I'm trying to create a text box where you can save the text and load it as a .txt file.
However, it keeps telling me that the textbox does not have a definition for those 2.
Here's the code of the program I'm trying to write:
private void saveToolStripMenuItem_Click(object sender, EventArgs e)
{
if (saveFileDialog1.ShowDialog() == DialogResult.OK)
textBox1.SaveFile(saveFileDialog1.FileName);
}
private void openToolStripMenuItem_Click(object sender, EventArgs e)
{
if (openFileDialog1.ShowDialog() == DialogResult.OK)
textBox1.LoadFile(openFileDialog1.FileName);
}
}
for save file:
SaveFileDialog saveFileDialog1 = new SaveFileDialog();
saveFileDialog1.ShowDialog();
File.WriteAllText(saveFileDialog1.FileName, textBox1.Text);
for load file to textBox:
OpenFileDialog openFileDialog1 = new OpenFileDialog ();
openFileDialog1.ShowDialog();
textBox1.Text = File.ReadAllText(openFileDialog1.FileName);
Related
I am able to run the codes for my FolderBrowserDialog and it will show me the entire path for the selected folder.
Are there any ways where the text box will only display the folder name instead of the whole path?
Now it is showing "C:\Users\Computer\Desktop\Flight"
I would only want to display the "Flight" folder.
private void button1_Click(object sender, EventArgs e)
{
FolderBrowserDialog FBD = new FolderBrowserDialog();
if (FBD.ShowDialog() == DialogResult.OK)
{
textBox1.Text = FBD.SelectedPath;
}
}
Thanks in advance.
Try this:
private void button1_Click(object sender, EventArgs e)
{
FolderBrowserDialog FBD = new FolderBrowserDialog();
if (FBD.ShowDialog() == DialogResult.OK)
{
textBox1.Text = new DirectoryInfo(FBD.SelectedPath).Name;
}
}
You'll need to make sure you have using System.IO; included as well.
I am creating a text editor and i am stuck on the SaveFileDialog window opening
and asking to overwrite the current file open.
I have seen all the similar questions asked like this on SO but none have been able to help me. I have even tried the code from this question: "Saving file without dialog" Saving file without dialog
I got stuck on my program having a problem with FileName.
Here is the code i have currently
namespace Text_Editor
{
public partial class Form1 : Form
{
public Form1()
{
InitializeComponent();
}
private void newToolStripMenuItem_Click(object sender, EventArgs e)
{
richTextBox1.Clear();
}
private void openToolStripMenuItem_Click(object sender, EventArgs e)
{
OpenFileDialog open = new OpenFileDialog();
open.Filter = "Text Files (.txt)|*.txt|All Files (*.*)|*.*";
open.Title = "Open File";
open.FileName = "";
if (open.ShowDialog() == DialogResult.OK)
{
this.Text = string.Format("{0}", Path.GetFileNameWithoutExtension(open.FileName));
StreamReader reader = new StreamReader(open.FileName);
richTextBox1.Text = reader.ReadToEnd();
reader.Close();
}
}
private void exitToolStripMenuItem_Click(object sender, EventArgs e)
{
Application.Exit();
}
private void saveToolStripMenuItem_Click(object sender, EventArgs e)
{
SaveFileDialog save = new SaveFileDialog();
save.Filter = "Text Files (.txt)|*.txt|All Files (*.*)|*.*";
save.Title = "Save File";
save.InitialDirectory = Environment.GetFolderPath(Environment.SpecialFolder.MyDocuments);
if (save.ShowDialog() == DialogResult.OK)
{
StreamWriter writer = new StreamWriter(save.FileName);
writer.Write(richTextBox1.Text);
writer.Close();
}
}
private void saveAsToolStripMenuItem_Click(object sender, EventArgs e)
{
SaveFileDialog saving = new SaveFileDialog();
saving.InitialDirectory = Environment.GetFolderPath(Environment.SpecialFolder.MyDocuments);
saving.Filter = "Text Files (.txt)|*.txt|All Files (*.*)|*.*";
saving.Title = "Save As";
saving.FileName = "Untitled";
if (saving.ShowDialog() == DialogResult.OK)
{
StreamWriter writing = new StreamWriter(saving.FileName);
writing.Write(richTextBox1.Text);
writing.Close();
}
}
}
}
So my question is how can i modify my code so that i can save a file currently open without having the SaveFileDialog box opening everytime?
