I have a disabled button, and it does not receive clicks when I use EVT_BUTTON on it. Is there a way to receive clicks even when it has been Disabled()?
The whole point of disabling a button is so that the EVT_BUTTON event is not fired. I'm sure you could create create an ugly hack using EVT_LEFT_DOWN and detecting where the mouse is in your app as a workaround, but why bother? This is intended behavior.
Perhaps wxpython has a mechanism similar to pygtk.
In pygtk you create a input-only (that is transparent) window over the widget you want to get clicks for and get your clicks there.
Related
I have a custom shelf in Maya with buttons built using the 'shelfButton' command. I would like to include short descriptions of what the button does when the mouse cursor hovers over the button. I have tried a couple of the available flags, like 'annotation,' but so far I have not gotten anything useful.
shelfButton command inside Maya has no flag for the desired behavior. Any string passed with the annotation flag will be displayed on Maya's native help box (At the buttom of the Maya UI). If you really want the behavior you may have to look into Qt.
I have an app in PyQt with a few buttons to shift between modes of the application.
I have the clicked() signals of the buttons linked to the appropriate methods. My problem is that there are other ways to change modes (for instance, loading settings will automatically move the user to their default mode), and I'm using the QPushButtons as indicators of the "active mode". This was previously accomplished by having two icons for each button, one for the button being off and the other for the button being on. It's all been designed in QtDesigner, so clicking on one button turns it on (and turns the other buttons off) and changes the icons appropriately. And when the mouse is released, that button stays on.
The button stays on until another button is pressed.
I'm trying to figure out how I can change a button from "on" to "off" without the user actually pressing the button, so I can change modes appropriately.
The Qt docs make reference to the property I'm looking for but I can't find any more details than the fact that these states exist:
The most important modes or states are:
Available or not (grayed out, disabled).
Standard push button, toggling push button or menu button.
On or off (only for toggling push buttons).
Default or normal. The default button in a dialog can generally be "clicked" using the
Enter or Return key.
Auto-repeat or not.
Pressed down or not.
(http://qt-project.org/doc/qt-5/QPushButton.html)
To be more specific, I'm looking for a way to see the state of a QPushButton; to see whether it is "On" or "Off", and I'm looking for a way to change that state.
EDIT: I found the appropriate method QPushButton.isOn() but the problem is that it's in Qt3. (I'm using Qt5, where this method no longer exists). Clearly it's obsolete now, would anyone happen to know what replaced it?
http://doc.qt.digia.com/3.2/qpushbutton.html#isOn
QPushButton inherits from QAbstractButton and and therefore has the following methods: isChecked, setChecked and isCheckable, setCheckable. This way you can convert the button into a toggle button and ask/set the state. There is also setAutoRepeat and autoRepeat which controls the auto repeat programmatically.
Especially instead of isOn use isChecked.
Furthermore it also inherits from QWidget which has methods isEnabled and setEnabled. With this you can activate/de-active the button which is shown by graying out the button as well as by prohibiting clicks on the button.
Basically just study the documentation for QAbstractButton and QWidget to see how you can programmatically interact with a QPushButton to enable/disable it.
I am trying to implement a feature such that if the user havn't interact with the GUI for some time X, I will stop some functionality of the GUI.
Currently I have a time stamp set up so that if any of the button is not clicked in X seconds, the GUI will terminate some functionality
button1.triggered.connect(keep_alive)
button2.triggered.connect(keep_alive)
....
buttonN.triggered.connect(keep_alive)
As you can see, this is not really elegant, and doesn't scale as the button increases. Therefore I am currently investigating another method such that I monitor the mouse clicks
mouse = app.mouseButtons()
if mouse != Qtcore.Qt.NoButton:
#keep_alive
I think this is a little hacky, but it will work for the functionality I envisioned, however, I do not know how to insert this to the execution loop of the QT.
Any suggestions will be appreciated
You must intercept the mouse events by reimplementing the mousePressEvent.
http://qt-project.org/doc/qt-4.8/qwidget.html#mousePressEvent
To make sure that it won't affect your other functionalities you'll need to propagate it to the parent widget. Read more details: https://www.qt.io/blog/2006/05/27/mouse-event-propagation
I would proceed by implementing it in the main window and make sure all mouse events are propagated to it.
In a pygame application window, the minimize, resize and close buttons are present. Is there a way to disable the close(X) button?
I don't think there is, because some window managers don't give you the ability to remove the close button. But you can write an event handler such that the close button does whatever you want, including nothing.
Why do you want to prevent the user from closing? If it's just a matter that you would rather provide an in-game "quit" button that confirms and/or saves before quitting, you can perform the same task when the user hits the close button.
Just for the record, another option would be to pass the following argument to the set_mode() method call:
pygame.display.set_mode(..., flags = pygame.NOFRAME)
This however makes the whole frame go away, including the top strip to move the window around and the other buttons, such as minimize, so it's rather overkill for just getting rid of the X button.
I would like to create an application that has 3-4 frames (or windows) where each frame is attached/positioned to a side of the screen (like a task bar). When a frame is inactive I would like it to auto hide (just like the Windows task bar does; or the dock in OSX). When I move my mouse pointer to the position on the edge of the screen where the frame is hidden, I would like it to come back into focus.
The application is written in Python (using wxPython for the basic GUI aspects). Does anyone know how to do this in Python? I'm guessing it's probably OS dependent? If so, I'd like to focus on Windows first.
I don't do GUI programming very often so my apologies if this makes no sense at all.
As far as I know, there's nothing built in for this.
When the window is hidden, do you want it completely invisible or can a border of a few pixels be showing? That would be an easy way to get a mouse hover event. Otherwise you might have to use something like pyHook to get system-wide mouse events to know when to expand your window.
The events EVT_ENTER_WINDOW and EVT_LEAVE_WINDOW might also be useful here to know when the user has entered/left the window so you can expand/collapse it.
Expanding/collapsing can just be done by showing/hiding windows or resizing them. Standard window functions, nothing fancy.
By the way, you might want to use wx.ClientDisplayRect to figure out where to position your window. That will give you a rectangle of the desktop that does NOT include the task bar or any other toolbars the user has, assuming you want to avoid overlapping with those things.
Personally, I would combine the EVT_ENTER_WINDOW and EVT_LEAVE_WINDOW that FogleBird mentioned with a wx.Timer. Then whenever it the frame or dialog is inactive for x seconds, you would just call its Hide() method.
I think you could easily just make a window that is the same size as the desktop then do some while looping for an inactivity variable based on mouse position, then thread off a timer for loop for the 4 inactivity variables. I'd personally design it so that when they reach 0 from 15, they change size and position to become tabular and create a button on them to reactivate. lots of technical work on this one, but easily done if you figure it out