I have created some code that saves my work to a text file, I was just wondering is there some way so that when I click 'Save' I can read in the text from richTextBox1and set it as the default file name, still with the 'txt' default file extension.
e.g. When I click 'save' the folder dialog comes up and asks you to name your file, just as it would if you were using Word for example, I want that box to already have the text from my richTextBox1 in.
Thanks.
private void saveToolStripMenuItem_Click(object sender, EventArgs e)
{
SaveFileDialog save = new SaveFileDialog();
save.InitialDirectory = "C:\\To-Do-List";
save.Filter = "Text Files (*.txt)|*.txt";
save.DefaultExt = ".txt";
DialogResult result = save.ShowDialog();
if (result == DialogResult.OK)
{
using (StreamWriter SW = new StreamWriter(save.FileName))
{
SW.WriteLine(richTextBox1.Text);
SW.WriteLine(richTextBox2.Text);
SW.WriteLine(richTextBox3.Text);
SW.WriteLine(richTextBox4.Text);
SW.WriteLine(richTextBox5.Text);
SW.Close();
}
}
Just set the FileName property on your SaveFileDialog.
Add
save.FileName = String.Format("{0}.txt", richTextBox1.Text);
Before you call ShowDialog.
Related
Still new to C#, snipping some code around to write a simple application and learning while doing.
I have an xml file that needs to be ingested and set as a variable so I can have it just insert words from that text file into different text fields.
private void button1_Click(object sender, EventArgs e)
{
// Displays an OpenFileDialog so the user can select a datafeed.
OpenFileDialog openFileDialog1 = new OpenFileDialog();
openFileDialog1.Filter = "Datafeed File|*.dfx5";
openFileDialog1.Title = "Select a dfx5";
// Show the Dialog.
// If the user clicked OK in the dialog and
if (openFileDialog1.ShowDialog() == System.Windows.Forms.DialogResult.OK)
{
// Assign the file as a variable.
}
}
How would I make that file a variable so that I can read from it?
Thank you in advanced. Google-ing didn't return anything helpful
How would I make that file a variable so that I can read from it?
Well, the file name that was selected is a property of the OpenFileDialog object:
string fileName;
// Show the Dialog.
// If the user clicked OK in the dialog and
if (openFileDialog1.ShowDialog() == System.Windows.Forms.DialogResult.OK)
{
// Assign the file as a variable.
fileName = openFileDialog1.FileName;
}
What you do with that file name at that point is up to you.
If you want to store the Filename in a variable, the other answers are what you are looking for.
To me it sounds like you need to actually read the content of the file.
If that's what you want, the following snippet (provided by Microsoft) should do:
try
{ // Open the text file using a stream reader.
using (StreamReader sr = new StreamReader(openFileDialog1.FileName))
{
// Read the stream to a string, and write the string to the console.
String line = sr.ReadToEnd();
Console.WriteLine(line);
}
}
catch (Exception e)
{
Console.WriteLine("The file could not be read:");
Console.WriteLine(e.Message);
}
This way line will contain the XML data you need. How you proceed from there is up to you.
If the file is pure XML, then I'd be inclined to do something like this:
using System.Xml.Linq;
private XDocument _xmlPayload;
private void button1_Click(object sender, EventArgs e)
{
// Displays an OpenFileDialog so the user can select a datafeed.
OpenFileDialog openFileDialog1 = new OpenFileDialog();
openFileDialog1.Filter = "Datafeed File|*.dfx5";
openFileDialog1.Title = "Select a dfx5";
// Show the Dialog.
// If the user clicked OK in the dialog and
var dialogResult = openFileDialog1.ShowDialog();
if (dialogResult == System.Windows.Forms.DialogResult.OK)
{
//Get file path from dialog
var filePath = openFileDialog1.FileName;
//load xml
using(var stream = File.OpenRead(filePath))
{
_xmlPayload = XDocument.Load(stream);
}
}
}
Then it's up to you how you work with the XML.
openFileDialog1.FileName should have the returned filename from the dialog. if multiselect is enabled I think its openFileDialog1.FileNames
How to make user control in Windows application c#
I need make attachments files in form but I need use an user control when click
button browse and choose the files or image add user control in form ?
Add a button on the form and use OpenFileDialog, like that:
private void buttonGetFile_Click(object sender, EventArgs e)
{
OpenFileDialog dialog = new OpenFileDialog();
dialog.Filter = "Text files | *.txt"; // file types, that will be allowed to upload
dialog.Multiselect = false; // allow/deny user to upload more than one file at a time
if (dialog.ShowDialog() == DialogResult.OK) // if user clicked OK
{
String path = dialog.FileName; // get name of file
using (StreamReader reader = new StreamReader(new FileStream(path, FileMode.Open), new UTF8Encoding())) // do anything you want, e.g. read it
{
// ...