I do understand that it has something to do with the fact that i'm calling .ShowDialog but i don't know how to modify it.
When opening the file, save the FileName in a form-level variable or property.
Now while saving the file, you can use this FileName instead of getting it from a FileOpenDialog.
First declare a variable to hold filename at form level
// declare at form level
private string FileName = string.Empty;
When opening a file, save the FileName in this variable
private void openToolStripMenuItem_Click(object sender, EventArgs e)
{
OpenFileDialog open = new OpenFileDialog();
open.Filter = "Text Files (.txt)|*.txt|All Files (*.*)|*.*";
open.Title = "Open File";
open.FileName = "";
if (open.ShowDialog() == DialogResult.OK)
{
// save the opened FileName in our variable
this.FileName = open.FileName;
this.Text = string.Format("{0}", Path.GetFileNameWithoutExtension(open.FileName));
StreamReader reader = new StreamReader(open.FileName);
richTextBox1.Text = reader.ReadToEnd();
reader.Close();
}
}
And when doing SaveAs operation, update this variable
private void saveAsToolStripMenuItem_Click(object sender, EventArgs e)
{
SaveFileDialog saving = new SaveFileDialog();
saving.InitialDirectory = Environment.GetFolderPath(Environment.SpecialFolder.MyDocuments);
saving.Filter = "Text Files (.txt)|*.txt|All Files (*.*)|*.*";
saving.Title = "Save As";
saving.FileName = "Untitled";
if (saving.ShowDialog() == DialogResult.OK)
{
// save the new FileName in our variable
this.FileName = saving.FileName;
StreamWriter writing = new StreamWriter(saving.FileName);
writing.Write(richTextBox1.Text);
writing.Close();
}
}
The save function can then be modified like this:
private void saveToolStripMenuItem_Click(object sender, EventArgs e)
{
if (string.IsNullOrEmpty(this.FileName))
{
// call SaveAs
saveAsToolStripMenuItem_Click(sender, e);
} else {
// we already have the filename. we overwrite that file.
StreamWriter writer = new StreamWriter(this.FileName);
writer.Write(richTextBox1.Text);
writer.Close();
}
}
In the New (and Close) function, you should clear this variable
private void newToolStripMenuItem_Click(object sender, EventArgs e)
{
// clear the FileName
this.FileName = string.Empty;
richTextBox1.Clear();
}
Create a new string variable in your class for example
string filename = string.empty
and then
private void saveToolStripMenuItem_Click(object sender, EventArgs e)
{
if(string.IsNullOrEmpty(filename)) {
//Show Save filedialog
SaveFileDialog save = new SaveFileDialog();
save.Filter = "Text Files (.txt)|*.txt|All Files (*.*)|*.*";
save.Title = "Save File";
save.InitialDirectory = Environment.GetFolderPath(Environment.SpecialFolder.MyDocuments);
if (save.ShowDialog() == DialogResult.OK)
{
filename = save.FileName;
}
}
StreamWriter writer = new StreamWriter(filename);
writer.Write(richTextBox1.Text);
writer.Close();
}
The SaveFileDialog now only opens if fileName is null or empty
You will have to store the fact that you have already saved the file, e.g. by storing the file name in a member variable of the Form class you have. Then use an if to check whether you have already saved your file or not, and then either display the SaveFileDialog using ShowDialog() (in case you haven't) or don't and continue to save to the already defined file name (stored in your member variable).
Give it a try, do the following:
Define a string member variable, call it _fileName (private string _fileName; in your class)
In your saveToolStripMenuItem_Click method, check if it's null (if (null == _fileName))
If it is null, continue as before (show dialog), and after getting the file name, store it in your member variable
Refactor your file writing code so that you either get the file name from the file dialog (like before), or from your member variable _fileName
Have fun, C# is a great language to program in.
