I am making a basic word processor in c# for training. Now I am making the open file part of it. I can open a text file without issue. But when I open the open file dialog and then cancel it, it crashes :/
private void openToolStripMenuItem_Click(object sender, EventArgs e)
{
openFileDialog1.ShowDialog();
var OpenFile = new System.IO.StreamReader(openFileDialog1.FileName);
getRichTextBox().Text = OpenFile.ReadToEnd();
}
I know it is because the streamreader has nothing to read but I am not sure how to solve this.
thanks in advance!
Edit: Thank you! it worked perfectly :)
You need to check the result of the dialog:
private void openToolStripMenuItem_Click(object sender, EventArgs e)
{
if (openFileDialog1.ShowDialog() == DialogResult.OK) {
using (var openFile = new StreamReader(openFileDialog1.FileName)) {
getRichTextBox().Text = OpenFile.ReadToEnd();
}
}
}
I've also added a using statement to ensure that your file is closed when you're done reading it.
You can simplify the code even further by simply using File.ReadAllText instead of messing around with StreamReader.
getRichTextBox().Text = File.ReadAllText(openFileDialog1.FileName);
(Thanks #keyboardP)
if (openFileDialog1.ShowDialog() == DialogResult.OK)
{
var OpenFile = new System.IO.StreamReader(openFileDialog1.FileName);
getRichTextBox().Text = OpenFile.ReadToEnd();
}
Related
I'm trying to make a save option on my menu strip that should save whats drawn on the picture Box, currently I'm making in with SaveFileDialog but in the end I am saving a file which I cannot open with any photo viewer.
Here's the code:
private void saveAsToolStripMenuItem_Click(object sender, EventArgs e)
{
if (saveFileDialog1.ShowDialog() == DialogResult.OK)
{
using (Stream s = File.Open(saveFileDialog1.FileName, FileMode.CreateNew))
using(StreamWriter swr=new StreamWriter(s))
{
swr.Write(pbCanvas.Image);
}
}
}
https://pastebin.com/mU8TXPgW
Maybe this would work:
pictureBox.Image.Save(#"FilePath",ImageFormat.Jpeg);
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
Okay so i wanted to know how i would take something that is in a textBox, then i press a button, the contents of the textBox will be saved to a file location, then when i load the .exe back up, the contents will reappear in the textBox.
This is what i have so far
private void button3_Click(object sender, EventArgs e)
{
File.WriteAllText(#"C:\Application.txt", textBox1.Text);
}
^To Write it to a file location, I have tried this multiple times but it doesnt seem to want to make the file on my C:.
private void Form1_Load(object sender, EventArgs e)
{
try
{
textBox1.Text = File.ReadAllText(#"C:\Application.txt", Encoding.ASCII);
}
catch
{
}
^To Load the file then inject it back into the textbox it came from
Any and all help is appreciated,
Thanks.
You likely get an exception when trying to write to your C drive because it requires administrative access. Try running Visual Studio as Administrator (therefor the app will run as admin when kicked off from VS) or try writing to another location. Your code is all fine. The Encoding.ASCII bit is unnecessary though and I recommend removing it (more than likely that's not the encoding you will write the file in).
Trying to write directly to the C: drive can cause problems.
Try writing to a location that you definitely have write access to. You could use the ApplicationData directory (for application files unique to the current user), or use SpecialFolder.MyDocuments if you prefer.
private string applicationFilePath = Path.Combine(Environment.GetFolderPath(
Environment.SpecialFolder.ApplicationData), "Application.txt");
private void button3_Click(object sender, EventArgs e)
{
File.WriteAllText(applicationFilePath, textBox1.Text);
}
private void Form1_Load(object sender, EventArgs e)
{
textBox1.Text = File.ReadAllText(applicationFilePath, Encoding.ASCII);
}
I would do something like this:
using System.IO;
private void button3_Click(object sender, EventArgs e)
{
using (var stream = new FileStream(#"C:\Application.txt", FileMode.Create))
using (var writer = new StreamWriter(stream, Encoding.ASCII))
writer.Write(textBox1.Text);
}
private void Form1_Load(object sender, EventArgs e)
{
using (var stream = new FileStream(#"C:\Application.txt", FileMode.Open))
using (var reader = new StreamReader(stream, Encoding.ASCII))
textBox1.Text = reader.ReadToEnd();
}
I think this method gives you more control on your content. Try to explore the contents of FileMode enum, and make sure to add your using System.IO; directive.
