Develop Reference

If statement to filter file extensions (C#)

I want to proceed in creating a program that does different events with different file extensions, with files opened with . I understand things like Path.GetExtension, and these answers: Refactoring if-else statements checking for different file extensions
C# custom file extension filter
Can someone explain me a straight forward how to create a button "If Statement" meant at executing say, if a ".jpeg" file is detected as the selected file's extension, while "else" may bring up message box, like "Improper File type uploaded."
Here is cut of the code where action takes place:
private void BtnMediaPlayer_Click(object sender, EventArgs e)
{
// Open File Dialog
OpenFileDialog open = new OpenFileDialog();
// Image Filters
open.Filter = "Image Files(*.jpg; *.jpeg; *.gif; *.bmp)|*.jpg; *.jpeg; *.gif; *.bmp";
if (open.ShowDialog() == DialogResult.OK)
{
// display image in picture box
pictureBox1.Image = new Bitmap(open.FileName);
}
}
As can be seen, the program opens an Open File Dialog that then is supposed to filter for certain files, and then Open them within the pciture box, but how do I rewrite it so that, "if (open.ShowDialog() == DialogResult.OK)" or the surrounding code produces different results (executions based on the If Statements) based on the file type(s)?
TL DR
How to write If Statements for Different files uploaded?

private void BtnMediaPlayer_Click(object sender, EventArgs e)
{
var ofd = new OpenFileDialog();
ofd.Filter = OpenFileImageFilter;
if (ofd.ShowDialog() == DialogResult.OK)
{
if (IsValidImageFile(ofd.FileName))
{
// display image in picture box
pictureBox1.Image = new Bitmap(open.FileName);
}
else
{
MessageBox.Show("Invalid file type selected.");
}
}
}
// array of image formats so we can use it in OpenFileDialog filter
// and in our IsValidImageFile method
private string[] ImageFormats => new[] { ".jpg", ".jpeg", ".gif", ".bmp" };
// helper method to turn array of image extensions into OpenFileDialog filter string
private string OpenFileImageFilter
{
get
{
string fileExts = "";
foreach (string s in ImageFormats)
{
fileExts += $"*{s};";
}
return $"Image Files({fileExts})|{fileExts}";
}
}
// reusable method that uses the ImageFormats property to check file extension
private bool IsValidImageFile(string filename)
{
if (string.IsNullOrEmpty(filename)) return false;
string extension = Path.GetExtension(filename);
return ImageFormats.Contains(extension);
}

How to open a file in C# using FileOpenDialog

I am trying to open a file by pressing a button (a label to be more exact but it works just the same way)
For some reason when the FileDialog opens and I select the file and press open it doesnt open the file it only closes the FileDialog
private void selectLbl_Click(object sender, EventArgs e)
{
OpenFileDialog ofd = new OpenFileDialog();
ofd.InitialDirectory = "c:\\";
ofd.Filter = "Script files (*.au3)|*.au3";
ofd.RestoreDirectory = true;
ofd.Title = ("Select Your Helping Script");
if (ofd.ShowDialog() == DialogResult.OK)
{
ofd.OpenFile(); //Not sure if there is supposed to be more here
}
}

ofd.OpenFile();
is returning the content of the file as Stream of bytes like described here. If you want to open the file like you described it, use
if (ofd.ShowDialog() == DialogResult.OK)
{
System.Diagnostics.Process.Start(ofd.FileName);
}
So your selected file starts with the associated application.

ofd.OpenFile() opens the file selected by the user as a Stream that you can use to read from the file.
What you do with that stream depends on what you are trying to achieve. For example, you can read and output all lines:
if (ofd.ShowDialog() == DialogResult.OK)
{
using (TextReader reader = new StreamReader(ofd.OpenFile()))
{
string line;
while((line = t.ReadLine()) != null)
Console.WriteLine(line);
}
}
Or if it is an xml file you can parse it as xml:
if (ofd.ShowDialog() == DialogResult.OK)
{
using(XmlTextReader t = new XmlTextReader(ofd.OpenFile()))
{
while (t.Read())
Console.WriteLine($"{t.Name}: {t.Value}");
}
}

The OpenFileDialog is not a dialog that opens the file. It only communicates with the operator with a dialog that is commonly shown if a program needs to know what file to open. So it is only a dialog box, not a file opener.
You decide what to do if the user pressed ok or cancel:
private void selectLbl_click(object sender, ...)
{
using (OpenFileDialog ofd = new OpenFileDialog())
{
ofd.InitialDirectory = "c:\\";
ofd.Filter = "Script files (*.au3)|*.au3";
ofd.RestoreDirectory = true;
ofd.Title = ("Select Your Helping Script");
var dlgResult = ofd.ShowDialog(this);
if (dlgResult == DialogResult.OK)
{ // operator pressed OK, get the filename:
string fullFileName = ofd.FileName;
ProcessFile(fullFileName);
}
}
}
if (ofd.ShowDialog() == DialogResult.OK)
{
ofd.OpenFile(); //Not sure if there is supposed to be more here
}

how to open the content of a folder in to different rich text boxes?