First, extract method from saveAsToolStripMenuItem_Click: what if you want add up a popup menu, speed button? Then just implement
public partial class Form1: Form {
// File name to save text to
private String m_FileName = "";
private Boolean SaveText(Boolean showDialog) {
// If file name is not assigned or dialog explictly required
if (String.IsNullOrEmpty(m_FileName) || showDialog) {
// Wrap IDisposable into using
using (SaveFileDialog dlg = new SaveFileDialog()) {
dlg.Filter = "Text Files (.txt)|*.txt|All Files (*.*)|*.*";
dlg.Title = "Save File";
dlg.FileName = m_FileName;
if (dlg.ShowDialog() != DialogResult.OK)
return false;
m_FileName = dlg.FileName;
}
}
File.WriteAllText(m_FileName, richTextBox1.Text);
this.Text = Path.GetFileNameWithoutExtension(m_FileName);
return true;
}
private void saveAsToolStripMenuItem_Click(object sender, EventArgs e) {
// SaveAs: always show the dialog
SaveText(true);
}
private void saveToolStripMenuItem_Click(object sender, EventArgs e) {
// Save: show the dialog when required only
SaveText(false);
}
...
}
I'm using C# web browser and I want to save some of pages when a key is pressed but when I use SaveFileDialgue a window pop ups and asks for destination! Is there any way to save it directly to a specific destination without prompting me?
here it is my code:
private void btnSubmit_Click(object sender, EventArgs e)
{
browser.Navigate("https://www.google.com/");
SaveFileDialog sfd = new SaveFileDialog();
sfd.Filter = " TEXT File |*.txt";
if (sfd.ShowDialog() == DialogResult.OK)
{
browser.SaveDocument(sfd.FileName);
}
}
remove the following lines
SaveFileDialog sfd = new SaveFileDialog();
sfd.Filter = " TEXT File |*.txt";
if (sfd.ShowDialog() == DialogResult.OK)
{
browser.SaveDocument(sfd.FileName);
}
add
browser.SaveDocument("c:\\outputfile.txt");
You can read all the page using
private void BtnSaveHtml_Click(object sender, EventArgs e)
{
// webBrowser1.ShowSaveAsDialog();
string html;
html = webBrowser1.DocumentText.ToString();
}
and with the help of stream writer you save all the data where you want to save.
I need to open the bitmap image in the window form using open file dialog (I will load it from drive). The image should fit in the picture box.
Here is the code I tried:
private void button1_Click(object sender, EventArgs e)
{
var dialog = new OpenFileDialog();
dialog.Title = "Open Image";
dialog.Filter = "bmp files (*.bmp)|*.bmp";
if (dialog.ShowDialog() == DialogResult.OK)
{
var PictureBox1 = new PictureBox();
PictureBox1.Image(dialog.FileName);
}
dialog.Dispose();
}
You have to create an instance of the Bitmap class, using the constructor overload that loads an image from a file on disk. As your code is written now, you're trying to use the PictureBox.Image property as if it were a method.
Change your code to look like this (also taking advantage of the using statement to ensure proper disposal, rather than manually calling the Dispose method):
private void button1_Click(object sender, EventArgs e)
{
// Wrap the creation of the OpenFileDialog instance in a using statement,
// rather than manually calling the Dispose method to ensure proper disposal
using (OpenFileDialog dlg = new OpenFileDialog())
{
dlg.Title = "Open Image";
dlg.Filter = "bmp files (*.bmp)|*.bmp";
if (dlg.ShowDialog() == DialogResult.OK)
{
PictureBox PictureBox1 = new PictureBox();
// Create a new Bitmap object from the picture file on disk,
// and assign that to the PictureBox.Image property
PictureBox1.Image = new Bitmap(dlg.FileName);
}
}
}
Of course, that's not going to display the image anywhere on your form because the picture box control that you've created hasn't been added to the form. You need to add the new picture box control that you've just created to the form's Controls collection using the Add method. Note the line added to the above code here:
private void button1_Click(object sender, EventArgs e)
{
using (OpenFileDialog dlg = new OpenFileDialog())
{
dlg.Title = "Open Image";
dlg.Filter = "bmp files (*.bmp)|*.bmp";
if (dlg.ShowDialog() == DialogResult.OK)
{
PictureBox PictureBox1 = new PictureBox();
PictureBox1.Image = new Bitmap(dlg.FileName);
// Add the new control to its parent's controls collection
this.Controls.Add(PictureBox1);
}
}
}
Works Fine.
