pyqt4 mousePressEvent not called (no widget?) - python

I have just wrote a little example and i can't manage to make it run.
from PyQt4 import QtGui, QtCore
import sys
class Drawer(QtGui.QWidget):
def __init__(self, parent=None):
super(Drawer, self).__init__(parent)
self.setStyleSheet("QWidget { background-color: %s }" % QtGui.QColor(99, 0, 0).name())
def mousePressEvent(self, event):
print 'mouse pressed'
self.update();
class MyApp(QtGui.QMainWindow):
def __init__(self, parent=None):
QtGui.QWidget.__init__(self, parent)
self.drawer = Drawer(self)
if __name__ == '__main__':
app = QtGui.QApplication(sys.argv)
myapp = MyApp()
myapp.show()
sys.exit(app.exec_())
widget is not shown (no color, the window is gray) and if i press the mouse no print..
where is my error?
SOLVED: as qiao just point me in a comment, my error is the way to add a widget in the qt4 kind of scenegraph. i thought i had to call the parent in init and that's all. This is not enogh, i have to add a QLayout and add the childs in it (this is quite obvious: the method addWidget is written only in QLayout and not in QWidget and having a scenegraph system without the possibility of adding new child is quite weird)

You have to set the central widget of the main window to be the drawer. Otherwise the drawer will not attached to the main window.
class MyApp(QtGui.QMainWindow):
def __init__(self, parent=None):
QtGui.QWidget.__init__(self, parent)
self.drawer = Drawer(self)
self.setCentralWidget(self.drawer)
After the above fix, you will see mouse press event working properly.
As for the color, setting the stylesheet of QMainWindow is fine, so is setting Drawer to be another widget(like QLineEdit). I don't know what's the matter here.

Related

PyQt4 How keep a QWidget always on top?

I'm using PyQt4 with python 2.7 on Windows7
I have a QWidget that I want to stay above the QMainWindow while clicking on the main. The idea is that the main will contain a series of 'edit' buttons that will open the edit widget, while the edit widget refreshes with info contained on the main. I don't particularly care if either is "always on top" as long as the widget stays in front of the main window.
There are a couple of questions that address this topic, but I'm not seeing an answer that works for my specific use case. One deals in widgets but provides an answer only for the app main window (widget stays above other environmental windows but falls behind the application main window when clicking on the main), and the other addresses Qt generally but not a pythonic example:
PyQt: Always on top
How keep a QWidget always on top?
Here is the code I have so far:
from PyQt4 import QtCore, QtGui
class Ui_MainWindow(QtGui.QMainWindow):
def __init__(self):
QtGui.QMainWindow.__init__(self, None, QtCore.Qt.WindowStaysOnTopHint)
self.setWindowTitle("MainWindow")
self.resize(400, 300)
class Ui_Widget(QtGui.QWidget):
def __init__(self):
QtGui.QWidget.__init__(self, None, QtCore.Qt.WindowStaysOnTopHint)
self.setWindowTitle("Widget")
self.resize(400, 300)
if __name__ == "__main__":
import sys
app = QtGui.QApplication(sys.argv)
MainWindow = Ui_MainWindow()
MainWindow.show()
Widget = Ui_Widget()
Widget.show()
sys.exit(app.exec_())
If you want Ui_Widget to always be on top of Ui_MainWindow, Ui_Widget must be a child of Ui_MainWindow and the flag Qt::Dialog must be activated as shown below:
from PyQt4 import QtCore, QtGui
class Ui_MainWindow(QtGui.QMainWindow):
def __init__(self):
QtGui.QMainWindow.__init__(self)
self.setWindowTitle("MainWindow")
self.resize(400, 300)
class Ui_Widget(QtGui.QWidget):
def __init__(self, parent=None):
QtGui.QWidget.__init__(self, parent)
self.setWindowFlags(self.windowFlags() | QtCore.Qt.Dialog)
self.setWindowTitle("Widget")
self.resize(400, 300)
self.move(200, 150)
if __name__ == "__main__":
import sys
app = QtGui.QApplication(sys.argv)
MainWindow = Ui_MainWindow()
MainWindow.show()
Widget = Ui_Widget(MainWindow)
Widget.show()
sys.exit(app.exec_())

