Develop Reference

WPF How to get second clicked file path for Single Instance App

I created my WPF single instance app by using the Microsoft.VisualBasic dll method. However I'm facing some difficulty to get the file path for second clicked file which associated with my app.
For example, I have two file "First.my" and "Second.my". When I click on file "First.my" it will launch my app and pop up message box to show "First.my" file path. Since my app is single instance app, when I click on file "Second.my" it should show the file path for "Second.my" but it still showing the file path for "First.my"..
Does anyone know how to pass the associate file path in single instance app?
Below is my code:
class WindowsFormsApp : Microsoft.VisualBasic.ApplicationServices.WindowsFormsApplicationBase
{
private App _wpfApp;
public WindowsFormsApp()
{
IsSingleInstance = true;
}
protected override bool OnStartup(Microsoft.VisualBasic.ApplicationServices.StartupEventArgs e)
{
MessageBox.Show("First File");
//Get 1st click file path
GetFilePath();
_wpfApp = new App();
_wpfApp.Run();
return false;
}
protected override void OnStartupNextInstance(Microsoft.VisualBasic.ApplicationServices.StartupNextInstanceEventArgs e)
{
MessageBox.Show("Second File");
//Get 2nd click file path
GetFilePath();
}
protected void GetFilePath()
{
if (AppDomain.CurrentDomain.SetupInformation.ActivationArguments.ActivationData != null &&
AppDomain.CurrentDomain.SetupInformation.ActivationArguments.ActivationData.Length > 0)
{
var filePath = AppDomain.CurrentDomain.SetupInformation.ActivationArguments.ActivationData[0];
var uri = new Uri(filePath);
MessageBox.Show(uri.LocalPath);
}
}
}

How to move wpf application into minimize tray at Window Start-up C#?

I have created setup of my application using Windows Installer.
Now I want to Start application at Windows Start-Up and move it system minimize tray as i don't want to display GUI(View) at Windows Start-Up.
I have searched in Google and i found to use Registry key But that is not enough for me as i also want to move to system minimize tray and application run.
My purpose to do it is, user do not feels annoying when application starts every time when he/she starts system.
Can anyone have answer?
Thanks..

In your application, add an event handler for the FrameworkElement.Loaded event. In that handler, add the following code:
WindowState = WindowState.Minimized;
This will minimise the application when it starts.
To start the application when the computer starts, you'll need to add your program into Windows Scheduler and set it to run at startup. You can find out more on the Schedule a task page at MSDN.

You also have to set this property to remove it from the taskbar
ShowInTaskbar= false;

Maybe this answer is late, but I still want to write it down to help those who haven't found solutions yet.
Firstly you need to add a function to minimize your app to tray when it autostarts as system startup.
In your App.xaml file, change the original StartupUri=... to Startup="App_Startup" as below. App_Startup is your function name and can be changed.
<Application x:Class="Yours.App"
xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
Startup="App_Startup">
In your App.xaml.cs file. Add the function below:
public partial class App : Application
{
private void App_Startup(object sender, StartupEventArgs e)
{
// Process command line args
var isAutoStart = false;
for (int i = 0; i != e.Args.Length; ++i)
{
if (e.Args[i] == "/AutoStart")
{
isAutoStart = true;
}
}
// Create main application window, starting minimized if specified
MainWindow mainWindow = new MainWindow();
if (isAutoStart)
{
mainWindow.WindowState = WindowState.Minimized;
}
mainWindow.OnAutoStart();
}
}
In your MainWindow.xaml.cs, add a function as below:
public void OnAutoStart()
{
if (WindowState == WindowState.Minimized)
{
//Must have this line to prevent the window start locatioon not being in center.
WindowState = WindowState.Normal;
Hide();
//Show your tray icon code below
}
else
{
Show();
}
}
Then you should set you app utostart as system start.
Now if you have a switch to decide whether you app to autostart as system start, you can just add the function below as your switch status changed event function.
private void SwitchAutoStart_OnToggled(object sender, RoutedEventArgs e)
{
const string path = #"SOFTWARE\Microsoft\Windows\CurrentVersion\Run";
var key = Registry.CurrentUser.OpenSubKey(path, true);
if (key == null) return;
if (SwitchAutoStart.IsOn)
{
key.SetValue("Your app name", System.Reflection.Assembly.GetExecutingAssembly().Location + " /AutoStart");
}
else
{
key.DeleteValue("Your app name", false);
}
}
If you want to automatically start the application for all users on Windows startup, just replace the forth line with
RegistryKey key = Registry.LocalMachine.OpenSubKey(path, true);
^_^

Restricting access to certain folders with a FolderBrowserDialog

I want to restrict what folder a person can choose to set their default save path in my app. Is there a class or method which would allow me to check access rights and either limit the user's options or show an error once they have made their selection. Is FileSystemSecurity.AccessRightType a possibility?