I'm trying to open a file dialog and open the files inside a folder into different rich text boxes? but i'm not sure what else I would need to add? could you please help out a new blood.
if (openFileDialog1.ShowDialog() == System.Windows.Forms.DialogResult.OK)
{
richTextBox1.Text = File.ReadAllText(openFileDialog1.FileName);
tabPage1.Text = openFileDialog1.SafeFileName;
}

If you want to allow your user to select a folder and then open the first 5 files present in that folder each one in a different richtextbox then you don't need the OpenFileDialog, but the FolderBrowserDialog
// First prepare two list with the richtextboxes and the tabpages
List<RichTextBox> myBoxes = new List<RichTextBox>()
{ richTextBox1, richTextBox2, richTextBox3, richTextBox4, richTextBox5 };
List<TabPage> myPages = new List<TabPage>()
{ tabPage1, tabPage2, tabPage3, tabPage4, tabPage5};
// Now open the folderbrowser dialog
// (see link above for some of its properties)
FolderBrowserDialog fbd = new FolderBrowserDialog();
if(fbd.ShowDialog() == DialogResult.OK)
{
int i = 0;
foreach(string file in Directory.GetFiles(fbd.SelectedPath))
{
myBoxes[i].Text = File.ReadAllText(file);
myPages[i].Text = Path.GetFileName(file);
i++;
// Added a warning if the folder contains more than 5 files
if(i >= 5)
{
MessageBox.Show("Too many files in folder, only 5 loaded");
break;
}
}
}

OpenFileDialog is to open but one file by default. Try to change its MultiSelect property to true. Something like this will do:
openFileDialog1.Multiselect = true;
if (openFileDialog1.ShowDialog() == System.Windows.Forms.DialogResult.OK) {
for (int i = 0; i < openFileDialog1.FileNames.Length; ++i) {
RichTextBox rtb = Controls.Cast<Control>().Single(x => x.Name == "richTextBox" + (i + 1).ToString()) as RichTextBox;
rtb.Text = File.ReadAllText(openFileDialog1.FileNames[i]);
}
tabPage1.Text = openFileDialog1.SafeFileName; //again, I wonder what you want to do with this. If needed be, consider to update this dynamically too
}
Old answer:
openFileDialog1.Multiselect = true; //important: set this to true
richTextBox1.Text = ""; //and you may want to reset this every time
if (openFileDialog1.ShowDialog() == System.Windows.Forms.DialogResult.OK) {
foreach(var filename in openFileDialog1.FileNames) //get file names here
richTextBox1.Text += File.ReadAllText(filename); //you may want to add enter per file
tabPage1.Text = openFileDialog1.SafeFileName; //but I wonder what you want to do with this....?
}

display file instead of RSC version

Whenever I try to open a custom file to a textbox or something which will display code. it never works, I'm not sure what I am doing wrong.
I want my program to display what is inside the file when I open it, I have this below:
private void button1_Click(object sender, EventArgs e)
{
//Show Dialogue and get result
Stream myStream = null;
OpenFileDialog openFileDialog1 = new OpenFileDialog();
openFileDialog1.InitialDirectory = "c:\\";
openFileDialog1.Filter = "rbt files (*.rbt)|*.rbt|All files (*.*)|*.*";
openFileDialog1.FilterIndex = 2;
openFileDialog1.RestoreDirectory = true;
if (openFileDialog1.ShowDialog() == DialogResult.OK)
{
try
{
if ((myStream = openFileDialog1.OpenFile()) != null)
{
using (myStream)
{
File.WriteAllText("", CodeBox.Text);
}
}
}
catch (Exception ex)
{
MessageBox.Show("RBT7 file open");
}
}
}
It only displays the RBT7 in a messagebox which is not what I want, I want the file to open and display its information to some sort of textbox which displays code.

Please read the documentation for File.WriteAllText.
The first parameter:
path: The file to write to.
You're passing it "". That is not a path. Are you trying to write all the text from the file into CodeBox.Text or write all the text from CodeBox.Text into a file?
In your comment, you indicate the former. Try this:
string[] lines = System.IO.File.ReadAllLines(#"your file path");
foreach (string line in lines)
{
CodeBox.Text += line;
}
You haven't shown the code for CodeBox so I can't guarantee the results of this.

Try this:
Replace this code
if ((myStream = openFileDialog1.OpenFile()) != null)
{
using (myStream)
{
File.WriteAllText("", CodeBox.Text);
}
}
with this
{
CodeBox.Text = File.ReadAllText(openFileDialog1.FileName);
}

How to save last folder in openFileDialog?

