Hello im having trouble with file dialog
Standard code, but it's freezing while selecting large files (600MB+ xls file that i need to first reformat to xlsb, then stream data from, but that is not important here).
What i wanted to achieve is to hide dialog when Open button is clicked, so that i could display Loading Message.
Yet im not sure how can i achieve that. Can i somehow subscribe to Open button?
OpenFileDialog openFile = new OpenFileDialog() { DereferenceLinks = false };
openFile.Title = "Select Supplier List";
openFile.Filter = "Excel files (*.*xls*)|*.*xls*";
try
{
if (openFile.ShowDialog() == true)
{
/* Takes long time to go after selection */
ViewModel.ReportPath = openFile.FileName;
ViewModel.FileTrueName = Path.GetFileName(ViewModel.ReportPath);
}
}
catch
{
ViewModel.ReportPath = null;
ViewModel.FileTrueName = null;
}
Wrapping code with task, helped :)
Some changes happened during searching for answer, based on comments.
Task.Run(() =>
{
OpenFileDialog openFile = new OpenFileDialog();
openFile.Title = "Select Supplier List";
openFile.Filter = "Excel Files(*.xls; *.xlsx; *.xlsb; *.xlsm)| *.xls; *.xlsx; *.xlsb; *.xlsm";
try
{
if (openFile.ShowDialog() == true)
{
ViewModel.ReportPath = openFile.FileName;
ViewModel.FileTrueName = Path.GetFileName(ViewModel.ReportPath);
}
}
catch
{
ViewModel.ReportPath = null;
ViewModel.FileTrueName = null;
}
});
Related
Im trying to use the openFileDialog to load in 2 different files that goes to 2 different area. I trying doing 2 ways, first i try adding an if statement to see which file they select and use that expression, but it gives me an error.
OpenFileDialog opendi = new OpenFileDialog();
opendi.Filter = "xml|*.xml";
if (opendi.ShowDialog() == DialogResult.OK)
{
if (opendi.FileName == Hotel_Filename)
{
hotelListData = HotelList.HotelLoadFile(opendi.FileName);
lblStatus.Text = "Success";
}
}
else
{
rmtp = roomtypedata.LoadFile(opendi.FileName);
lblStatus.Text = "Success";
}
and if i do this way , it works, but i have to load the files in the right order, or else if i load the 2nd one first , i get an error. so is there a better method to this?
OpenFileDialog opendi = new OpenFileDialog();
opendi.Filter = "xml|*.xml";
if (opendi.ShowDialog() == DialogResult.OK)
{
hotelListData = HotelList.HotelLoadFile(opendi.FileName);
lblStatus.Text = "Success";
}
if (opendi.ShowDialog() == DialogResult.OK)
{
rmtp = roomtypedata.LoadFile(opendi.FileName);
lblStatus.Text = "Success";
}
Ok, the problem seems to be with checking the file name. You want to compare only the file name,not all the path. Try this:
Hotel_Filename = "hotels.xml"
....
for (int=0;i<2;i++)
{
OpenFileDialog opendi = new OpenFileDialog();
opendi.Filter = "xml|*.xml";
if (opendi.ShowDialog() == DialogResult.OK)
{
if (Path.GetFileName(opendi.FileName) == Hotel_Filename)
{
hotelListData = HotelList.HotelLoadFile(opendi.FileName);
lblStatus.Text = "Success";
}
else
{
rmtp = roomtypedata.LoadFile(opendi.FileName);
lblStatus.Text = "Success";
}
}
}
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
}
My issue is that I keep seeing a recurring theme with trying to allow my Notepad clone to save a file. Whenever I try to save a file, regardless of the location on the hard disk, the UnauthorizedAccess Exception continues to be thrown. Below is my sample code for what I've done, and I have tried researching this since last night to no avail. Any help would be greatly appreciated.
