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;
}
Related
Here's some code I'm working on:
private void Form1_Load(object sender, EventArgs e)
{
}
private void openFileDialog1_FileOk(object sender, CancelEventArgs e)
{
}
private void button1_Click(object sender, EventArgs e)
{
{
OpenFileDialog openFileDialog1 = new OpenFileDialog
{
InitialDirectory = #"C:\",
Title = "Add a PDF",
CheckFileExists = true,
CheckPathExists = true,
DefaultExt = "pdf",
Filter = "pdf files (*.pdf)|*.pdf",
FilterIndex = 2,
RestoreDirectory = true,
ReadOnlyChecked = true,
ShowReadOnly = true
};
if (openFileDialog1.ShowDialog() == DialogResult.OK)
{
textBox1.Text = openFileDialog1.FileName;
string myFile = textBox1.Text;
Console.WriteLine(myFile);
}
}
}
private void button2_Click(object sender, EventArgs e)
{
System.IO.File.Move(myFile, #"C:\testing\records\file.pdf");
}
}
}
So anyway toward the bottom at button2, I'm trying to set up a few things. I want to add a button that saves the file using the System.IO line there. But when I add the button, I can't get it to work properly. The "myFile" variable doesn't seem to be declared anymore. I'm sure this is probably the messiest code anyone will paste on here today, but a lot of it was auto-genned by Visual Studio and I'm afraid to clean it up because I'm not 100% sure what some of this stuff is. I have tried cutting and pasting the button stuff up nearer to the myFile variable declaration since it's private and maybe that's why it doesn't know what it means anymore. But when I move it up there, I get a different error regarding the "private" at the beginning of the button call.
The issue here is that you're declaring your myFile variable within the scope of the if statement and so when that if statement block exits the myFile variable goes out of scope and no longer exists.
To get it to work you need to move the creation of the myFile variable to the class level.
public class Form1 : Form
{
private string myFile;
public Form1()
{
InitializeComponent();
}
// Other code snipped out
}
Now modify your button1_Click code so that instead of creating an instance of myFile:
string myFile = textBox1.Text;
You just populate the private variable you declared in the main part of the code:
myFile = textBox1.Text;
So the if statement looks like this:
if (openFileDialog1.ShowDialog() == DialogResult.OK)
{
textBox1.Text = openFileDialog1.FileName;
myFile = textBox1.Text;
Console.WriteLine(myFile);
}
You need a little modification to button2_Click:
private void button2_Click(object sender, EventArgs e)
{
// Check that myFile has some text and isn't null.
if (string.IsNullOrWhitespace(myFile))
return;
// Check that the file exists before attempting to move it.
if (File.Exists(myFile))
System.IO.File.Move(myFile, #"C:\testing\records\file.pdf");
}
I have created a button that opens a text file through openDialogBox1 and display the content in a listBox. Then I wanted to create another button to close this file, but I couldn't figure out what code to use. Can anyone help me please? Thank you.
Here is a sample of my openFile button code :
private void openButton_Click(object sender, EventArgs e)
{
try
{
StreamReader inputFile;
string File;
OpenFileDialog openFileDialog1 = new OpenFileDialog();
openFileDialog1.DefaultExt = "txt";
openFileDialog1.Filter = "Text Files (*txt)|*txt";
if (openFileDialog1.ShowDialog() == System.Windows.Forms.DialogResult.OK)
{
inputFile = File.OpenText(openFileDialog1.FileName);
fileListBox.Items.Clear();
while (!inputFile.EndOfStream)
{
File = inputFile.ReadLine();
fileListBox.Items.Add(File);
}
inputFile.Close();
}
}
catch
{
MessageBox.Show("Invalid File!");
}
}
Declare StreamReader inputFile; globally so it is available in the second button click.
However, the file should be opened and closed as soon as you are done with it to release resources. Plus, you don't know, the second button may never get clicked.
I have a set-up a Windows form that on completion creates a .txt document containing imputed data for the user on one form and then that form can then be opened into a richTextBox on another form using a ComboBox as a selection tool.
