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Python Shell restart after while loop with PyQt5

actually I try to do a displayer of html page. I have some of list of html that I need to display, each page must stay visible while x secondes.
But, after the first page is display, the application crash and the python shell restart in consequence.
On my mind, I would create the window to display the page and close the application because after I will try to display png/jpg, so I need to close app to use Pygame for pictures and re-build the app to display html page after.
My list looking for it :
html page/html page/picture/html page/picture/picture
So I have build a sample code to test in while boucle the displayer :
from PyQt5 import QtWidgets, QtWebEngineWidgets, QtCore
import sys
continuer = True
while continuer:
print("Application created")
# Create application
app = QtWidgets.QApplication(sys.argv)
# Add window
win = QtWidgets.QWidget()
win.setWindowTitle('My first rendering')
# Add layout
layout = QtWidgets.QVBoxLayout()
win.setLayout(layout)
# Create QWebView
view = QtWebEngineWidgets.QWebEngineView()
view.setUrl(QtCore.QUrl('https://google.com'))
# Add QWebView to the layout
layout.addWidget(view)
# Show window, run app
win.show()
QtCore.QTimer.singleShot(7*1000, win.close)
QtCore.QTimer.singleShot(7*1000, app.quit)
print("View displayed")
# While loop
app.exec_()
print('Close Application')
print("End While Loop")
Result after executing
It's probably sys.argv in app var the mistake, but I'm new in Python so I don't know how to fix the problem.

The problem is that QApplication is not necessarily eliminated and therefore you would be creating more than one QApplication what Qt forbids, a better solution is to rethink it verifying that if it does not exist then create a new one:
import sys
from PyQt5 import QtCore, QtWidgets, QtWebEngineWidgets
continuer = True
while continuer:
print("Application created")
# Create application
app = QtWidgets.QApplication.instance()
if app is None:
app = QtWidgets.QApplication(sys.argv)
# Add window
win = QtWidgets.QWidget()
win.setWindowTitle("My first rendering")
# Create QWebEngineView
view = QtWebEngineWidgets.QWebEngineView()
view.setUrl(QtCore.QUrl("https://google.com"))
# Add layout
layout = QtWidgets.QVBoxLayout(win)
win.setLayout(layout)
# Add QWebView to the layout
layout.addWidget(view)
# Show window, run app
win.show()
QtCore.QTimer.singleShot(7 * 1000, win.close)
QtCore.QTimer.singleShot(7 * 1000, app.quit)
print("View displayed")
# While loop
app.exec_()
print("Close Application")
print("End While Loop")

PyQt Creates Seperate Window For Button

Hi recently I started experimenting with the PyQt5 GUI library since Tkinter doesn't look very modern or nice. I am trying to create a very simple window with a button on it, but for some reason it opens a new window for the button. How can I solve this? Here is the code:
from PyQt5 import QtWidgets
app = QtWidgets.QApplication([])
window = QtWidgets.QWidget()
window.setGeometry(0, 0, 500, 300)
window.setWindowTitle("UI")
window.show()
btn = QtWidgets.QPushButton("Test")
btn.resize(100,100)
btn.move(100,100)
btn.show()
(app.exec_())
I am aware that I should be using classes for the GUI, but I'd like to learn the basics before I start doing that.

For a widget like the QPushButton to be part of a window, the widget must be fulfilled:
be child of the window or
be a child of some child from the window or
it is part of a layout that belongs to the window.
In your case QPushButton does not meet any of it so it will be a new window.So it can be solved using the 1 or 3 rule:
import sys
from PyQt5 import QtWidgets
if __name__ == '__main__':
app = QtWidgets.QApplication(sys.argv)
window = QtWidgets.QWidget()
window.setGeometry(0, 0, 500, 300)
window.setWindowTitle("UI")
btn = QtWidgets.QPushButton("Test", window)
btn.resize(100,100)
btn.move(100,100)
window.show()
sys.exit(app.exec_())
Or:
import sys
from PyQt5 import QtWidgets
if __name__ == '__main__':
app = QtWidgets.QApplication(sys.argv)
window = QtWidgets.QWidget()
window.setGeometry(0, 0, 500, 300)
window.setWindowTitle("UI")
btn = QtWidgets.QPushButton("Test")
lay = QtWidgets.QHBoxLayout(window)
lay.addWidget(btn)
window.show()
sys.exit(app.exec_())