//located at base class level
private const string fileFilter = "Text Files|*.txt|All Files|*.*";
private string currentPath = Environment.GetFolderPath(Environment.SpecialFolder.MyDocuments);
private void MenuFileSaveAs_Click(object sender, RoutedEventArgs e)
{
SaveFileDialog sfd = new SaveFileDialog();
sfd.DefaultExt = "*.txt";
sfd.Filter = fileFilter;
sfd.AddExtension = true;
sfd.InitialDirectory = currentPath;
sfd.RestoreDirectory = true;
sfd.OverwritePrompt = true;
sfd.ShowDialog();
try
{
System.IO.File.WriteAllText(currentPath,TxtBox.Text,Encoding.UTF8);
}
catch (ArgumentException)
{
// Do nothing
}
catch(UnauthorizedAccessException)
{
MessageBox.Show("Access Denied");
}
}
Change the following lines.
...
if (sfd.ShowDialog() != true)
return;
try
{
using (var stream = sfd.OpenFile())
using (var writer = new StreamWriter(stream, Encoding.UTF8))
{
writer.Write(TxtBox.Text);
}
}
...
I hope it helps you.
You need to get the correct path context and file object from the dialog box once the user has hit 'ok'. Namely verify the user actually hit ok and then use the OpenFile property to see what their file selection is:
if (sfd.ShowDialog.HasValue && sfd.ShowDialog)
{
if (sfd.OpenFile() != null)
{
// convert your text to byte and .write()
sfd.OpenFile.Close();
}
}
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)
When I click the button to open a file the OpenFileDialog box opens twice. The first box only opens the image files and the second box text files. If the user decides to close out the first box without choosing a file the second box pops up as well. I'm not sure what I am over looking on this issue. Any help would be appreciated. Thanks!
OpenFileDialog of = new OpenFileDialog();
of.Filter = "All Image Formats|*.jpg;*.png;*.bmp;*.gif;*.ico;*.txt|JPG Image|*.jpg|BMP image|*.bmp|PNG image|*.png|GIF Image|*.gif|Icon|*.ico|Text File|*.txt";
if (of.ShowDialog() == System.Windows.Forms.DialogResult.OK)
{
try
{
image1.Source = new BitmapImage(new Uri(of.FileName));
// enable the buttons to function (previous page, next page, rotate left/right, zoom in/out)
button1.IsEnabled = false;
button2.IsEnabled = false;
button3.IsEnabled = false;
button4.IsEnabled = false;
button5.IsEnabled = true;
button6.IsEnabled = true;
button7.IsEnabled = true;
button8.IsEnabled = true;
}
catch (ArgumentException)
{
// Show messagebox when argument exception arises, when user tries to open corrupted file
System.Windows.Forms.MessageBox.Show("Invalid File");
}
}
else if(of.ShowDialog() == System.Windows.Forms.DialogResult.OK)
{
try
{
using (StreamReader sr = new StreamReader(of.FileName))
{
textBox2.Text = sr.ReadToEnd();
button1.IsEnabled = true;
button2.IsEnabled = true;
button3.IsEnabled = true;
button4.IsEnabled = true;
button5.IsEnabled = true;
button6.IsEnabled = true;
button7.IsEnabled = true;
button8.IsEnabled = true;
}
}
catch (ArgumentException)
{
System.Windows.Forms.MessageBox.Show("Invalid File");
}
}
You're calling ShowDialog() twice:
if (of.ShowDialog() == System.Windows.Forms.DialogResult.OK)
{
//...
}
else if(of.ShowDialog() == System.Windows.Forms.DialogResult.OK)
Just call it once, and save the results:
var dialogResult = of.ShowDialog();
if (dialogResult == System.Windows.Forms.DialogResult.OK)
{
//...
}
// Note that the condition was the same!
else if(dialogResult != System.Windows.Forms.DialogResult.OK)
Edit:
If you want the second dialog to only show when the first is handled, you can do:
var dialogResult = of.ShowDialog();
if (dialogResult == System.Windows.Forms.DialogResult.OK)
{
//...
// Do second case here - Setup new options
dialogResult = of.ShowDialog(); //Handle text file here
}
You are calling ShowDialog both in the if and else if condition. That method is responsible for showing the dialog and only has the nice side effect of also telling you what the user clicked.
Store the method's result in a variable and check that in your if..elseif conditions instead.