I would like to set two things for submenus in PyQt:
not open submenus when hovered
no delay when submenus get opened by a click
I suppose that I had to modify the hover behavior of the QAction object that is returned by the menuAction() method of the submenu object - but how to do this?
There is a QStyle::SH_Menu_SubMenuPopupDelay setting mentioned in the docs that is maybe what I need for the second but I also have no clue how to set this in PyQt.
My basic menu example:
#!/usr/bin/env python
from PyQt4 import QtGui
from PyQt4 import QtCore
class MainWindow(QtGui.QMainWindow):
def __init__(self, parent=None):
super(MainWindow, self).__init__(parent)
self.menubar = QtGui.QMenuBar(self)
self.setMenuBar(self.menubar)
self.menuFile = QtGui.QMenu(self.menubar, title='File')
self.menubar.addAction(self.menuFile.menuAction())
self.submenu = QtGui.QMenu(self.menuFile, title='Submenu')
self.menuFile.addAction(QtGui.QAction(self, text="First"))
self.menuFile.addAction(self.submenu.menuAction())
self.menuFile.addAction(QtGui.QAction(self, text="Third"))
self.submenu.addAction(QtGui.QAction(self, text="First"))
self.submenu.addAction(QtGui.QAction(self, text="Second"))
self.submenu.addAction(QtGui.QAction(self, text="Third"))
if __name__ == '__main__':
import sys
app = QtGui.QApplication(sys.argv)
mw = MainWindow()
mw.show()
sys.exit(app.exec_())
Related
I'm now using PyQt5 in python to develop a GUI for my program. I have added some QAction to the toolbar, like start, pause and stop icon. However, I don't know how to add a QSlider to the toolbar in my mainwindow. I prefer to add the QSlider to the position which is marked by a red rectangle. Thanks!
QToolBar has the addWidget() method that allows you to add widgets:
from PyQt5 import QtCore, QtWidgets
class MainWindow(QtWidgets.QMainWindow):
def __init__(self, parent=None):
super(MainWindow, self).__init__(parent)
toolbar = self.addToolBar("toolbar")
toolbar.addAction("start")
toolbar.addAction("pause")
toolbar.addAction("stop")
self.slider = QtWidgets.QSlider(QtCore.Qt.Horizontal)
toolbar.addWidget(self.slider)
if __name__ == '__main__':
import sys
app = QtWidgets.QApplication(sys.argv)
w = MainWindow()
w.show()
sys.exit(app.exec_())
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_())
I'm trying to use the stackWidget to show different widgets .
However when I pressed the item on the listWdget , stackWidget below will not be displayed.
Here is my code,and I don't know where is wrong.
#!/usr/bin/python
# music_1.py
import sys
from PyQt4 import QtGui, QtCore
from music_ui import Ui_Form
class music(QtGui.QDialog):
def __init__(self, parent=None):
QtGui.QDialog.__init__(self, parent)
self.ui = Ui_Form()
self.ui.setupUi(self)
self.ui.listWidget.insertItem(0,("Warm"))
self.ui.listWidget.insertItem(1,("Funny"))
self.ui.listWidget.insertItem(2,("Terror"))
self.ui.stackedWidget=QtGui.QStackedWidget()
warm=Warm()
funny=Funny()
terror=Terror()
self.ui.stackedWidget.addWidget(warm)
self.ui.stackedWidget.addWidget(funny)
self.ui.stackedWidget.addWidget(terror)
self.connect(self.ui.listWidget,QtCore.SIGNAL("currentRowChanged(int)"),self.ui.stackedWidget,QtCore.SLOT("setCurrentIndex(int)"))
class Warm(QtGui.QWidget):
def __init__(self,parent=None):
super(Warm,self).__init__(parent)
w1=QtGui.QPushButton("w1")
w2=QtGui.QPushButton("w2")
buttonLayout=QtGui.QHBoxLayout()
buttonLayout.addStretch(1)
buttonLayout.addWidget(w1)
buttonLayout.addWidget(w2)
class Funny(QtGui.QWidget):
def __init__(self,parent=None):
super(Funny,self).__init__(parent)
f1=QtGui.QPushButton("f1")
f2=QtGui.QPushButton("f2")
f3=QtGui.QPushButton("f3")
buttonLayout=QtGui.QHBoxLayout()
buttonLayout.addStretch(1)
buttonLayout.addWidget(f1)
buttonLayout.addWidget(f2)
buttonLayout.addWidget(f3)
class Terror(QtGui.QWidget):
def __init__(self,parent=None):
super(Terror,self).__init__(parent)
t1=QtGui.QPushButton("t1")
t2=QtGui.QPushButton("t2")
buttonLayout=QtGui.QHBoxLayout()
buttonLayout.addStretch(1)
buttonLayout.addWidget(t1)
buttonLayout.addWidget(t2)
if __name__ == "__main__":
app = QtGui.QApplication(sys.argv)
myapp = music()
myapp.show()
sys.exit(app.exec_())
I never worked with UI Files but this is my guess:
self.ui.stackedWidget=QtGui.QStackedWidget()
here you create a QStackedWidget which gets assigned to the class. The problem is that it is not added to the UI that is getting displayed but just to the instance of the Ui_Form. What you have to do is either:
Add the QStackedWidget in the UI Designer and erase the line above
or
Add the QStackedWidget to one of your layouts
which can be done like this:
self.ui.mySuperCoolLayout.addWidget(self.ui.stackedWidget)
I have created a form using PyQt4 which has a push button. On this push button I want to call another python script which looks like this:
File1.py:
import sys
from PyQt4 import QtCore, QtGui
from file1_ui import Ui_Form
class MyForm(QtGui.QMainWindow):
def __init__(self, parent=None):
