OpenFile(String,FileAccess) Метод (FileManager)

Vintasoft.Data Namespace / FileManager Class / OpenFile(String,FileAccess) Method

Синтаксис Exceptions Ремарки Требования Смотрите также

В этом разделе

OpenFile(String,FileAccess) Метод (FileManager)

В этом разделе

Opens an existing file with the specified access mode.

Синтаксис

'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

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
)

Parameters

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

Return Value

File stream.

Исключения

Исключение	Описание
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.

Ремарки

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

Требования

Целевые платформы: .NET 10; .NET 9; .NET 8; .NET 7; .NET 6; .NET Framework 4.8, 4.7, 4.6, 4.5, 4.0, 3.5

Смотрите также

Справочник

FileManager Class
Члены типа FileManager

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