}
}
}
I want to create a file in a directory selected by the user and named it by the user input.
I tried FolderBrowserDialog but it didn't prompt me to give a file name:
FolderBrowserDialog fbd = new FolderBrowserDialog();
DialogResult result = fbd.ShowDialog();
string path = fbd.SelectedPath;
//string FileName; then concatenate it with the path to create a new file
how can I do that?
You want to create a new file in a folder, so you should:
ask the user to select a folder (with FolderBrowserDialog)
offer the user a way to type a file name, with an input field (separate from the folder dialog)
Then you concat those 2 infos to get your full file name.
Or you can use SaveFileDialog and check if the file already exists when the user has selected a file (with a File.Exists...). There is a property for displaying an alert when the file does not exists, but not on the other side.
So when you got the DialogResult, use File.Exists and you can alert the user.
Sample for this solution:
In this sample (I hope without errors, cannot test right now):
- I open the saveFileDialog on a button called SaveButton with the SaveButton_Click click method
- I have a SaveFileDialog component on my form, called saveFileDialog1. On this component, the event FileOK is associated to my saveFileDialog1_FileOk method
private void SaveButton_Click(object sender, EventArgs e)
{
// Set your default directory
saveFileDialog1.InitialDirectory = #"C:\";
// Set the title of your dialog
saveFileDialog1.Title = "Save file";
// Do not display an alert when the user uses a non existing file
saveFileDialog1.CheckFileExists = false;
// Default extension, in this sample txt.
saveFileDialog1.DefaultExt = "txt";
if (saveFileDialog1.ShowDialog() == DialogResult.OK)
{
// DO WHAT YOU WANT WHEN THE FILE AS BEEN CHOSEN
}
}
// This method handles the FileOK event. It checks if the file already exists
private void saveFileDialog1_FileOk(object sender, System.ComponentModel.CancelEventArgs e)
{
if (File.Exists(saveFileDialog1.FileName))
{
// The file already exists, the user must select an other file
MessageBox.Show("Please select a new file, not an existing one");
e.Cancel = true;
}
}
When I open a file using this code
if (ofd.ShowDialog() == DialogResult.OK)
text = File.ReadAllText(ofd.FileName, Encoding.Default);
A window appear and ask me to choose file (The File Name is blank as you can see on the image)
If I press second time the button Open to open a file the File Name show the path of the previous selected file (see on image) How I can clear this path every time he press Open button?
You are probably using the same instance of an OpenFileDialog each time you click the button, which means the previous file name is still stored in the FileName property. You should clear the FileName property before you display the dialog again:
ofd.FileName = String.Empty;
if (ofd.ShowDialog() == DialogResult.OK)
text = File.ReadAllText(ofd.FileName, Encoding.Default);
try this:
ofd.FileName = String.Empty;
You need to reset the filename.
openFileDialog1.FileName= "";
Or
openFileDialog1.FileName= String.Empty()
you can simply add this line before calling ShowDialog():
ofd.FileName = String.Empty;
To clear just the filename (and not the selected path), you can set the property FileName to string.Empty.
private void button1_Click(object sender, EventArgs e)
{
openFileDialog1.ShowDialog();
}
private void openFileDialog1_FileOk(object sender, CancelEventArgs e)
{
label1.Text = sender.ToString();
}
What about this one.
I have a quick question, how do you save a file in a different format like in "save as"
so far i got this
private void saveToolStripMenuItem_Click(object sender, EventArgs e)
{
//this saves the file as a text or richtext.
saveFileDialog1.Filter = ("RichText*.rtf; )|*.rtf; |TextDocs *.txt;|*.txt");
saveFileDialog1.FilterIndex = 2;
//this gives the title of the savefiledialog.
saveFileDialog1.Title = "save file";
//this prompts the user if they want to overwrite an existing file.
saveFileDialog1.OverwritePrompt = true;
//gets the input made by the savefiledialog.
if (saveFileDialog1.ShowDialog() == System.Windows.Forms.DialogResult.OK)
{
//saves the file.
richTextBox1.SaveFile(saveFileDialog1.FileName,
//saves the text in the richbox
RichTextBoxStreamType.RichText);
I want to be able to save as ether a rtf or a txt format. thanks.
Use filename different name and pass it to SaveFile with the read content buffer from origianl file.