QtGui.QWidget.__init__(self, parent)
self.ui = Ui_Form()
self.ui.setupUi(self)
if __name__ == "__main__":
app = QtGui.QApplication(sys.argv)
myapp = MyForm()
myapp.show()
sys.exit(app.exec_())
File1_ui.py
from PyQt4 import QtCore, QtGui
try:
_fromUtf8 = QtCore.QString.fromUtf8
except AttributeError:
_fromUtf8 = lambda s: s
class Ui_Form(object):
def setupUi(self, Form):
Form.setObjectName(_fromUtf8("Form"))
Form.resize(400, 300)
self.pushButton = QtGui.QPushButton(Form)
self.pushButton.setGeometry(QtCore.QRect(120, 200, 95, 20))
self.pushButton.setObjectName(_fromUtf8("pushButton"))
self.retranslateUi(Form)
QtCore.QObject.connect(self.pushButton, QtCore.SIGNAL(_fromUtf8("clicked()")), Form.close)
QtCore.QMetaObject.connectSlotsByName(Form)
def retranslateUi(self, Form):
Form.setWindowTitle(QtGui.QApplication.translate("Form", "Form", None, QtGui.QApplication.UnicodeUTF8))
self.pushButton.setText(QtGui.QApplication.translate("Form", "Close", None, QtGui.QApplication.UnicodeUTF8))
File2.py
import sys
from PyQt4 import Qt
from taurus.qt.qtgui.application import TaurusApplication
app = TaurusApplication(sys.argv)
panel = Qt.QWidget()
layout = Qt.QHBoxLayout()
panel.setLayout(layout)
from taurus.qt.qtgui.panel import TaurusForm
panel = TaurusForm()
model = [ 'test/i1/1/%s' % p for p in props ]
panel.setModel(model)
panel.show()
sys.exit(app.exec_())
File1_ui.py is created from the Qtdesigner and then I am using File1.py to execute it.So File2.py when executed alone opens up a panel and displays few attributes.I want this script to be called on the button click in the first form(file1.py) which I created using Qtdesigner.Could you let me know how I could achieve this functionality.Thanks.
You will need to make some modifications to File2.py to make the appropriate calls depending on whether it is running standalone or not. When you are launching the script via File1.py there will already be a QApplication instance with event loop running, so trying to create another and run its event loop will cause problems.
Firstly, move the core part of your script into its own function. This will allow you to easily call it from File1.py. You can then handle the case where the script is running standalone and needs to create a QApplication instance and start its event loop. (I am not familiar the the taurus library you are using, but you can probably substitute TaurusApplication for QtGui.QApplication)
File2.py:
import sys
from PyQt4 import QtCore, QtGui
def runscript():
panel = QtGui.QWidget()
layout = QtGui.QHBoxLayout(panel)
return panel # Must return reference or panel will be deleted upon return
if __name__ == '__main__':
app = QtGui.QApplication(sys.argv)
panel = runscript()
panel.show()
sys.exit(app.exec_())
Assuming your files are in the same directory you can simply write import File2 and use File2.runscript() to run your code. You then just need to connect the function to your pushbuttons clicked() signal to run it. The only problem here is that the reference to the QWidget returned from the runscript() function will be lost (and the object deleted) if you connect directly to runscript(). For this reason I created a method launch_script() which saves a reference in MyForm.
File1.py:
import sys
from PyQt4 import QtCore, QtGui
from file1_ui import Ui_Form
import File2
class MyForm(QtGui.QMainWindow):
def __init__(self, parent=None):
QtGui.QWidget.__init__(self, parent)
self.ui = Ui_Form()
self.ui.setupUi(self)
# This is a bit of a hack.
self.ui.pushButton.clicked.connect(self.launch_script)
def launch_script(self):
self.panel = File2.runscript()
self.panel.show()
if __name__ == "__main__":
app = QtGui.QApplication(sys.argv)
myapp = MyForm()
myapp.show()
sys.exit(app.exec_())
I don't use Qt Designer, so I don't know the correct way to go about connecting the signal to launch_script(). The code I have written should work, but obviously violates OOP principles and is dependent on the name of the pushbutton widget assigned by the software.
Please can someone tell me what im doing wrong here with respect to calling pwTxt.text.
#!/usr/bin/python
import sys
from PyQt4 import QtCore, QtGui
from mainwindow import Ui_MainWindow
class MyForm(QtGui.QMainWindow):
def __init__(self, parent=None):
QtGui.QWidget.__init__(self, parent)
self.ui = Ui_MainWindow()
self.ui.setupUi(self)
def on_pwExtract_pressed(self):
print self.pwTxt.text
if __name__ == "__main__":
app = QtGui.QApplication(sys.argv)
myapp = MyForm()
myapp.show()
sys.exit(app.exec_())
The line print self.pwTxt.text fails because it can't find the widget, pwTxt is a QLineEdit defined on the main window. I just made it in QTDesigner and generated python code with pyuic4.
How do I correctly reference other widgets on the same window, in this case I just want to get the text from a QLineEdit named pwTxt when the QPushButton pwExtract is pressed.
Thanks a lot.
Try:
print self.ui.pwTxt.text()