I am having a problem with a windows form coded in Visual Studio using C# as part of a web development course. When I click on the menu item ('Import') the file dialogue box opens twice.
The first dialogue seems incorrect as it does not have the filter or title applied. Then the second dialogue box which immediately opens afterward has the correct filter and title applied.
Obviously it is only meant to open once to enable me to read the selected file and add it to the Listbox. This works, but only once I have selected the file twice.
Thanks in advance for your help.
private void importToolStripMenuItem_Click(object sender, EventArgs e)
{
openFileDialog1.ShowDialog();
openFileDialog1.Filter = "Text|*.txt";
openFileDialog1.Title = "Choose a file to import";
if (openFileDialog1.ShowDialog() == DialogResult.Cancel)
{
MessageBox.Show("Oh. No file selected!");
}
else
{
string usersFile = openFileDialog1.FileName;
string[] lines = File.ReadAllLines(usersFile);
foreach (string line in lines)
{
groceryList.Items.Add(line);
}
}
Start of your code you have
private void importToolStripMenuItem_Click(object sender, EventArgs e)
{
openFileDialog1.ShowDialog();
this is unneccessary, the ShowDialog will show a dialog with the filters not correctly set up as your next lines have:
openFileDialog1.Filter = "Text|*.txt";
openFileDialog1.Title = "Choose a file to import";
Then you ShowDialog again.
Therefore, just remove the first
openFileDialog1.ShowDialog();
and you should be fine.
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
So!
A user right clicks any file on their computer and a custom made option opens up my program, on Form_load I ask the user to choose a file however, I want that file to be the file they opened the program with.
private void Form1_Load(object sender, System.EventArgs e)
{
DialogResult result = openFileDialog1.ShowDialog();
if (result == DialogResult.OK)
{
source = openFileDialog1.FileName;
}
else
{
Application.Exit();
}
}
How could I achieve this?
When your program is associated with any file extension, Windows will pass the file name via command line parameter, so you can examine them:
private void Form1_Load(object sender, System.EventArgs e) {
String[] data = Environment.GetCommandLineArgs();
// 0th parameter is executable itself, we want the 1st parameter
if (data.Length > 1)
openFileDialog1.FileName = data[1];
if (openFileDialog1.ShowDialog() == DialogResult.OK) {
source = openFileDialog1.FileName;
}
else
Application.Exit();
}
If you asking about 'Send to' menu in Windows context menu -> You should put your application in special folder as described here: How-to geek: Customize the Send To Menu in Windows 10, 8, 7, or Vista and you will receive the full path including file name as first parameter in args in function Main()
First, I am using Visual Studio 2013 and coding in C# to develop a Windows Form Application. I have added the "System.IO" namespace.
I need to show a directory path in a textbox when selected by the user.
The code works correctly to where the user selects a folder from a popup
and presses the OK button, which then displays the number of files within
that folder -- but the folder path does NOT get displayed as I desired.
Code looks like this:
private void button1_Click(object sender, EventArgs e)
{
//
// This event handler was created by clicking the button in the application GUI.
//
DialogResult button1_Click = folderBrowserDialog1.ShowDialog();
if (button1_Click == DialogResult.OK)
{
//
// The user selected a folder and pressed the OK button.
// A message pops up and identifies the number of files found within that folder.
//
string[] files = Directory.GetFiles(folderBrowserDialog1.SelectedPath);
MessageBox.Show("Files found: " + files.Length.ToString(), "Message");
}
}
private void textBox1_TextChanged(object sender, EventArgs e)
{
string path;
path = folderBrowserDialog1.SelectedPath;
// folderBrowserDialog1.ShowDialog(); // NOT SURE ABOUT USING THIS!
textBox1.Text = path;
}
You could just add this to the end of your button1_Click method (inside the if block):
textBox1.Text = folderBrowserDialog1.SelectedPath;
I've created a log in panel in which I've used Transparent group box(with user name text box and password text box), and used a wallpaper on background, now I've Used a link label on this log in panel by clicking on which the user can change the background wallpaper of the log in panel.
Means when user clicks on the link label (lnklblChangeBackGround) with text "Click Here to Change Background" open dialogue box will open and user can select the Wallpaper from here and then by clicking on OK or Select the wallpaper will be assigned to background of the log in panel
Can any one help me out that
how can i open a open dialogue box by clicking on the link label
how can i assign a select wallpaper to the background of my log in panel
Note: I'm Creating this using VS 2010 using C#. and it's a desktop App and i'm using winform here.
At first you have to add an Event (LinkClicked) to your Link Label.
Just place this code here to open a file dialog.
private String getPicture()
{
string myPic = string.Empty;
OpenFileDialog openFileDialog1 = new OpenFileDialog();
openFileDialog1.Filter = "jpg (*.jpg)|*.jpg|png (*.png)|*.png";
if (openFileDialog1.ShowDialog() == DialogResult.OK)
myPic = openFileDialog1.FileName;
return myPic;
}
You can edit the filter to avoid the user choosing images, which is not supported in your opinion.
With this code below you can set the Background Image of your pictureBox
private void setBackground(String picture)
{
pictureBox1.Image = null;
pictureBox1.Image = Image.FromFile(picture);
}
And the final version would look like this
private void linkLabel1_LinkClicked(object sender, LinkLabelLinkClickedEventArgs e)
{
String myFile = getPicture();
setBackground(myFile);
}
if this is too much code or too complicated for you, then you can just put it all in one function like this
private void linkLabel1_LinkClicked(object sender, LinkLabelLinkClickedEventArgs e)
{
string myPic = string.Empty;
OpenFileDialog openFileDialog1 = new OpenFileDialog();
openFileDialog1.Filter = "jpg (*.jpg)|*.jpg|png (*.png)|*.png";
if (openFileDialog1.ShowDialog() == DialogResult.OK)
myPic = openFileDialog1.FileName;
pictureBox1.Image = null;
pictureBox1.Image = Image.FromFile(myPic);
}
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.