The problem I am having is that the ComboBox does not refresh the directory listings where the .txt documents are saved after a new .txt has been created and so the user has to restart the program before it shows up in the ComboBox listing, wondering how to solve this. Possibly force the ComboBox to refresh the listings onClick of a button?
Form with ComboBox selection method on:
public Default()
{
InitializeComponent();
string[] files = Directory.GetFiles(#"C:\Modules");
foreach (string file in files)
ModuleSelectorComboBox.Items.Add(Path.GetFileNameWithoutExtension(file));
}
private void moduleToolStripMenuItem_Click(object sender, EventArgs e)
{
NewModule newmodule = new NewModule();
newmodule.Show();
}
private void ModuleSelectorComboBox_SelectedValueChanged(object sender, EventArgs e)
{
richTextBox1.Clear(); //Clears previous Modules Text
string fileName = (string)ModuleSelectorComboBox.SelectedItem;
string filePath = Path.Combine(#"C:\Modules\", fileName + ".txt");
if (File.Exists(filePath))
richTextBox1.AppendText(File.ReadAllText(filePath));
else
MessageBox.Show("There's been a problem. Please restart the program. \nError 1", "Error 1", //error 1 is file deleted while the program is running
MessageBoxButtons.OK,
MessageBoxIcon.Exclamation,
MessageBoxDefaultButton.Button1);
}
To add I want to avoid the using of the Dialog save/open file method and is why I am using the ComboBox to do this.
Thanks in advance.
The form to create new .txt document (I don't see this as essentially needed I have just added it for reference):
private void button5_Click(object sender, EventArgs e)
{
RichTextBox newbox = new RichTextBox();
{
String Saved_Module = Path.Combine("C:\\Modules", txtModuleName.Text + ".txt");
newbox.AppendText(txtModuleName.Text + "\n" + ModuleDueDate.Text + "\n" + txtModuleInfo.Text + "\n" + txtModuleLO.Text);
newbox.SaveFile(Saved_Module, RichTextBoxStreamType.PlainText);
Directory.CreateDirectory(Path.Combine(#"C:\Modules", txtModuleName.Text));
this.Close();
}
}
First of all, encapsulate the logic of the combobox population in a method.
public Default()
{
InitializeComponent();
LoadComboBox();
}
void LoadComboBox()
{
ModuleSelectorComboBox.Items.Clear();
string[] files = Directory.GetFiles(#"C:\Modules");
foreach (string file in files)
ModuleSelectorComboBox.Items.Add(Path.GetFileNameWithoutExtension(file));
}
I suggest to open the form as a dialog form using ShowDialog method
and set the DialogResult to DialogResult.OK in the button5_Click method.
private void button5_Click(object sender, EventArgs e)
{
RichTextBox newbox = new RichTextBox();
String Saved_Module = Path.Combine("C:\\Modules", txtModuleName.Text + ".txt");
newbox.AppendText(txtModuleName.Text + "\n" + ModuleDueDate.Text + "\n" + txtModuleInfo.Text + "\n" + txtModuleLO.Text);
newbox.SaveFile(Saved_Module, RichTextBoxStreamType.PlainText);
Directory.CreateDirectory(Path.Combine(#"C:\Modules", txtModuleName.Text));
this.DialogResult = DialogResult.OK;
this.Close();
}
And then, based on the DialogResult load or do not load the combobox's items in the moduleToolStripMenuItem_Click method.
private void moduleToolStripMenuItem_Click(object sender, EventArgs e)
{
NewModule newmodule = new NewModule();
newmodule.ShowDialog();
if (result == DialogResult.OK)
{
LoadComboBox();
}
}
Update:
If you don't want to use a Dialog, you can subscribe to an FormClosing event and update the combobox in a handler to the event.
private void moduleToolStripMenuItem_Click(object sender, EventArgs e)
{
NewModule newmodule = new NewModule();
newmodule.FormClosing += F_FormClosing
newmodule.Show();
}
private void F_FormClosing(object sender, FormClosingEventArgs e)
{
LoadComboBox();
}
I suggest breaking out LoadComboBox as Valentin suggests, but using a FileSystemWatcher instantiated and disposed with the form to monitor the Modules directory and call LoadComboBox on any create/delete.