How do I make my application store the last path opened in openFileDialog and after new opening restore it?
OpenFileDialog openFileDialog1 = new OpenFileDialog();
openFileDialog1.Filter = "txt files (*.txt)|*.txt|All files (*.*)|*.*";
if (openFileDialog1.ShowDialog() == DialogResult.OK)
{
acc_path = openFileDialog1.FileName;
Settings.Default.acc_path = acc_path;
foreach (string s in File.ReadAllLines(openFileDialog1.FileName))
{
accs.Enqueue(s);
}
label2.Text = accs.Count.ToString();
}

This is the easiest way: FileDialog.RestoreDirectory.

After changing Settings you have to call
Settings.Default.Save();
and before you open the OpenFileDialog you set
openFileDialog1.InitialDirectory = Settings.Default.acc_path;

I think it would be enough for you to use SetCurrentDirectory to ste the current directory for the OS. So on the next dialog opening it would pick that path.
Or simply save path into some variable of your application and use
FileDialog.InitialDirectory property.

The following is all you need to make sure that OpenFileDialog will open at the directory the user last selected, during the lifetime off your application.
OpenFileDialog OpenFile = new OpenFileDialog();
OpenFile.RestoreDirectory = false;

I find that all you have to do is NOT set the initial directory and the dialog box remembers your last save/open location. This remembers even after the application is closed and reopened. Try this code with the initial directory commented out. Many of the suggestions above will also work but if you are not looking for additional functionality this is all you have to do.
private void button1_Click(object sender, EventArgs e)
{
Stream myStream = null;
OpenFileDialog openFileDialog1 = new OpenFileDialog();
//openFileDialog1.InitialDirectory = "c:\\";
openFileDialog1.Filter = "txt files (*.txt)|*.txt|All files (*.*)|*.*";
openFileDialog1.FilterIndex = 2;
if (openFileDialog1.ShowDialog() == DialogResult.OK)
{
try
{
if ((myStream = openFileDialog1.OpenFile()) != null)
{
using (myStream)
{
// Insert code to read the stream here.
}
}
}
catch (Exception ex)
{
MessageBox.Show("Error: Could not read file from disk. Original error: " + ex.Message);
}
}
}

I know this is a bit of an old thread, but I was not able to find a solution I liked to this same question so I developed my own. I did this in WPF but it should work almost the same in Winforms.
Essentially, I use an app.config file to store my programs last path.
When my program starts I read the config file and save to a global variable. Below is a class and function I call when my program starts.
public static class Statics
{
public static string CurrentBrowsePath { get; set; }
public static void initialization()
{
ConfigurationManager.RefreshSection("appSettings");
Configuration config = ConfigurationManager.OpenExeConfiguration(ConfigurationUserLevel.None);
CurrentBrowsePath = ConfigurationManager.AppSettings["lastfolder"];
}
}
Next I have a button that opens the file browse dialog and sets the InitialDirectory property to what was stored in the config file. Hope this helps any one googling.
private void browse_Click(object sender, RoutedEventArgs e)
{
OpenFileDialog open_files_dialog = new OpenFileDialog();
open_files_dialog.Multiselect = true;
open_files_dialog.Filter = "Image files|*.jpg;*.jpeg;*.png";
open_files_dialog.InitialDirectory = Statics.CurrentBrowsePath;
try
{
bool? dialog_result = open_files_dialog.ShowDialog();
if (dialog_result.HasValue && dialog_result.Value)
{
string[] Selected_Files = open_files_dialog.FileNames;
if (Selected_Files.Length > 0)
{
ConfigWriter.Update("lastfolder", System.IO.Path.GetDirectoryName(Selected_Files[0]));
}
// Place code here to do what you want to do with the selected files.
}
}
catch (Exception Ex)
{
MessageBox.Show("File Browse Error: " + Environment.NewLine + Convert.ToString(Ex));
}
}

You can use the InitialDirectory property : http://msdn.microsoft.com/fr-fr/library/system.windows.forms.filedialog.initialdirectory.aspx
OpenFileDialog openFileDialog1 = new OpenFileDialog();
openFileDialog1.Filter = "txt files (*.txt)|*.txt|All files (*.*)|*.*";
openFileDialog1.InitialDirectory = previousPath;
if (openFileDialog1.ShowDialog() == DialogResult.OK)
{
previousPath = Path.GetDirectoryName(openFileDialog1.FileName);
acc_path = openFileDialog1.FileName;
Settings.Default.acc_path = acc_path;
foreach (string s in File.ReadAllLines(openFileDialog1.FileName))
{
accs.Enqueue(s);
}
label2.Text = accs.Count.ToString();
}

if your using
Dim myFileDlog As New OpenFileDialog()
then you can use this to restore the last directory
myFileDlog.RestoreDirectory = True
and this to not
myFileDlog.RestoreDirectory = False
(in VB.NET)