C# File Open(String, FileMode, FileAccess)
Description
File Open(String, FileMode, FileAccess) Opens a FileStream
on the specified path, with the specified mode and access.
Syntax
File.Open(String, FileMode, FileAccess) has the following syntax.
public static FileStream Open(
string path,/* w w w. ja va 2 s. co m*/
FileMode mode,
FileAccess access
)
Parameters
File.Open(String, FileMode, FileAccess) has the following parameters.
path- The file to open.mode- A FileMode value that specifies whether a file is created if one does not exist, and determines whether the contents of existing files are retained or overwritten.access- A FileAccess value that specifies the operations that can be performed on the file.
Returns
File.Open(String, FileMode, FileAccess) method returns An unshared FileStream that provides access to the specified file, with
the specified mode and access.
Example
The following example opens a file with read-only access.
/* w w w . j a v a2s . c om*/
using System;
using System.IO;
using System.Text;
class Test
{
public static void Main()
{
string filePath = @"c:\temp\MyTest.txt";
using (FileStream fs = File.Create(filePath))
{
Byte[] info = new UTF8Encoding(true).GetBytes("from java2s.com.");
fs.Write(info, 0, info.Length);
}
using (FileStream fs = File.Open(filePath, FileMode.Open, FileAccess.Read))
{
byte[] b = new byte[1024];
UTF8Encoding temp = new UTF8Encoding(true);
while (fs.Read(b,0,b.Length) > 0)
{
Console.WriteLine(temp.GetString(b));
}
try
{
fs.Write(b,0,b.Length);
}
catch (Exception e)
{
Console.WriteLine("Writing was disallowed, as expected: {0}", e.ToString());
}
}
}
}