[C#] How to use File.Open Method

Byantonhyip

Mar 3, 2019 #c# file open dialog, #c# launch file, #c# open file from path, #c# open file with default program, #filemode c#, #filestream c#, #how to open a text file in c#, #read file c#

Chia sẻ

The Open method opens a FileStream on the specified file in the specified file mode.

FileStream fs = fi.Open(FileMode.Open, FileAccess.Write);

Here is the complete code sample.

string fileName = @"C:fileNametxt";
FileInfo fi = new FileInfo(fileName);
// If file does not exist, create file
if (!fi.Exists)
{
	//Create the file.
	using (FileStream fs = fi.Create())
	{
		Byte[] info = new UTF8Encoding(true).GetBytes("File Start");
		fs.Write(info, 0, info.Length);
	}
}
try
{
	using (FileStream fs = fi.Open(FileMode.Open, FileAccess.Write))
	{
		Byte[] info = new UTF8Encoding(true).GetBytes("Add more text");
		fs.Write(info, 0, info.Length);
	}
}
catch (Exception Ex)
{
	Console.WriteLine(Ex.ToString());
}

NOTE:
A File must be opened using an IO resource before it can be read or write to. A file can be opened to read and/or write purpose. The FileInfo class provides four methods to open a file.

+ Open
+ OpenRead
+ OpenText
+ OpenWrite