessed.

Return Value

File stream.

Exceptions

Exception	Description
System.IO.FileNotFoundException	Thrown if file must be opened but file is not found.
System.InvalidOperationException	Thrown if file must be created but file exists already.
System.TimeoutException	Thrown if file is locked.

Remarks

The method will create new file if fileAccess is set to System.IO.FileAccess.Write; otherwise, the method will open a file.

Requirements

Target Platforms: .NET9; .NET 8; .NET 7; .NET 6; .NET Framework 4.8, 4.7, 4.6, 4.5, 4.0, 3.5

Reference

FileManager Class
FileManager Members

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