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How to use openfiledialog to open any file as text in C#?

I am writing a winforms program in C# that uses the openfiledialog. I would like it to be able to take the file that the user selected and open it as text, regardless of the file type.
I tried it like this:
if (openFileDialog1.ShowDialog() == DialogResult.OK)
{
textBox1.Text = Process.Start("notepad.exe", openFileDialog1.ToString()).ToString();
}
However, that didn't work and I'm not sure if I"m even on the right track.

You should use this code:
First add this namespace :
using System.IO;
Then add this codes to your function:
OpenFileDialog openFileDialog = new OpenFileDialog();
openFileDialog.Filter = "txt files (*.txt)|*.txt|All files (*.*)|*.*";
if (openFileDialog.ShowDialog()== DialogResult.OK)
{
textBox1.Text = File.ReadAllText(openFileDialog.FileName);
}

To open the file using notepad, you need to pass the file name as second parameter of Start method. For example:
using (var ofd = new OpenFileDialog())
{
if(ofd.ShowDialog()== DialogResult.OK)
{
System.Diagnostics.Process.Start("notepad.exe", ofd.FileName);
}
}
Also if for any reason while knowing that not all file contents are text, you are going to read the file content yourself:
using (var ofd = new OpenFileDialog())
{
if(ofd.ShowDialog()== DialogResult.OK)
{
var txt = System.IO.File.ReadAllText(ofd.FileName);
}
}

What you are doing at the moment is starting a Process with the argument openFileDialog1.ToString(), calling ToString() on the process and setting this as text in the TextBox. If the path was valid, the result would probably be something like "System.Diagnostics.Process". But since you use openFileDialog1.ToString() as a path, your application probably crashes with a file not found error.
To get the selected file of an OpenFileDialog, use openFileDialog1.FileName. (See the docs here)
What I think you actually want to do, is read from the file and write its contents as text to the TextBox. To do this, you need a StreamReader, like so:
if (openFileDialog1.ShowDialog() == DialogResult.OK)
{
using(var reader = new StreamReader(openFileDialog1.FileName))
{
textBox1.Text = reader.ReadToEnd();
}
}
This way, you open the file with the StreamReader, read its contents and then assign them to the text box.
The using statement is there because a StreamReader needs to be disposed of after you're done with it so that the file is no longer in use and all resources are released. The using statement does this for you automatically.

How can I write and read a text doucument in a folder?

string curetn = Environment.CurrentDirectory;
string path = curetn.ToString() + #"\DATA\SaveGame.txt";
Console.WriteLine(path);
TextReader tr = new StreamReader(path);
Hello, I am making a text-adventure, and I do not like having all my save files, and mp3 file in the same place as my application. I would like for the files to be in a folder. I want to be able to use StreamWriter and StreamReader, to be able to write and read files that are in a folder. This file is also in a distributable folder, not just in the Visual Studios Projects folders. I have tried everything I can, and this is what I have. I also have one of these for StreamWriter. Please help!
Edit:
The thing that does not work, is that it does not read the lines, and assigns them to a variable. I have it in a try-catch, and it catches, and displays the error message that I wrote.

If you are looking for simply read and write lines from file you can try this
using (StreamReader sr = new StreamReader(path))
{
while (!sr.EndOfStream)
{
sr.ReadLine();
}
}
string s;
using (StreamWriter sw = new StreamWriter(path))
{
sw.WriteLine(s);
}

So basically what you want to do is read the text file:
string data[] = File.ReadAllLines(path); // Read the text file.
var x = data[1]; // Replace the '1' with the line number you want.
Console.WriteLine(x);
This is a good way to read the text file, I think it's better than opening a stream.
You can also write to it, so every time you want to save, just do this:
// When you want to write:
File.WriteAllText(path, "");
File.AppendAllText(path, "Add a data line" + Environment.NewLine); // Environment.NewLine adds a line.
Keep appending text to the file for the data you need.

