I have openfiledialog that reading user image address with file info and load it in textbox
I want to have another button in order to open image address (that already saved in textbox)
how to code this button at wpf ? I know i should use process.start but no idea !
Microsoft.Win32.OpenFileDialog dlg = new Microsoft.Win32.OpenFileDialog();
tbl_Moshtari tt = new tbl_Moshtari();
dlg.FileName = "pic-file-name"; // Default file name
dlg.DefaultExt = ".jpg"; // Default file extension
dlg.Filter = "JPEG(.jpeg)|*.jpeg | PNG(.png)|*.png | JPG Files (*.jpg)|*.jpg|GIF Files (*.gif)|*.gif"; // Filter files by extension
Nullable<bool> result = dlg.ShowDialog();
if (result == true)
{
//// picbox.Source = new BitmapImage(new Uri(dlg.FileName, UriKind.Absolute));
//bitmapImage = new BitmapImage();
//bitmapImage.BeginInit();
//bitmapImage.StreamSource = System.IO.File.OpenRead(dlg.FileName);
//bitmapImage.EndInit();
////now, the Position of the StreamSource is not in the begin of the stream.
//picbox.Source = bitmapImage;
FileInfo fi = new FileInfo(dlg.FileName);
string filename = dlg.FileName;
txt_picaddress.Text = filename;
System.Windows.MessageBox.Show("Successfully done");
}
This second button i have
private void btn_go_Click(object sender, RoutedEventArgs e)
{
Microsoft.Win32.OpenFileDialog dlg = new Microsoft.Win32.OpenFileDialog();
//FileInfo fi = new FileInfo(dlg.FileName);
string filename = dlg.FileName;
Process.Start(filename);
}
This isnt working for me .
Process.Start() should open up the image as long as filename is an absolute path to the file. With that being said, where in your btn_go_Click method are you actually opening up the dialog to get the file name? dlg.FileName returns an empty string if you don't show the dialog in which case Process.Start() fails.
If the file name needs to come from the previous dialog, you shouldn't create a new dialog; instead, change
Process.Start(filename)
to
Process.Start(txt_picaddress.Text)
Of course, you need to do some input verification to make sure the path is correct (unless the textbox is read-only).
Also, consider setting a breakpoint on string filename = dlg.FileName; to make sure it has the correct path to the file if it's still not working.
To open and highlight the file in Windows Explorer:
string filename = txt_picaddress.Text;
ProcessStartInfo pInfo =
new ProcessStartInfo("explorer.exe", string.Format("/Select, {0}", filename));
Process.Start(pInfo);
In your second code sample, you created a new instance of openFileDialog, you need instead to use the previous instance of the openFileDialog that is holding the correct image filename:
if you create the first openFileDialog in the window constructor you can do this:
private void btn_go_Click(object sender, RoutedEventArgs e)
{
string filename = this.dlg.FileName;
Process.Start(filename);
}
hope this helps, this is what i can say given the code you provided.
You don't need an OpenFileDialog in btn_go_Click if you want to use the path in your textbox:
private void btn_go_Click(object sender, RoutedEventArgs e)
{
string filename = txt_picaddress.Text;
Process.Start(filename);
}
Related
I need help with this code. I want to create basic image convert program but this program is not working? What am I doing wrong. Thanks for answers.
private void button1_Click(object sender, EventArgs e)
{
OpenFileDialog file = new OpenFileDialog();
file.ShowDialog();
string DosyaYolu = file.FileName;
string DosyaAdi = file.SafeFileName;
if (file.ShowDialog() == DialogResult.OK)
{
System.Drawing.Image image = System.Drawing.Image.FromFile(DosyaYolu);
image.Save(DosyaYolu, System.Drawing.Imaging.ImageFormat.Png);
}
You choose the wrong target path to save the new image to. Also you invoked the ShowDialog() twice, which is not necessary. The following code will save the new file with the same name but a different extension.
var dialog = new OpenFileDialog();
if (dialog.ShowDialog() == DialogResult.OK)
{
string sourceFile = dialog.FileName;
string targetFile = Path.ChangeExtension(sourceFile, "png");
Image image = Image.FromFile(sourceFile);
image.Save(targetFile, ImageFormat.Png);
}
I have 2 system a c# winform and a php and their database is stored in a single database mysql now my problem is storing pictures...., in my html I save my pictures in htdocs/"foldername"/productimages now what I want in my C# winform was get the location path of the picture from open dialogue and copy that picture to the specific folder which is the htdocs/"foldername"/productimages how do i do that
my code
string picloc;
private void UpdBtn_Click(object sender, EventArgs e)
{
dlg.Filter = "JPG Files(*.jpg)|*.jpg|PNG Files(*.png)|*.png|ALL Files(*.*)|*.*";
dlg.Title = "Select Thumbnail";
if (dlg.ShowDialog() == DialogResult.OK)
{
// Result();
picloc = dlg.FileName.ToString();
pic1.ImageLocation = picloc;
}
}
so How do I copy the file picture from the string picloc to the specific loc?
You can use File.Copy to do this. Something like the following should manage the copy for you:
string picloc;
string new_loc;
private void UpdBtn_Click(object sender, EventArgs e)
{
dlg.Filter = "JPG Files(*.jpg)|*.jpg|PNG Files(*.png)|*.png|ALL Files(*.*)|*.*";
dlg.Title = "Select Thumbnail";
if (dlg.ShowDialog() == DialogResult.OK)
{
// Result();
picloc = dlg.FileName.ToString();
pic1.ImageLocation = picloc;
File.Copy(picloc, new_loc); // new_loc being the new location for the file.
}
}
This assumes you already know the location for the copy of the file. If you want to give the user the choice, do so via a SaveFileDialog to get the new location (as a string) and then perform the copy.