Python PyQt4 open from QDialog new QWidget window

By pressing a QPushButton in my QDialog window I want to open a new QWidget window.
My code:
from PyQt4 import QtGui
import sys
class MainWindow(QtGui.QWidget):
def __init__(self, parent=None):
super(MainWindow, self).__init__(parent)
self.setWindowTitle("Main Window")
class FirstWindow(QtGui.QDialog):
def __init__(self, parent=None):
super(FirstWindow, self).__init__(parent)
self.createWindow()
def createWindow(self):
btn = QtGui.QPushButton('Open New Window', self)
btn.move(10, 10)
self.openNewWindow = MainWindow(self)
btn.clicked.connect(self.openMainWin)
self.setGeometry(250,250, 150,50)
self.setWindowTitle("First Window")
self.show()
def openMainWin(self):
self.openNewWindow.show()
if __name__ == "__main__":
app = QtGui.QApplication(sys.argv)
firstwin = FirstWindow()
sys.exit(app.exec_())
When I run the code nothing happens by pressing the button.
But when I change the class from
class MainWindow(QtGui.QWidget) to
class MainWindow(QtGui.QDialog) or class MainWindow(QtGui.QMainWindow)
it works!
What am I doing wrong?! Please assist me.
When you instantiate MainWindow you pass in a parent. Qwidget only makes a new window if you don't specify a parent.
This is of course deliberate. If QWidgets with parents were shown in new windows, then you could never build a GUI. Imagine having every widget in it's own window!
QMainWindow and QDialog are specifically designed to both have a parent, and create a new window. You should use them.

How to detect any mouse click on PySide Gui?

I am trying implement a feature such that when a mouse is clicked on the gui, a function is triggered
Below is my mouse click detection, it doesn't work when I click on any part of the gui
from PySide.QtCore import *
from PySide.QtGui import *
import sys
class Main(QWidget):
def __init__(self, parent=None):
super(Main, self).__init__(parent)
layout = QHBoxLayout(self)
layout.addWidget(QLabel("this is the main frame"))
layout.gui_clicked.connect(self.anotherSlot)
def anotherSlot(self, passed):
print passed
print "now I'm in Main.anotherSlot"
class MyLayout(QHBoxLayout):
gui_clicked = Signal(str)
def __init__(self, parent=None):
super(MyLayout, self).__init__(parent)
def mousePressEvent(self, event):
print "Mouse Clicked"
self.gui_clicked.emit("emit the signal")
a = QApplication([])
m = Main()
m.show()
sys.exit(a.exec_())
This is my goal
Mouseclick.gui_clicked.connect(do_something)
Any advice would be appreciated
Define mousePressEvent inside Main:
from PySide.QtCore import *
from PySide.QtGui import *
import sys
class Main(QWidget):
def __init__(self, parent=None):
super(Main, self).__init__(parent)
layout = QHBoxLayout(self)
layout.addWidget(QLabel("this is the main frame"))
def mousePressEvent(self, QMouseEvent):
#print mouse position
print QMouseEvent.pos()
a = QApplication([])
m = Main()
m.show()
sys.exit(a.exec_())
This can get complicated depending on your needs. In short, the solution is an eventFilter installed on the application. This will listen the whole application for an event. The problem is "event propagation". If a widget doesn't handle an event, it'll be passed to the parent (and so on). You'll see those events multiple times. In your case, for example QLabel doesn't do anything with a mouse press event, therefore the parent (your main window) gets it.
If you actually filter the event (i.e. you don't want the original widget to respond to the event), you won't get that problem. But, I doubt that this is your intent.
A simple example for just monitoring:
import sys
from PySide import QtGui, QtCore
class MouseDetector(QtCore.QObject):
def eventFilter(self, obj, event):
if event.type() == QtCore.QEvent.MouseButtonPress:
print 'mouse pressed', obj
return super(MouseDetector, self).eventFilter(obj, event)
class MainWindow(QtGui.QWidget):
def __init__(self, parent=None):
super(MainWindow, self).__init__(parent)
layout = QtGui.QHBoxLayout()
layout.addWidget(QtGui.QLabel('this is a label'))
layout.addWidget(QtGui.QPushButton('Button'))
self.setLayout(layout)
if __name__ == '__main__':
app = QtGui.QApplication(sys.argv)
mouseFilter = MouseDetector()
app.installEventFilter(mouseFilter)
main = MainWindow()
main.show()
sys.exit(app.exec_())
You can see that, clicking on the QLabel will give you something like:
mouse pressed <PySide.QtGui.QLabel object at 0x02B92490>
mouse pressed <__main__.MainWindow object at 0x02B92440>
Because, QLabel receives the event and since it doesn't do anything with it, it's ignored and passed to the parent (MainWindow). And it's caught by the filter/monitor again.
Clicking on the QPushButton doesn't have any problem because it uses that event and does not pass to the parent.
PS: Also note that this can cause performance problems since you are inspecting every single event in the application.

