Input/Output 2: OpenFileDialog
Written by: maffelu , 2009-04-16 21:23:50
The Basics
Sometimes you want to change textfiles in a program. This tutorial will create a program that opens a textfile and let you alter it.
It will look like this:
A user will be able to import a textfile and alter it.
A little list of the three controls involved:
- txtMain = The textbox that holds the text
- cmdBrowse = The button that says browse
- cmdAlterFile = the button that says Alter file
Getting into it
This is what the code looks like:
using System;
using System.Windows.Forms;
using System.IO;
namespace InputOutput
{
public partial class MainForm : Form
{
//We'll use this path later on to hold the current path of the file
public string ourPath;
public MainForm()
{
InitializeComponent();
}
void CmdBrowseClick(object sender, EventArgs e)
{
//Create a OpenFileDialog instance to use
OpenFileDialog openDialog = new OpenFileDialog();
//We select what types are permitted to open
openDialog.Filter = "Text Files (*.txt)|*.txt";
//A title for the dialogbox
openDialog.Title = "Open textfile";
//If the user press cancel then end this void
if (openDialog.ShowDialog() == DialogResult.Cancel) {
return;
}
//Create a filestream
FileStream fStr;
try {
//Set filestream to the result of the pick of the user
fStr = new FileStream(openDialog.FileName, FileMode.Open, FileAccess.Read);
//Create a streamreader, sr, to read the file
StreamReader sr = new StreamReader(fStr);
//While the end of the file has not been reached...
while (sr.Peek() >= 0) {
//Create a 'line' that contains the current line of the textfile
string line = sr.ReadLine();
//Enter the text into the main textbox together with a new line
txtMain.Text += line + Environment.NewLine;
}
//Close the file so other modules can access it
sr.Close();
//Set public variable ourPath to the current path
ourPath = Path.GetDirectoryName(openDialog.FileName);
ourPath += Path.GetFileName(openDialog.FileName);
//If something goes wrong, tell the user
} catch (Exception) {
MessageBox.Show("Error opening file", "Error message");
}
}
void CmdAlterFileClick(object sender, EventArgs e)
{
//Open a streamwriter using public string ourPath
StreamWriter sw = new StreamWriter(ourPath);
//Write the textbox values into the textfile
sw.Write(txtMain.Text);
//Close the textfile
sw.Close();
}
}
}
Allright, so the important things here are:
Line 12: We create a public variable where we hold the path to the file we've opened. The string variable gets data from line 52-53.
Line 37: We create a filestream that handles the selected file. It opens a read the file so that we can enter the data into the textbox.
lines 40-49: We open the filestream with a streamreader and read the file into the textbox.
lines 61-70: We re-write the textfile to what's in the textbox.
So, what we've done here is created a program that can open and edit textfiles, sort of like a basic notepad.
Advanced
These examples shows you how to open a filestream for reading, writing and creating files:
Open an existing file for read and write:
FileStream fileStream = new FileStream(@"c:\file.txt", FileMode.Open);
Open existing file for reading:
FileStream fileStream = new FileStream(@"c:\file.txt", FileMode.Open, FileAccess.Read);
Open existing file for writing:
FileStream fileStream = new FileStream(@"c:\file.txt", FileMode.Open, FileAccess.Write);
Open file for writing(appending) and, if the file doesn't exist, creates it:
FileStream fileStream = new FileStream(@"c:\file.txt", FileMode.Append);
Create a new file and open it for read and write and if the file exists, overwrites it:
FileStream fileStream = new FileStream(@"c:\file.txt", FileMode.Append);
Create a new file and open it for read and write. Throws an exception if the file exists:
FileStream fileStream = new FileStream(@"c:\file.txt", FileMode.CreateNew);
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Maffelu
vignesh
2011-10-29 06:02:30
hi
If i open a noteped file using opendialogbox means,how to open as it as notepad...pls help me