FileDialog select target file? - c#

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()

Related

Using textbox to gather user's destination path for exporting a CSV in C# Windows application?

I'm looking for an option that allows a user to input their destination folder (Usually copy/Paste) into a text box on a Windows Application, and append the file's name with a specific name. However, I'm stuck as to how to accomplish this.
//Exporting to CSV.
string folderPath = MessageBox.Show(textBox1_TextChanged);
File.WriteAllText(folderPath + "DIR_" + (DateTime.Now.ToShortDateString()) + ".csv", csv);
So it can look like: C:/DIR_9132017.csv, or Documents/DIR_9132017.csv, depending on what the user inputs into the textbox. I have nothing in my textbox code section at the moment, if that also gives a clearer picture about the situation.
Any help will be greatly appreciated. Thank you!
You can either use FolderBrowserDialog or can just copy/paste the path and create the directory.
Using FolderBrowserDialog
Step1 : Create Browse button (so that the user can choose the directory)
Step2 : Create Export button (Place the code to write the file here)
private void browseButton_Click(object sender, EventArgs e)
{
if (folderBrowserDialog1.ShowDialog() == DialogResult.OK)
{
var folderPath = folderBrowserDialog1.SelectedPath;
textBox1.Text = folderPath;
}
}
private void exportToCsvButton_Click(object sender, EventArgs e)
{
var path = textBox1.Text;
var file = Directory.CreateDirectory(path);
var filename = "DIR_" + (DateTime.Now.ToShortDateString()) + ".csv";
File.WriteAllText(Path.Combine(path, "test.csv"), "content");
}
Using Copy/Paste
Step1 : Create Export button (User copy pastes the path. System create the directory and writes the file)
private void exportToCsvButton_Click(object sender, EventArgs e)
{
var path = textBox1.Text;
var file = Directory.CreateDirectory(path);
var filename = "DIR_" + (DateTime.Now.ToShortDateString()) + ".csv";
File.WriteAllText(Path.Combine(path, "test.csv"), "content");
}
You would use a FolderBrowserDialog for that. After adding it to your form, you can use it like this:
public void ChooseFolder()
{
if (folderBrowserDialog1.ShowDialog() == DialogResult.OK)
{
folderPath = folderBrowserDialog1.SelectedPath;
}
}
Source: https://msdn.microsoft.com/en-us/library/aa984305(v=vs.71).aspx
You haven't specified whether it's WinForms or WPF, so I'm using WPF, but is should be the same in WinForms.
To select the folder, you could use FolderBrowseDialog as follows. This code should be placed inside a button or some other similar control's Click event.
if (folderBrowse.ShowDialog() == DialogResult.OK)
{
txtPath.Text = folderBrowse.SelectedPath;
}
txtPath being a TextBox your selected path displayed in. You can substitute it with a simple string variable if you don't want to use a TextBox.
And if you want the user to be able to drag-drop a folder into a TextBox, you can create a TextBox control, and in it's PreviewDragOver and Drop events, you can do the following.
private void txtPath_PreviewDragOver(object sender, DragEventArgs e)
{
e.Handled = true;
}
private void txtPath_Drop(object sender, DragEventArgs e)
{
string[] files = (string[])e.Data.GetData(DataFormats.FileDrop);
if (files != null && files.Length != 0)
txtPath.Text = files[0];
}
Using both above in combination user has the ability to either select the folder he/she wants by clicking a button, drag-and-drop a folder into the TextBox, or copy-paste the folder path into the TextBox.

C# OpenFileDialog opening twice

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.

How do I show a directory path in a textbox when selected by the user?

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;

Executing another Application with selected file using a button (C# Programming)

To start, I'm a newbie with C# Programming and I've tried googling for solutions about my problem but it seems that i cannot find one, or simply too unlucky or too blind to spot one. I'm using Microsoft Visual Studio 2005.
Anyways. I was assigned to modify/create an automated test environment input application. The said application already has a function to run/start a CANoe program with a pre-defined file or if it's already running, stop the program.
private void button1_Click(object sender, EventArgs e)
{
// Execute CANoe(Obtain CANoe application objectg)
mApp = new CANoe.Application();
mMsr = (CANoe.Measurement)mApp.Measurement;
try
{
mApp.Open("C:\\Users\\uidr3024\\Downloads\\SRLCam4T0_Validation_ControlTool\\cfg\\SVT_SRLCam4T0_025B.cfg", true, true);
}
catch (System.Exception ex)
{
System.Console.WriteLine(ex.Message);
}
}
private void button2_Click(object sender, EventArgs e)
{
// Finish CANoe
if (mApp != null) {
mApp.Quit();
}
// Release the object
fnReleaseComObject(mMsr);
fnReleaseComObject(mApp);
}
What I wanted to do now is to have an OpenFileDialog box that will show a selection of files and user will be able to browse and select any file to start the CANoe program with the selected file and not just the file path that's been input in the code along the "mApp.Open()" syntax. I've tried this:
private void button5_Click_1(object sender, EventArgs e)
{
OpenFileDialog openFileDialog1 = new OpenFileDialog();
openFileDialog1.InitialDirectory = #"C:\Users\uidr3024\Downloads\SRLCam4T0_Validation_ControlTool\cfg";
openFileDialog1.Title = "Browse Configuration Files";
openFileDialog1.CheckFileExists = true;
openFileDialog1.CheckPathExists = true;
openFileDialog1.Filter = "CANalyzer/CANoe Configuration (*.cfg)|*.cfg |All files (*.*)|*.*";
openFileDialog1.FilterIndex = 1;
openFileDialog1.RestoreDirectory = true;
if (openFileDialog1.ShowDialog() == DialogResult.OK)
{
textBox1.Text = openFileDialog1.FileName;
}
}
I've tried this code that I often see in the Web and in the Tutorials but I don't know how to incorporate it with the button for running the CANoe program so that when user clicks the Open button from the Dialog Box, the file path would be shown in the textbox(optional) and/or when the user then clicks the Start CANoe, CANoe program would start with the selected .cfg file.
Am I making any sense here? Or am I doing the right thing here?
Btw, I found these... And I'm using a CANoe library for these, ofc.
#region "***** CANoe Object definition *****"
private CANoe.Application mApp = null; // CANoe Application CANoeƒAƒvƒŠƒP[ƒVƒ‡ƒ“
private CANoe.Measurement mMsr = null; // CANoe Mesurement function CANoe‘ª’è‹#”\
private CANoe.Variable mSysVar = null; // System variable ƒVƒXƒeƒ€•Ï”
private CANoe.Variable mSysVar_start = null; // System variable ƒVƒXƒeƒ€•Ï”
#endregion
I think you have done most of the hard work, unless I've missed something I think you just need something like this in your button1_Click method:
if( textBox1.Text != String.Empty && System.IO.File.Exists(textBox1.Text) )
{
// The textbox has a filename in it, use it
mApp.Open(textBox1.Text, true, true);
}
else
{
// The user hasn't selected a config file, launch with default
mApp.Open("C:\\Users\\uidr3024\\Downloads\\SRLCam4T0_Validation_ControlTool\\cfg\\SVT_SRLCam4T0_025B.cfg", true, true);
}

c# Openfiledialog

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.

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