I have an NSPanel. It's purpose is to sit on top of everything in all spaces and it does that just fine.
I don't want it to ever become firstResponder and take focus from another app.
I sub classed NSPanel like so
class SomePanelClass: NSPanel {
override var acceptsFirstResponder: Bool{
return false
}
}
When the window shows or I drag it around the app I was in still appears to have focus and it's name is in the menu but keystrokes do not register in that app until I click back into it's window.
Is there something else I can do to prevent my app and panel from stealing input?
You probably want to override canBecomeKeyWindow on your Panel subclass to return NO.
- (BOOL) canBecomeKeyWindow {
return NO;
}
Related
I have a window with some NSTextFields. When I click in one and edit the value and press return, I want the focus to go back to what it was before. I don't want the blue ring around the text field and I don't want further keystrokes going to that text field. I would have thought this would happen automatically.
I tried these, and none of them work
sender.resignFirstResponder()
sender.window?.makeFirstResponder(nil)
InspectorWindowController.window?.makeFirstResponder(nil)
AnotherWindowController.window?.becomeFirstResponder()
I'm doing these at the end of my IBAction associated with the text field. Maybe I have to do it from somewhere else?
Thanks
I figured this out. I guess the sent action is happening on another thread. So you have to call makeFirstResponder using Dispatch async.
DispatchQueue.main.async { //omg
sender.window?.makeFirstResponder(nil)
}
I needed to dismiss first responder in my SwiftUI macOS app and here what I found working in a way I need:
func controlTextDidEndEditing(_ obj: Notification) {
DispatchQueue.main.async {
guard let window = self.textField.window else {
return
}
// https://stackoverflow.com/questions/5999148/how-to-determine-whether-an-nssearchfield-nstextfield-has-input-focus
// We need to make sure that our text field is still first responder.
guard let textView = window.firstResponder as? NSTextView,
textView.delegate === self.textField else {
return
}
window.makeFirstResponder(nil)
}
}
I am currently writing a tvOS app. I've been detecting and overriding the menu button with tapRecognizer to switch between storyboards and other functions. My issue is when I am on my home screen and press menu it does not exit the app. Instead it remembers the last function I used when overriding the menu button and performs that function. Any thoughts on how to clear the tapRecognizer? Or a function that will exit the app?
I'm overriding the menu button with
in Storyboard1
tapRecognizer = [[UITapGestureRecognizer alloc]initWithTarget:self action:#selector(home)];
tapRecognizer.allowedPressTypes = #[[NSNumber numberWithInteger:UIPressTypeMenu]];
[self.view addGestureRecognizer:tapRecognizer];
in my home subroutine I send the user back to my home page storyboard. But from then on the menu button will not exit the app but send me back to storyboard1.
thanks,
SW
Instead of using your own gesture recognizer, override pressesBegan:
override func pressesBegan(presses: Set<UIPress>, withEvent event: UIPressesEvent?) {
if(presses.first?.type == UIPressType.Menu) {
// handle event
} else {
// perform default action (in your case, exit)
super.pressesBegan(presses, withEvent: event)
}
}
If you are using UIGestureRecognizer instead of responding to presses, all you need to do is to disable the recognizer:
tapRecognizer.enabled = NO;
So if no recognizer with UIPressTypeMenu is listening, tvOS suspends the app and displays the home screen. (I've tested this)
You have to override 2 methods to prevent exiting app by pressing Menu button.
Here is ready-to-use template:
override func pressesBegan(presses: Set<UIPress>, withEvent event: UIPressesEvent?) {
for press in presses {
switch press.type {
case .Menu:
break
default:
super.pressesBegan(presses, withEvent: event)
}
}
}
override func pressesEnded(presses: Set<UIPress>, withEvent event: UIPressesEvent?) {
for press in presses {
switch press.type {
case .Menu:
//Do some staff there!
self.menuButtonPressed()
default:
super.pressesEnded(presses, withEvent: event)
}
}
}
If you overwrite the menu button, the app won't be accepted:
EDIT: You can overwrite, but the menu button has to work as a back button to homescreen from the entry point of the app.
10.1 Details
The Menu button on the Siri Remote does not behave as expected in your
app.
Specifically, when the user launches the app and taps the Menu button
on the Siri remote, the app does not exit to the Apple TV Home screen.
Next Steps
Please revise your app to ensure that the Siri remote buttons behave
as expected and comply with the Apple TV Human Interface Guidelines.