I have a TextBox named textbox1 and a Button named button1.
When I click on button1 I want to browse my files to search only for image files (type jpg, png, bmp...).
And when I select an image file and click Ok in the file dialog I want the file directory to be written in the textbox1.text like this:
textbox1.Text = "C:\myfolder\myimage.jpg"
Something like that should be what you need
private void button1_Click(object sender, RoutedEventArgs e)
{
// Create OpenFileDialog
Microsoft.Win32.OpenFileDialog dlg = new Microsoft.Win32.OpenFileDialog();
// Set filter for file extension and default file extension
dlg.DefaultExt = ".png";
dlg.Filter = "JPEG Files (*.jpeg)|*.jpeg|PNG Files (*.png)|*.png|JPG Files (*.jpg)|*.jpg|GIF Files (*.gif)|*.gif";
// Display OpenFileDialog by calling ShowDialog method
Nullable<bool> result = dlg.ShowDialog();
// Get the selected file name and display in a TextBox
if (result == true)
{
// Open document
string filename = dlg.FileName;
textBox1.Text = filename;
}
}
var ofd = new Microsoft.Win32.OpenFileDialog() {Filter = "JPEG Files (*.jpeg)|*.jpeg|PNG Files (*.png)|*.png|JPG Files (*.jpg)|*.jpg|GIF Files (*.gif)|*.gif"};
var result = ofd.ShowDialog();
if (result == false) return;
textBox1.Text = ofd.FileName;
Goal:
Save a picture from the harddrive into my WPF application. The picture should be available if copying the WPF application. The address to the picture located in the WPF application should be saved in the database.
Problem:
How should I do it in a course of action?
private void btnBrowse_Click(object sender, RoutedEventArgs e)
{
Microsoft.Win32.OpenFileDialog dlg = new Microsoft.Win32.OpenFileDialog();
dlg.Filter = "jpg files (*.jpg)|*.jpg|gif files (*.gif)|*.gif|jpeg files (*.jpeg)|*.jpeg";
Nullable<bool> result = dlg.ShowDialog();
if (result == true)
{
// Open document
string filename = dlg.FileName;
txtPicture.Text = filename;
BitmapImage myBitmapImage = new BitmapImage(new Uri(dlg.FileName, UriKind.Absolute));
string sss = myBitmapImage.Format.ToString();
string asd = dlg.SafeFileName.ToString();
}
}
There are several image encoders available. A simple example for PNG files can be found here, or a more complete sample here. The same concept applies for the other supported image file types.
I'm new to programming and I'm starting to create a simple notepad, with only 4 buttons (Open, Save, New and Font).
If I open or save I'm getting an error:
This is my code:
//Declare save as a new SaveFileDailog
SaveFileDialog save = new SaveFileDialog();
//Declare filename as a String equal to the SaveFileDialog's FileName
String filename = save.FileName;
//Declare filter as a String equal to our wanted SaveFileDialog Filter
String filter = "Text Files|*.txt|All Files|*.*";
//Set the SaveFileDialog's Filter to filter
save.Filter = filter;
//Set the title of the SaveFileDialog to Save
save.Title = "Save";
//Show the SaveFileDialog
if (save.ShowDialog(this) == DialogResult.OK)
{
//Write all of the text in txtBox to the specified file
System.IO.File.WriteAllText(filename, textBox1.Text);
}
else
{
//Return
return;
}//Declare save as a new SaveFileDailog
SaveFileDialog save = new SaveFileDialog();
//Declare filename as a String equal to the SaveFileDialog's FileName
String filename = save.FileName;
//Declare filter as a String equal to our wanted SaveFileDialog Filter
String filter = "Text Files|*.txt|All Files|*.*";
//Set the SaveFileDialog's Filter to filter
save.Filter = filter;
//Set the title of the SaveFileDialog to Save
save.Title = "Save";
//Show the SaveFileDialog
if (save.ShowDialog(this) == DialogResult.OK)
{
//Write all of the text in txtBox to the specified file
System.IO.File.WriteAllText(filename, textBox1.Text);
}
else
{
//Return
return;
}
Any idea? Thanks and regards
ooopss I forgot to write the error sorry about that:
Here is the error:
"Error: ArgumentException was unhandled.
Empty path name is not legal"
I get this if I open a text file. Then it highlighted this line code:
textBox1.Text=System.IO.File.ReadAllText(filename,System.Text.Encoding.Default);
And if I save nothing happens.
Thanks
I expect you should be reading the filename after the user has used the dialog:
System.IO.File.WriteAllText(save.FileName, textBox1.Text);
Also - SaveFileDialog is IDisposable, so you should be "using" it...
using (SaveFileDialog save = new SaveFileDialog())
{
// your code that involves "save"
}
Try moving the line
String filename = save.FileName;
inside the IF block.
You are assigning to filename before the SaveDialog's property is set by the user.
You need to understand that this line does not create a permanent link between your filename variable and the FileName property of the dialog.
You get the filename from a SaveFileDialog after you call ShowDialog. You are setting filename beforehand.
Well, it looks like you're saving with a blank filename - this changes during the call to .ShowDialog(), so it doesnt help that you've retrieved it beforehand.
You just need to pull out .FileName again after .ShowDialog.
//To declare private int docno
//To declare private string filename
//To declare private bool modified=false;
if (modify == true)
{
//this.Text = filename;
filename = saveFileDialog1.FileName;
sw = new StreamWriter(filename);
sw.Write(textBox1.Text);
sw.Close();
//modify = false;
}
else
{
saveFileDialog1.FileName = "Untitled" + docno.ToString() + ".txt";
dresult = saveFileDialog1.ShowDialog();
docno++;
}