Use pyqt4 to create GUI that runs python script

I have written a python script that manipulates data and then saves it into an excel file with the following function:
def saveData():
table = pd.DataFrame(data)
writer = pd.ExcelWriter('manipulatedData.xlsx')
table.to_excel(writer, 'sheet 1')
writer.save()
I also have the following super basic/minimalistic gui window:
from PyQt4 import QtGui
app = QtGui.QApplication(sys.argv)
window = QtGui.QWidget()
window.setGeometry(500,500,500,500)
window.setWindowTitle("PyQt Test!")
window.setWindowIcon(QtGui.QIcon('pythonLogoSmall.png'))
window.show()
sys.exit(app.exec_())
What I really would like to know is this:
1) How can I let the user of my gui select the location + name for the manipulatedData file?
2) How can I add a "run" button, that starts my python script and manipulates the data, before it saves the manipulated data to the desired location.

PyQt4 has QLineEdit() and QPushButton()
from PyQt4 import QtGui
import sys
# --- functions ---
def my_function(event=None):
print 'Button clicked: event:', event
print linetext.text()
# run your code
# --- main ---
app = QtGui.QApplication(sys.argv)
window = QtGui.QWidget()
# add "layout manager"
vbox = QtGui.QVBoxLayout()
window.setLayout(vbox)
# add place for text
linetext = QtGui.QLineEdit(window)
vbox.addWidget(linetext)
# add button
button = QtGui.QPushButton("Run", window)
vbox.addWidget(button)
# add function to button
button.clicked.connect(my_function)
window.show()
sys.exit(app.exec_())

Open window application in current window

I want to open a game application replacing the menu window.
So I have a game application which I can start with:
subprocess.call(["mygameprogram", "argument one"])
This opens up another window, but I want to replace the qt window with the game window without closing the menu window.
This is my MainWindow class:
class MainWindow(QMainWindow):
def __init__(self):
super().__init__()
self.initMenu()
self.show()
def initMenu(self):
# load ui file
uic.loadUi("somefile.ui", self)
# add pushbutton
btn = QPushButton()
# add click listener
btn.clicked.connect(self.startLevel)
# add button to layout
self.layout_buttons.addWidget(btn)
def startLevel(self):
# open game, put the right code in here :D
pass
if __name__ == "__main__":
app = QApplication(sys.argv)
ex = MainWindow()
sys.exit(app.exec_())
Is there any solution for that? And if not, what should I do instead?
Thanks in advance

Quiting the print dialog-pyqt

I defined a print function using QPrinter and QDialog. However, when I launch the printer dialog and then press cancel, the entire mainwindow goes into not responding mode. I have tried to do QtGui.QPrintDialog.close() but does not work.
Code:
import sys
from PyQt4 import QtCore
from PyQt4 import QtGui
class QButton(QtGui.QWidget):
def __init__(self, parent=None):
QtGui.QWidget.__init__(self, parent)
self.button = QtGui.QPushButton('Button', self)
self.name='me'
self.button.clicked.connect(self.calluser)
def calluser(self):
Appli=QtGui.QApplication(sys.argv)
printer= QtGui.QPrinter()
doc=QtGui.QTextDocument("Set local variables in this printing slot." )
dialog = QtGui.QPrintDialog(printer)
dialog.setModal(True)
dialog.setWindowTitle("Print Document" )
if dialog.exec_() == True:
doc.print_(printer)
# dialog.addEnabledOption(QAbstractPrintDialog.PrintSelection)
def demo_QButton():
app = QtGui.QApplication(sys.argv)
tb = QButton()
tb.show()
app.exec_()
if __name__=='__main__':
demo_QButton()

You created a new application in the calluser method. Delete or comment the line:
Appli=QtGui.QApplication(sys.argv)
and try again. I think this time your main window will remain responsive.