automate openFileDialog process in c#

I have tried this code it would open the file dialog to the correct location and there is only one xml file which needs to selected ( where i need to select it and click on open ) instead of selecting the file and click open to process the file is there any way to disable the open button on open file dialog. Here my xml file changes everyday. i have given *.xml but gives me a error Illegal characters in path.. my file format is this.
lborough vehicles_in 2014-06-05.xml == this changes everyday according to date.
Without clicking on open how to select the file.
OpenFileDialog dialog = new OpenFileDialog();
dialog.Filter = "XML Files (*.xml)|*.xml";
string initPath = Path.GetFullPath("C:/Users/IT-Administrator/Desktop/LUVS/");
dialog.InitialDirectory = Path.GetFullPath(initPath);
tblVehicles = new DataTable();
dv = new DataView(tblVehicles);
if (dialog.ShowDialog() == DialogResult.OK)
{
if (dialog.FileName.Length > 0)
{
//Load Schema and Vehicle_In XML file
tblVehicles.ReadXmlSchema(Path.Combine(applicationFolder, "vehicles_in.xsd"));
tblVehicles.ReadXml(dialog.FileName);
this.dataGridView1.DataSource = tblVehicles;
this.dataGridView1.AllowUserToAddRows = false;
this.dataGridView1.ReadOnly = true;
**Update**
I have tried this can you tell me how to open the file from Directory.get files at runtime
string[] filePaths = Directory.GetFiles(#"C:\Users\IT-Administrator\Desktop\LUVS/", "*.xml", SearchOption.AllDirectories);
FileStream stream = File.Open(#"C:\Users\IT-Administrator\Desktop\LUVS*.xml",
FileMode.Open);
tblVehicles = new DataTable();
dv = new DataView(tblVehicles);
tblVehicles.ReadXmlSchema(Path.Combine(applicationFolder, "vehicles_in.xsd"));
tblVehicles.ReadXml(stream);

Your attempted solution at the end doesn't quite get it:
/* Gives you an array of file names */
string[] filePaths = Directory.GetFiles(#"C:\Users\IT-Administrator\Desktop\LUVS/", "*.xml", SearchOption.AllDirectories);
FileStream stream = File.Open(#"C:\Users\IT-Administrator\Desktop\LUVS*.xml",
FileMode.Open);
You aren't using the array, but instead just trying to open a wildcard path; You can't do that. File.Open only accepts a single file path.
Instead, try something more like this:
/* Gives you an array of file names */
string[] filePaths = Directory.GetFiles(#"C:\Users\IT-Administrator\Desktop\LUVS/", "*.xml", SearchOption.AllDirectories);
// Work with each file individually
foreach(var filePath in filePaths)
{
using(FileStream stream = File.Open(filePath, FileMode.Open))
{
tblVehicles = new DataTable();
dv = new DataView(tblVehicles);
tblVehicles.ReadXmlSchema(Path.Combine(applicationFolder, "vehicles_in.xsd"));
tblVehicles.ReadXml(stream);
// Do whatever you need to do with the data from this one file, then move on....
{
}

Is there any reason that you can't use Directory.GetFiles to get all files in a directory and use File.Open to get the file? Why do you want to do this with a FileDialog, if you don't want the FileDialog?
Update:
//Load Schema and Vehicle_In XML file
tblVehicles.ReadXmlSchema(Path.Combine(applicationFolder, "vehicles_in.xsd"));
// Get all XML files from the files directory
string[] filePaths = Directory.GetFiles(#"files\", "*.xml", SearchOption.AllDirectories);
// Read the first XML file in the files directory
tblVehicles.ReadXml(filePaths[0]);
Is this what you asked for?

As you want to select your file and inmediately work with it, you want a button behaviour. Best way here is to make your own UserControl showing files existing on your directory.
1. Get files from directory
2. Show dialog with buttons, every buttons asociated to its file.
3. On button click, close dialog and pass file to your method.

You could use SendKeys() but it's clunky, and if the user moves focus elsewhere you might end up sending the keystrokes to the wrong window.
openFileDialog isn't very customizable, so you might want to consider using openFolderDialog and append the known filename to the user selected directory.

Writing a Text File in memory and saving it with savefiledialog

I am trying to make a text file in memory, add some lines to it and at the end save the file in a text file. I can handle the savedialog part but I dont know how to get the text file from memory. Any help and tips will be appriciated.
What I am doing so far is:
//Initialize in memory text writer
MemoryStream ms = new MemoryStream();
TextWriter tw = new StreamWriter(ms);
tw.WriteLine("HELLO WORLD!");
tw.WriteLine("I WANT TO SAVE THIS FILE AS A .TXT FILE!);
please note
I will call tw.WriteLine() add more lines in different places so I want to save this at end of program (so this shouldent be wrapped between something like using{} )
UPDATE
StringBuilder seems to be a more reliable option for doing this! I get strange cut-outs in my text file when I do it using MemoryStream.
Thanks.