It may be help you...
it is swift code.
let menuPressRecognizer = UITapGestureRecognizer()
menuPressRecognizer.addTarget(self, action: #selector(YourViewController.menuButtonAction(_:)))
menuPressRecognizer.allowedPressTypes = [NSNumber(integer: UIPressType.Menu.hashValue)]
self.view.addGestureRecognizer(menuPressRecognizer)
As per Apple's documentation, for custom press handling, we should override all four of these methods-
- (void)pressesBegan:(NSSet<UIPress *> *)presses withEvent:(nullable UIPressesEvent *)event NS_AVAILABLE_IOS(9_0);
- (void)pressesChanged:(NSSet<UIPress *> *)presses withEvent:(nullable UIPressesEvent *)event NS_AVAILABLE_IOS(9_0);
- (void)pressesEnded:(NSSet<UIPress *> *)presses withEvent:(nullable UIPressesEvent *)event NS_AVAILABLE_IOS(9_0);
- (void)pressesCancelled:(NSSet<UIPress *> *)presses withEvent:(nullable UIPressesEvent *)event NS_AVAILABLE_IOS(9_0);
This is the official documentation from XCode:
Generally, all responders which do custom press handling should
override all four of these methods.
Your responder will receive either pressesEnded:withEvent or pressesCancelled:withEvent: for each
press it is handling (those presses it received in pressesBegan:withEvent:).
pressesChanged:withEvent: will be invoked for presses that provide
an analog value
(like thumbsticks or analog push buttons)
*** You must handle cancelled presses to ensure correct behavior in
your application. Failure to
do so is very likely to lead to incorrect behavior or crashes.
SwiftUI Seekers:
I don't know how much this answer helps, but just adding available actions in SwiftUI.
YourAnyView
.onExitCommand(perform: {
print("onExitCommand")
})
.onMoveCommand { direction in
print("onMoveCommand", direction)
}
REF:
onmovecommand
onexitcommand
I added a childWindow of a custom subclass of NSWindow to a parentWindow (also a custom subclass of NSWindow). The childWindow has the NSBorderlessWindowMask and canBecomeKeyWindow: is overridden to return YES and canBecomeMainWindow: to return NO.
The childWindow is set to resize with the parentWindow. So I want to create the illusion that the views of the childWindow are part of the parentWindow. The main idea is to arrange the document windows created by the document-based application within a main window to provide a tabbed interface (just like in a browser) to switch between the documents.
My problem is that whenever I click in one of the views of the childWindow, the parentWindow (the main window) looses focus and the traffic light buttons are getting greyed out. This is obviously contrary to what I want to achieve.
I found this answer:
Make NSView in NSPanel first responder without key window status
But even if I override isKeyWindow: (of the main window) to always return YES, the title bar gets greyed out nonetheless when I click into the childWindow.
I also tried to follow this advice:
http://www.cocoabuilder.com/archive/cocoa/143945-non-focused-child-window.html
But I'm not sure what "include the child window in its responder chain just ahead of its nextResponder" means. With canBecomeKeyWindow: to return NO (for the childWindow), the views within the child never can become key and are always greyed out.
Any clue what I am doing wrong?
One addition: Is it possible to make the views in the childWindow FirstResponder without giving the childWindow key-status?
I got this working by mimicking the behaviour of NSPopover. On investigation I found that a popover (which uses a private NSPanel subclass "_NSPopoverWindow") believed it was the key & main window, even though it is not the window returned from [NSApp keyWindow].
Create your own custom NSPanel subclass, attach it to the parent window, and then override the following methods as so:
- (BOOL)isKeyWindow {
return YES;
}
- (BOOL)isMainWindow {
return YES;
}
- (BOOL)canBecomeKeyWindow {
return YES;
}
- (BOOL)canBecomeMainWindow {
return YES;
}
- (void)makeKeyWindow {
[super makeKeyWindow];
[self.parentWindow makeKeyWindow];
}
- (void)makeMainWindow {
[super makeMainWindow];
[self.parentWindow makeMainWindow];
}
- (void)becomeKeyWindow {
[super becomeKeyWindow];
}
- (void)becomeMainWindow {
[super becomeMainWindow];
[self.parentWindow becomeMainWindow];
}
- (void)resignMainWindow {
}
- (void)resignKeyWindow {
}
I am designing an application without a title bar, however when I remove the title bar using interface builder in Xcode 4 , it causes the editable fields (the ones I tried are textView, and textField) not being editable, despite editable checked in there properties? why this happens and is there anyway to prevent it?