Here is my code;
private void button1_Click(object sender, EventArgs e)
{
string newFile =textBox1.Text;
string temp = newFile.Replace("YNATEST.", "");
SaveFileDialog a1 = new SaveFileDialog();
a1.FileName = "";
a1.Filter = "Text Files(*txt)|*.txt";
a1.DefaultExt = "txt";
a1.ShowDialog();
StreamWriter yazmaislemi = new StreamWriter(a1.FileName);
yazmaislemi.WriteLine(temp);
yazmaislemi.Close();
}
it is saving the text on Desktop but i want to save it to the following path:
C:\Users\esra.ur\Desktop\projee1
use save file dialog, so you can save your text in your specific directory
using System;
using System.ComponentModel;
using System.IO;
using System.Windows.Forms;
namespace WindowsFormsApplication30
{
public partial class Form1 : Form
{
public Form1()
{
InitializeComponent();
}
private void button1_Click(object sender, EventArgs e)
{
// When user clicks button, show the dialog.
saveFileDialog1.ShowDialog();
}
private void saveFileDialog1_FileOk(object sender, CancelEventArgs e)
{
// Get file name.
string name = saveFileDialog1.FileName;
// Write to the file name selected.
// ... You can write the text from a TextBox instead of a string literal.
File.WriteAllText(name, "test");
}
}
}
this code snippets is from this link http://www.dotnetperls.com/savefiledialog
I hope it will help
1) Wrap your show dialog to check the result.
if(a1.ShowDialog() == DialogResult.OK)
2) The SaveFileDialog has a property for setting an initial path. This is for the directory which will be shown when the dialog is first open. For the desktop you want to use the Environment.GetFolderPath like so.
a1.InitialDirectory = Environment.GetFolderPath(Environment.SpecialFolder.Desktop);
3) Try to separate concerns:
private string OutputFile {get;set;}
private void button1_Click(object sender, EventArgs e)
{
if(string.IsNullOrEmpty(this.OutputPath))
{
SaveFileDialog a1 = new SaveFileDialog();
a1.FileName = textBox1.Text;
a1.Filter = "Text Files(*txt)|*.txt";
a1.DefaultExt = "txt";
a1.InitialDirectory = Environment.GetFolderPath(Environment.SpecialFolder.Desktop);
if(a1.ShowDialog() == DialogResult.OK)
{
this.OutputFile = ai.FileName
}
}
this.SaveFile(this.OutputFile);
}
private void SaveFile(string FileName)
{
string newFile = FileName;
string temp = newFile.Replace("YNATEST.", "");
using(StreamWriter yazmaislemi = new StreamWriter(temp))
{
yazmaislemi.WriteLine(temp);
yazmaislemi.Close();
}
}
The SaveFileDialog object has a property called InitialDirectory, which is a string you can specify, for example
SaveFileDialog a1 = new SaveFileDialog();
a1.InitialDirectory = #"C:\Users\esra.ur\Desktop\projee1";
If this directory doesn't exist, it will default back to documents. Be careful about writing a file even if the user tries to cancel. Hope this helps?
In response to your comment, it sounds like you want to hard code the destination file name. This is dangerous as you can get an exception if the directory doesn't exist, but you can use the following: (I'm not sure what you want to do with the file name)
'string newFile = textBox1.Text;
string temp = newFile.Replace("YNATEST.", "");
StreamWriter yazmaislemi = new StreamWriter(#"C:\Users\esra.ur\Desktop\projee1\" + temp + ".txt");
yazmaislemi.WriteLine(temp);
yazmaislemi.Close();
In this case you don't need the SaveFileDialog at all. I think this is what you're asking for, but it's dangerous to code in this way.
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
}