Be careful not to confuse the using statement with the using directive.
Also, always remember to dispose/close your stream when done to ensure that the data has been flushed and that the file is no longer in use by your application. Here, the using statement does the job of disposing when the stream is no longer in use.
EDIT: As mentioned by the other posts, writing to the C: directory causes problems on newer operating systems due to Admininstrative Access restrictions. Make sure to write to different drives/folders that you definitely have access to.
// Current User
Path.Combine(Environment.GetFolderPath(Environment.SpecialFolder.ApplicationData), "Application.txt");
// All users
Path.Combine(Environment.GetFolderPath(Environment.SpecialFolder.CommonApplicationData), "Application.txt");
Our primary database application has a feature that exports either an Excel workbook or an Access mdb for specified work orders. Those files are then sent to our subcontractors to populate with the required data. I am building an application that connects to the files and displays the data for review prior to being imported into the primary database. Simple enough, right? Here’s my problem: the application opens a OpenFileDialog box for the user to select the file that will be the datasource for the session. That works perfectly. If I open a MessageBox after it, that box open up behind any other open windows. After that, they respond correctly. I only expect to be using MessageBoxes for error handling, but the problem is perplexing. Has anyone encountered this problem?
The MessageBoxes in the code are only to verify that the path is correct and to solve this problem; but, here’s my code:
private void SubContractedData_Load(object sender, EventArgs e)
{
string FilePath;
string ext;
OpenFileDialog ofd = new OpenFileDialog();
ofd.Filter = "Microsoft Access Databases |*.mdb|Excel Workbooks|*.xls";
ofd.Title = "Select the data source";
ofd.InitialDirectory = ElementConfig.TransferOutPath();
if (ofd.ShowDialog() == DialogResult.OK)
{
FilePath = ofd.FileName.ToString();
ext = FilePath.Substring((FilePath.Length - 3));
if (ext == "xls")
{
MessageBox.Show(FilePath);
AccessImport(FilePath);
}
else if (ext == "mdb")
{
MessageBox.Show(FilePath);
AccessImport(FilePath);
}
}
else
{
System.Windows.Forms.Application.Exit();
}
}
While it isn't advisable to use MessageBoxes to debug your code, I think the immediate problem is that you are doing this in the form's load event.
Try it like this:
protected override void OnShown(EventArgs e) {
base.OnShown(e);
// your code
}
Your problem is definitely the fact that you're trying to do this while loading the Form, because the form isn't yet displayed.
Another alternative would be moving this functionality out of the Load event and give the user a button to push or something like that.
In other words:
private void SubContractedData_Load(object sender, EventArgs e)
{
// unless you're doing something else, you could remove this method
}
Add a button that handles this functionality:
private void SelectDataSourceClick(object sender, EventArgs e)
{
OpenFileDialog ofd = new OpenFileDialog();
ofd.Filter = "Microsoft Access Databases |*.*|Excel Workbooks|*.xls";
ofd.Title = "Select the data source";
ofd.InitialDirectory = ElementConfig.TransferOutPath();
if (ofd.ShowDialog() == DialogResult.OK)
{
var FilePath = ofd.FileName.ToString();
var ext = Path.GetExtension(FilePath).ToLower();
switch (ext)
{
case ".xls":
MessageBox.Show(FilePath);
AccessImport(FilePath);
break;
case ".mdb":
MessageBox.Show(FilePath);
AccessImport(FilePath);
break;
default:
MessageBox.Show("Extension not recognized " + ext);
break;
}
}
else
{
System.Windows.Forms.Application.Exit();
}
}
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.