You have to subclass your window and overwrite the following methods:
- (BOOL)canBecomeKeyWindow {
// because the window is borderless, we have to make it active
return YES;
}
- (BOOL)canBecomeMainWindow {
// because the window is borderless, we have to make it active
return YES;
}
Updated for Swift 4 and general tips about how to do this in 2018:
The canBecomeKeyWindow and canBecomeMainWindow methods no longer exist on an NSWindow. They have since been replaced by stored properties called canBecomeKey and canBecomeMain. Because they are stored, if you want to override them you can do so by making them computed properties, like this:
override var canBecomeKey: Bool {
return true
}
override var canBecomeMain: Bool {
return true
}
I have an NSMenu popping out of an NSStatusItem using popUpStatusItemMenu. These NSMenuItems show a bunch of different links, and each one is connected with setAction: to the openLink: method of a target. This arrangement has been working fine for a long time. The user chooses a link from the menu and the openLink: method then deals with it.
Unfortunately, I recently decided to experiment with using NSMenuItem's setView: method to provide a nicer/slicker interface. Basically, I just stopped setting the title, created the NSMenuItem, and then used setView: to display a custom view. This works perfectly, the menu items look great and my custom view is displayed.
However, when the user chooses a menu item and releases the mouse, the action no longer works (i.e., openLink: isn't called). If I just simply comment out the setView: call, then the actions work again (of course, the menu items are blank, but the action is executed properly). My first question, then, is why setting a view breaks the NSMenuItem's action.
No problem, I thought, I'll fix it by detecting the mouseUp event in my custom view and calling my action method from there. I added this method to my custom view:
- (void)mouseUp:(NSEvent *)theEvent {
NSLog(#"in mouseUp");
}
No dice! This method is never called.
I can set tracking rects and receive mouseEntered: events, though. I put a few tests in my mouseEntered routine, as follows:
if ([[self window] ignoresMouseEvents]) { NSLog(#"ignoring mouse events"); }
else { NSLog(#"not ignoring mouse events"); }
if ([[self window] canBecomeKeyWindow]) { dNSLog((#"canBecomeKeyWindow")); }
else { NSLog(#"not canBecomeKeyWindow"); }
if ([[self window] isKeyWindow]) { dNSLog((#"isKeyWindow")); }
else { NSLog(#"not isKeyWindow"); }
And got the following responses:
not ignoring mouse events
canBecomeKeyWindow
not isKeyWindow
Is this the problem? "not isKeyWindow"? Presumably this isn't good because Apple's docs say "If the user clicks a view that isn’t in the key window, by default the window is brought forward and made key, but the mouse event is not dispatched." But there must be a way do detect these events. HOW?
Adding:
[[self window] makeKeyWindow];
has no effect, despite the fact that canBecomeKeyWindow is YES.
Add this method to your custom NSView and it will work fine with mouse events
- (void)mouseUp:(NSEvent*) event {
NSMenuItem* mitem = [self enclosingMenuItem];
NSMenu* m = [mitem menu];
[m cancelTracking];
[m performActionForItemAtIndex: [m indexOfItem: mitem]];
}
But i'm having problems with keyhandling, if you solved this problem maybe you can go to my question and help me a little bit.
Add this to your custom view and you should be fine:
- (BOOL)acceptsFirstMouse:(NSEvent *)theEvent
{
return YES;
}
I added this method to my custom view, and now everything works beautifully:
- (void)viewDidMoveToWindow {
[[self window] becomeKeyWindow];
}
Hope this helps!
I've updated this version for SwiftUI Swift 5.3:
final class HostingView<Content: View>: NSHostingView<Content> {
override func viewDidMoveToWindow() {
window?.becomeKey()
}
}
And then use like so:
let item = NSMenuItem()
let contentView = ContentView()
item.view = HostingView(rootView: contentView)
let menu = NSMenu()
menu.items = [item]
So far, the only way to achieve the goal, is to register a tracking area manually in updateTrackingAreas - that is thankfully called, like this:
override func updateTrackingAreas() {
let trackingArea = NSTrackingArea(rect: bounds, options: [.enabledDuringMouseDrag, .mouseEnteredAndExited, .activeInActiveApp], owner: self, userInfo: nil)
addTrackingArea(trackingArea)
}
Recently I needed to show a Custom view for a NSStatusItem, show a regular NSMenu when clicking on it and supporting drag and drop operations on the Status icon.
I solved my problem using, mainly, three different sources that can be found in this question.
Hope it helps other people.
See the sample code from Apple named CustomMenus
In there you'll find a good example in the ImagePickerMenuItemView class.
It's not simple or trivial to make a view in a menu act like a normal NSMenuItem.
There are some real decisions and coding to do.