I think your best option here would be to write to a StringBuilder, and when done, File.WriteAllText. If the contents are large, you might consider writing directly to the file in the first place (via File.CreateText(path)), but for small-to-medium files this should be fine.
var sb = new StringBuilder();
sb.AppendLine("HELLO WORLD!");
sb.AppendLine("I WANT TO SAVE THIS FILE AS A .TXT FILE!");
File.WriteAllText(path, sb.ToString());

Or, something nigh-on the same as #Marc's answer, but different enough that I think it's worth putting out there as a valid solution:
using (var writer = new StringWriter())
{
writer.WriteLine("HELLO WORLD!");
writer.WriteLine("I WANT TO SAVE THIS FILE AS A .TXT FILE!");
File.WriteAllLines(path, writer.GetStringBuilder().ToString());
}
Where path is a string representing a valid file system entry path, predefined by you somewhere in the application.

Assume your SaveFileDialog name is "dialog"
File.WriteAllBytes(dialog.FileName, Encoding.UTF8.GetBytes("Your string"));
or
var text = "Your string";
text += "some other text";
File.WriteAllText(dialog.FileName, text);
also in your own solution you can do this :
MemoryStream ms = new MemoryStream();
TextWriter tw = new StreamWriter(ms);
tw.WriteLine("HELLO WORLD!");
tw.WriteLine("I WANT TO SAVE THIS FILE AS A .TXT FILE!);
// just add this
File.WriteAllBytes(dialog.FileName, ms.GetBuffer());

Something like this.
Microsoft.Win32.SaveFileDialog dlg = new Microsoft.Win32.SaveFileDialog();
dlg.FileName = "Document"; // Default file name
dlg.DefaultExt = ".text"; // Default file extension
dlg.Filter = "Text documents (.txt)|*.txt"; // Filter files by extension
// Show save file dialog box
Nullable<bool> result = dlg.ShowDialog();
// Process save file dialog box results
if (result == true)
{
// Save document
using (FileStream file = File.CreateText(dlg.FileName)
{
ms.WriteTo(file)
}
}
I haven't worried about whether the file already exists but this should get you close.
You might need a ms.Seek(SeekOrgin.Begin, 0) too.

Another way of appending text to the end of a file could be:
if (saveFileDialog.ShowDialog() == DialogResult.OK) {
using (var writer = new StreamWriter(saveFileDialog.Filename, true)) {
writer.WriteLine(text);
}
}
supposing that text is the string you need to save into your file.
If you want to append new lines to that string in an easy way, you can do:
var sb = new StringBuilder();
sb.AppendLine("Line 1");
sb.AppendLine("Line 2");
and the resulting string will be sb.ToString()
If you already have a Stream object (in your example, a MemoryStream), you can do the same but replace the line:
using (var writer = new StreamWriter(saveFileDialog.Filename, true)) {
by
using (var writer = new StreamWriter(memoryStream)) {
Edit:
About wrapping the statements inside using:
Take in count that this is not a problem at all. In my first example, all you will have to do is to keep that StringBuilder object, and keep adding lines to it. Once you have what you want, just write the data into a text file.
If you are planning to write more than once to the text file, just clear the StringBuilder everytime you write, in order to not get duplicated data.

Formatting a text file, how to update the file after I finished parsing it?

How would I open a file, perform some regex on the file, and then save the file?
I know I can open a file, read line by line, but how would I update the actual contents of a file and then save the file?

The following approach would work regardless of file size, and will also not corrupt the original file in anyway if the operation would fail before it is complete:
string inputFile = Path.Combine(Environment.GetFolderPath(
Environment.SpecialFolder.MyDocuments), "temp.txt");
string outputFile = Path.Combine(Environment.GetFolderPath(
Environment.SpecialFolder.MyDocuments), "temp2.txt");
using (StreamReader input = File.OpenText(inputFile))
using (Stream output = File.OpenWrite(outputFile))
using (StreamWriter writer = new StreamWriter(output))
{
while (!input.EndOfStream)
{
// read line
string line = input.ReadLine();
// process line in some way
// write the file to temp file
writer.WriteLine(line);
}
}
File.Delete(inputFile); // delete original file
File.Move(outputFile, inputFile); // rename temp file to original file name

string[] lines = File.ReadAllLines(path);
string[] transformedLines = lines.Select(s => Transform(s)).ToArray();
File.WriteAllLines(path, transformedLines);
Here, for example, Transform is
public static string Transform(string s) {
return s.Substring(0, 1) + Char.ToUpper(s[1]) + s.Substring(2);
}

Open the file for read. Read all the contents of the file into memory. Close the file. Open the file for write. Write all contents to the file. Close the file.
Alternatively, if the file is very large:
Open fileA for read. Open a new file (fileB) for write. Process each line of fileA and save to fileB. Close fileA. Close fileB. Delete fileA. Rename fileB to fileA.

Close the file after you finish reading it
Reopen the file for write
Write back the new contents