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
Related
I'm developing a tool that processes an .fbx model and user input into a single file for use in a game. The code for when the user presses the "Import Model" button is as follows, and is similar for every button:
private void E_ImportModelButton_Click_1(object sender, EventArgs e)
{
E_model = null; // byte array where model is stored
E_SelectedFileLabel.Text = "No Model Selected"; // label on form
OpenFileDialog ofd = new OpenFileDialog();
ofd.Filter = "FBX Model (.fbx)|*.fbx";
ofd.Multiselect = false;
if (ofd.ShowDialog() == DialogResult.OK)
{
// adjusts variables for game file
string s = Path.GetDirectoryName(ofd.FileName);
E_model = File.ReadAllBytes(s);
E_SelectedFileLabel.Text = "File Selected: " + ofd.FileName;
}
}
The problem is, whenever I click OK, an UnauthorizedAccessException occurs. I have tried importing files from C:\Users\Owner\Downloads as well as C:\Users\Owner\Desktop and the C:\ drive itself, but it still occurs. What could I add to this code to gain access to these (and other) folders?
You are trying to read from directory via the method intended to read from a file:
string s = Path.GetDirectoryName(ofd.FileName);
E_model = File.ReadAllBytes(s);
Replace it with:
E_model = File.ReadAllBytes(ofd.FileName);
You can't ready the directory, you have to read a file:
string s = Path.GetDirectoryName(ofd.FileName);
E_model = File.ReadAllBytes(s);
Try adding the file name here
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 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.
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.