Load other windows when button clicked. PyQt

I am trying to call another window from a button click in python 2.7 using PyQt4. The code below opens the AddBooking dialog but immediately closes it. Im new to Gui programming, can somebody please tell me what is wrong with my code?
from PyQt4 import QtGui
from HomeScreen import Ui_HomeScreen
from AddBooking import Ui_AddBooking
import sys
class HomeScreen(QtGui.QWidget, Ui_HomeScreen):
def __init__(self, parent=None):
QtGui.QWidget.__init__(self, parent)
self.setupUi(self)
self.show()
self.Add_Booking_Button.clicked.connect(self.handleButton)
def handleButton(self):
AddBooking2()
class AddBooking2(QtGui.QWidget, Ui_AddBooking):
def __init__(self, parent=None):
QtGui.QWidget.__init__(self, parent)
self.setupUi(self)
self.show()
if __name__ == '__main__':
app = QtGui.QApplication(sys.argv)
window = HomeScreen()
window.show()
sys.exit(app.exec_())
Don't use multi-inheritance and neither call show function inside class initializer. The problem is that the object you are creating with AddBooking2() is a temporal and it's destroyed automatically when the function ends. So you need use some variable to reference that object something like:
addbooking = AddBooking2()
addbooking.show()
Also, since you are working with QtDesigner and pyuic4 tools you can make connections a little bit easier. Said that, your code can be modified:
from PyQt4 import QtGui
from PyQt4.QtCore import pyqtSlot
from HomeScreen import Ui_HomeScreen
from AddBooking import Ui_AddBooking
import sys
class HomeScreen(QtGui.QWidget):
def __init__(self, parent=None):
QtGui.QWidget.__init__(self, parent)
self.ui = Ui_HomeScreen()
self.ui.setupUi(self)
#pyqtSlot("")
def on_Add_Booking_Button_clicked(self): # The connection is carried by the Ui_* classes generated by pyuic4
addbooking = AddBooking2()
addbooking.show()
class AddBooking2(QtGui.QWidget):
def __init__(self, parent=None):
QtGui.QWidget.__init__(self, parent)
self.ui = Ui_AddBooking()
self.ui.setupUi(self)
if __name__ == '__main__':
app = QtGui.QApplication(sys.argv)
window = HomeScreen()
window.show()
sys.exit(app.exec_())
The dialog closes immediately because you are not keeping a reference to it, and so it will get garbage-collected as soon as it goes out of scope.
The simplest way to fix it would be to do something like this:
def handleButton(self):
self.dialog = AddBooking2()
self.dialog.show()
and you can also remove the self.show() lines from AddBooking2.__init__ and HomeScreen.__init__, which are redundant. Other than that, your code looks fine.

Hover issue in PyQt

I want to do hover. I saw an example and then write a script which will be use as I made program. I am facing one problem that hover only occur if you put mouse on the left corner of button. I want that it will happen for all the button that if i move cursor on button then it should change.
Here is my code:
from PyQt4 import QtGui, QtCore
from PyQt4.QtCore import pyqtSignal
import os,sys
class HoverButton(QtGui.QToolButton):
def enterEvent(self,event):
print("Enter")
button.setStyleSheet("background-color:#45b545;")
def leaveEvent(self,event):
button.setStyleSheet("background-color:yellow;")
print("Leave")
app = QtGui.QApplication(sys.argv)
widget = QtGui.QWidget()
button = QtGui.QToolButton(widget)
button.setMouseTracking(True)
buttonss = HoverButton(button)
button.setIconSize(QtCore.QSize(200,200))
widget.show()
sys.exit(app.exec_())
Is this what you're looking for
from PyQt4 import QtGui, QtCore
from PyQt4.QtCore import pyqtSignal
import os,sys
class Main(QtGui.QWidget):
def __init__(self, parent=None):
super(Main, self).__init__(parent)
layout = QtGui.QVBoxLayout(self) # layout of main widget
button = HoverButton(self)
button.setIconSize(QtCore.QSize(200,200))
layout.addWidget(button) # set your button to the widgets layout
# this will size the button nicely
class HoverButton(QtGui.QToolButton):
def __init__(self, parent=None):
super(HoverButton, self).__init__(parent)
self.setMouseTracking(True)
def enterEvent(self,event):
print("Enter")
self.setStyleSheet("background-color:#45b545;")
def leaveEvent(self,event):
self.setStyleSheet("background-color:yellow;")
print("Leave")
app = QtGui.QApplication(sys.argv)
main = Main()
main.show()
sys.exit(app.exec_())
In your code you had a button in a button and the nested button wasn't assigned to a QLayout widget. Although, I'm not sure why you're adding a button inside of a button. One thing that I've learned from working with GUI's is that it's so much easier if you modularize your code. Now you can take this custom button and apply it somewhere else.
You should use the stylesheet as
QToolButton:hover
{
background-color: rgb(175,175,175);
}
You probably want focus and blur, rather than enter and leave will only fire when the mouse actually enters or leaves the boundary of the button and will probably just be short duration impulses rather than toggles. focus and blur will toggle with the hover.

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