Try this,
private void addImageButton_Click(object sender, EventArgs e)
{
OpenFileDialog of = new OpenFileDialog();
//For any other formats
of.Filter = "Image Files (*.bmp;*.jpg;*.jpeg,*.png)|*.BMP;*.JPG;*.JPEG;*.PNG";
if (of.ShowDialog() == DialogResult.OK)
{
pictureBox1.ImageLocation = of.FileName;
}
}
You should try to:
Create the picturebox visually in form (it's easier)
Set Dock property of picturebox to Fill (if you want image to fill form)
Set SizeMode of picturebox to StretchImage
Finally:
private void button1_Click(object sender, EventArgs e)
{
OpenFileDialog dlg = new OpenFileDialog();
dlg.Title = "Open Image";
dlg.Filter = "bmp files (*.bmp)|*.bmp";
if (dlg.ShowDialog() == DialogResult.OK)
{
PictureBox1.Image = Image.FromFile(dlg.Filename);
}
dlg.Dispose();
}
private void button1_Click(object sender, EventArgs e)
{
OpenFileDialog open = new OpenFileDialog();
if (open.ShowDialog() == DialogResult.OK)
pictureBox1.Image = Bitmap.FromFile(open.FileName);
}
You, can also try like this, PictureBox1.Image = Image.FromFile("<your ImagePath>" or <Dialog box result>);
PictureBox.Image is a property, not a method. You can set it like this:
PictureBox1.Image = System.Drawing.Image.FromFile(dlg.FileName);
You can try the following:
private void button1_Click(object sender, EventArgs e)
{
OpenFileDialog fDialog = new OpenFileDialog();
fDialog.Title = "Select file to be upload";
fDialog.Filter = "All Files|*.*";
// fDialog.Filter = "PDF Files|*.pdf";
if (fDialog.ShowDialog() == DialogResult.OK)
{
textBox1.Text = fDialog.FileName.ToString();
}
}
It's simple. Just add:
PictureBox1.BackgroundImageLayout = ImageLayout.Zoom;
I am working on form . I want small window to pop up when I click button and to will select XML file of my choice from various folder.
I guess, this OpenFileDialog will help me.
private void button3_Click(object sender, EventArgs e)
{
/
OpenFileDialog OpenFileDialog1 = new OpenFileDialog();
openFileDialog1.Filter = " XML Files|*.xml";
openFileDialog1.InitialDirectory = #"D:\";
if (OpenFileDialog1.ShowDialog() == DialogResult.OK)
{
MessageBox.Show(filed.FileName.ToString());
}
}
I tried using following code but when I click on the button there window doesn't pop up.
I am not geting what mistake I have made.
What is the problem with that?
Thanks!
You cant just open the file dialog from a console app. You will have to workaround it with some setting to single thread apartment (STA).
[STAThread]
static void Main(string[] args)
{
MessageBox.Show("Test");
}
--EDIT--
Following works on click event:
OpenFileDialog f = new OpenFileDialog();
f.Filter = "XML Files|*.xml";
f.InitialDirectory = "D:\\";
if(f.ShowDialog() == DialogResult.OK)
{
MessageBox.Show(f.FileName);
}
You cant open file fialog in console app.
You say I have button, so this must be Win app, use
openFileDialog1.ShowDialog(); is missing
private void button3_Click(object sender, EventArgs e)
{
OpenFileDialog OpenFileDialog1 = new OpenFileDialog();
openFileDialog1.Filter = " XML Files|*.xml";
openFileDialog1.InitialDirectory = #"D:\";
openFileDialog1.ShowDialog();
// Get file name and use OpenFileDialog1.FileName or something like that
}