Since the FolderBrowserDialog is a rather closed control (it opens a modal dialog, does it stuff, and lets you know what the user picked), I don't think you're going to have much luck intercepting what the user can select or see. You could always make your own custom control, of course ;)
As for testing if they have access to a folder
private void OnHandlingSomeEvent(object sender, EventArgs e)
{
DialogResult result = folderBrowserDialog1.ShowDialog();
if(result == DialogResult.OK)
{
String folderPath = folderBrowserDialog1.SelectedPath;
if (UserHasAccess(folderPath))
{
// yay! you'd obviously do something for the else part here too...
}
}
}
private bool UserHasAccess(String folderPath)
{
try
{
// Attempt to get a list of security permissions from the folder.
// This will raise an exception if the path is read only or do not have access to view the permissions.
System.Security.AccessControl.DirectorySecurity ds =
System.IO.Directory.GetAccessControl(folderPath);
return true;
}
catch (UnauthorizedAccessException)
{
return false;
}
}
I should note that the UserHasAccess function stuff was obtained from this other StackOverflow question.

C# Passing Arguments Through Context Menu

so I made a program that creates a registry key on the pc, that adds an option to push your file to your device. (The program itself pushes files to your Android sd card.)
Everything works well, or at least should, once I figure out this issue. When I right-click on an item and choose my command (which is to open up with my program) my program just opens up, but no arguments are passed.
A month ago, I designed a program that opens when the user double clicks (just double click, not right-clicking through context menu) a file, it opened the specific program, and from there I was able to grab the arguments.
Why can't I from the context menu?
Thanks, sorry for lengthy post.
And here's part of my code:
private void Form1_Load(object sender, EventArgs e)
{
string c = textBox3.Text;
string[] args = System.Environment.GetCommandLineArgs();
try
{
string location = args[1];
MessageBox.Show(location);
//For testing purposes only
Properties.Settings.Default.thinglocation = location;
Properties.Settings.Default.Save();
Process();
}
catch
{
}
Here's the registry code. BTw string c is "C:\File push.exe" the location of my program.
string MenuName = "*\\shell\\NewMenuOption";
string Command = "*\\shell\\NewMenuOption\\command";
RegistryKey regmenu = null;
RegistryKey regcmd = null;
try
{
regmenu = Registry.ClassesRoot.CreateSubKey(MenuName);
if (regmenu != null)
regmenu.SetValue("", "Push to Android");
regcmd = Registry.ClassesRoot.CreateSubKey(Command);
if (regcmd != null)
regcmd.SetValue("", c +"%1");
}
catch (Exception ex)
{
MessageBox.Show(this, ex.ToString());
}
finally
{
if (regmenu != null)
regmenu.Close();
if (regcmd != null)
regcmd.Close();

I assume that you have created a registry entry which creates the context menu item which you are using to launch your program. Likelihood is that you need to add ' %1' after the name of the exe you want to launch. In the context menu registry entries %1 represents the selected file, so adding this should launch your program and provide the name of the selected file as the single argument.
the reason why it worked before with the double click is that windows will launch the default program for the file double clicked, with the double clicked file as the argument.

Did you try looking in the sender or the EventArgs parameters? I would think their might be something in there. That's just a guess though. If its just a double click then why would there be any arguments? Maybe I am just misunderstanding that point. Anyways, maybe that helps. If not maybe you could clarify a little bit where the arguments are supposed to come from.
Erik

C# Windows Forms: OpenFileDialog Strange Issues

This is kinda strange, let me try to explain it as best as possible:
When I create a new file and Save it, it saves correctly (test.xml).
When I make changes to this file and Save it, it saves correctly (to test.xml)
When I make changes again to this file or just choose Save As, it works correctly (newtest.xml)
However, when I do a file open, make changes to a file (test.xml) and click Save it is saving to (newtest.xml).
This is in my MainForm.cs
if (this.openEditorDialog1.ShowDialog(this) == DialogResult.OK && editForm != null)
{
editForm.Close();
editForm = new EditorForm(this);
editForm.OpenFile(this.openEditorDialog1.FileName);
editForm.Closing += new CancelEventHandler(EditorForm_Closing);
editForm.MdiParent = this;
editForm.Show();
}
private void biFileSave_Click(object sender, EventArgs e)
{
if (!editForm.HasFileName)
{
if (this.saveEditorDialog1.ShowDialog(this) == DialogResult.OK)
{
this.ActiveDiagram.SaveSoap(this.saveEditorDialog1.FileName);
editForm.FileName = this.saveEditorDialog1.FileName;
}
}
else
{
this.ActiveDiagram.SaveSoap(this.saveEditorDialog1.FileName);
}
This is in my EditorForm.cs
public void OpenFile(string strFileName)
{
diagramComponent.LoadSoap(mainForm.openEditorDialog1.FileName);
this.FileName = mainForm.openEditorDialog1.FileName;
this.tabControl1.SelectedTab = DiagramTab;
}
I'm sure it has to do with the what I'm doing in the EditoForm but I can't seem to figure it out.

else
{
this.ActiveDiagram.SaveSoap(this.saveEditorDialog1.FileName);
It looks like you want:
this.ActiveDiagram.SaveSoap(editForm.FileName);

It must have to do with mainForm.openEditorDialog1.FileName. Use a FileName property of the form that does the saving. When you open the file, set the fileName to mainForm.openEditorDialog1.FileName. When you SaveAs, set the FileName property there, too. This way, whenever the current file, changes you set the FileName property appropriately. Then, when it comes time to save the file, you always have the correct filename.
In summary, only use the .FileName property of the SaveAs dialog or the FileOpen dialog right after you use them.