Develop Reference

SaveFile and LoadFile issue

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);

opening a .txt in a windows form richtextbox

There is a question like this: Open a .txt file into a richTextBox in C#
But I need something a little different, I can open the txt in a richTextBox but I am using a button to make the file open. I want to skip the button and just have it load in the richTextBox.
Here is the button code, how do I move it to the richTextBox?:
private void button1_Click(object sender, EventArgs e)
{
OpenFileDialog openFileDialog1 = new OpenFileDialog();
openFileDialog1.FileName = "Changes.txt";
string strfilename = openFileDialog1.FileName;
string filetext = File.ReadAllText(strfilename);
richTextBox1.Text = filetext;
}
private void richTextBox1_TextChanged(object sender, EventArgs e)
{
}
By double clicking the rich I get richTextBox1_TextChanged:

How about the load event of the Form?
private void form_load(object sender, Eventargs e)
{
OpenFileDialog openFileDialog1 = new OpenFileDialog();
openFileDialog1.FileName = "Changes.txt";
string strfilename = openFileDialog1.FileName;
string filetext = File.ReadAllText(strfilename);
richTextBox1.Text = filetext;
}
I'm not sure what you want to do in the richTextBox1_TextChanged event, maybe you could provide more information regarding this.
Having seen your edit, your'e still better off using the button to open the dialog and selecting a file that way. More intuitive then selecting the RichTextBox to trigger the OpenFileDialog.

From what I understand, you are probably searching for the event RichTextBox_Click. In your form.cs design view, click your RichTextBox and press F4 to see its properties and click the thunderbolt icon to see its events. Search for "Click" and then double click it. That will create the click event in your class, and you can add your logic there.

modify your code with this. when dialogbox visible select the text file which you named here as "Changes.txt".
private void button1_Click(object sender, EventArgs e)
{
OpenFileDialog openFileDialog1 = new OpenFileDialog();
openFileDialog1.FileName = "Changes.txt";
openFileDialog1.ShowDialog();
string strfilename = openFileDialog1.FileName;
string filetext = File.ReadAllText(strfilename);
richTextBox1.Text = filetext;
}

Beginners help C# showing a filepath in text box

I'm new to programming and am using C# and Visual Studio Express 2012. I am creating a windows form and have inserted a button which runs open file dialog when clicked. I have a text box on the form that I'd like to have show the file path of the file that the user selected. I have found some code examples on this site but struggle to understand where they should be placed in the code structure as the examples are often standalone snippets. I hope its not too dumb a question!
Thanks in advance
Lee
The answer in case it's of use to anyone was.......
public partial class Form1 : Form
{
public Form1()
{
InitializeComponent();
}
private void button1_Click(object sender, EventArgs e)
{
using (FileDialog fileDialog = new OpenFileDialog())
{
if (DialogResult.OK == fileDialog.ShowDialog())
{
string filename = fileDialog.FileName;
textBox1.Text = fileDialog.FileName;
}
}
}
}

Your OpenFileDialog has property FileName that contains the path of the selected file, assign that to your TextBox.Text
if (openFileDialog.ShowDialog() == DialogResult.OK)
{
yourTextBox.Text = openFileDialog.FileName;
}

OpenFileDialog openFileDialog1 = new OpenFileDialog();
openFileDialog1.Filter = "txt files (*.txt)|*.txt|All files (*.*)|*.*";
if(openFileDialog1.ShowDialog() == DialogResult.OK)
textbox.text = openFileDialog1.FileName;

Filtering text files in C#

How do I just open files with the .txt extension, I want to my program to pop up an error message if the file is not a .txt file I want a code that can modify this code below
private void button1_Click(object sender, EventArgs e)
{
OpenFileDialog of = new OpenFileDialog();
of.ShowDialog();
textBox1.Text = of.FileName;
}
Can someone help let's say I want to put this loop
if fileextension is .txt then
OpenFileDialog of = new OpenFileDialog();
of.ShowDialog();
textBox1.Text = of.FileName;
else show error message(like can not open this file)

As I understood correctly, you want to see only txt files in your dialog?
If so, use Filter property.
OpenFileDialog of = new OpenFileDialog();
of.Filter = "Text files (*.txt)|*.txt";

Can use Path.GetExtension method for this
OpenFileDialog of = new OpenFileDialog();
if(of.ShowDialog() == DialogResult.OK)
{
if(Path.GetExtension(of.FileName).Equals("txt",
StringComparison.InvariantCultureIgnoreCase))
textBox1.Text = of.FileName;
}

You shouldn't allow all extensions if you only allow txt extension.
of.Filter = "Text Files|*.txt";
Will make the OpenFileDialog accept only txt extension files.

Load a bitmap image into Windows Forms using open file dialog

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;