I want to create a code which will display a window with a button, which clicked will create another window with some fields (like QLabel, QLineEdit, QSpinBox, etc.). However, I don't know how to create that popup window...
Here is my code:
#!/usr/bin/env python
# -*- coding: utf-8 -*-
import sys # Needed for PySide
from PySide.QtCore import *
from PySide.QtGui import *
class Form(QDialog):
def __init__(self, parent=None):
super(Form, self).__init__(parent)
# Create widgets
self.label1 = QLabel("Label1")
self.button_open = QPushButton("Open popup")
self.button = QPushButton("Go!")
self.qbtn = QPushButton('Quit')
# Create layout and add widgets
layout = QVBoxLayout()
layout.addWidget(self.label1)
layout.addWidget(self.button_open)
# Buttons layout
hbox_buttons = QHBoxLayout()
hbox_buttons.addStretch(1)
hbox_buttons.addWidget(self.button)
hbox_buttons.addWidget(self.qbtn)
# Main layout
layout.addStretch(1)
layout.addWidget(self.button_open)
layout.addLayout(hbox_buttons)
self.setLayout(layout)
# Add buttons slots
self.button_open.clicked.connect(self.popup)
self.button.clicked.connect(self.function_runner)
self.qbtn.clicked.connect(QCoreApplication.instance().quit)
def popup (self, parent=__init__):
new_win = # I wonder what should be here
if __name__ == '__main__':
# Create the Qt Application
app = QApplication(sys.argv)
# Create and show the form
form = Form()
form.show()
# Run the main Qt loop
sys.exit(app.exec_())
I don't know if this is the best way, but the one I could figure out over the night... I hope it will help someone who got stuck with a similar problem.
So, I (simply) created a separate code for that second window and called it with
from subprocess import call
call("./my_2nd_window_code.py")
Related
My computer doesn't have any way of letting me know if my NumLk is on or off, so I am trying to add an icon in my systray that will changed depending on the state of my NumLk. This .py will always be running when my computer is on.
So far I was able to mix 3 codes and I am able to display the icon in the systray but it doesn't get updated when the state of NumLk change. Actually if I press NumLk twice, I still get the same icon (the on one) and I get this error:
QCoreApplication::exec: The event loop is already running
File "\systray_icon_NumLk_on_off.py", line 21, in on_key_press
main(on)
File "\systray_icon_NumLk_on_off.py", line 46, in main
sys.exit(app.exec_())
SystemExit: -1
My code may not be the best way to do it, so any alternative is welcome! Here is what I came up so far:
#####get the state of NumLk key
from win32api import GetKeyState
from win32con import VK_NUMLOCK
#how to use: print(GetKeyState(VK_NUMLOCK))
#source: http://stackoverflow.com/questions/21160100/python-3-x-getting-the-state-of-caps-lock-num-lock-scroll-lock-on-windows
#####Detect if NumLk is pressed
import pyglet
from pyglet.window import key
window = pyglet.window.Window()
#source: http://stackoverflow.com/questions/28324372/detecting-a-numlock-capslock-scrlock-keypress-keyup-in-python
on=r'on.png'
off=r'off.png'
#window.event
def on_key_press(symbol, modifiers):
if symbol == key.NUMLOCK:
if GetKeyState(VK_NUMLOCK):
#print(GetKeyState(VK_NUMLOCK))#should be 0 and 1 but
main(on)
else:
main(off)
#window.event
def on_draw():
window.clear()
### display icon in systray
import sys
from PyQt5 import QtCore, QtGui, QtWidgets
#source: http://stackoverflow.com/questions/893984/pyqt-show-menu-in-a-system-tray-application - add answer PyQt5
class SystemTrayIcon(QtWidgets.QSystemTrayIcon):
def __init__(self, icon, parent=None):
QtWidgets.QSystemTrayIcon.__init__(self, icon, parent)
menu = QtWidgets.QMenu(parent)
exitAction = menu.addAction("Exit")
self.setContextMenu(menu)
def main(image):
app = QtWidgets.QApplication(sys.argv)
w = QtWidgets.QWidget()
trayIcon = SystemTrayIcon(QtGui.QIcon(image), w)
trayIcon.show()
sys.exit(app.exec_())
if __name__ == '__main__':
pyglet.app.run()
The reason for QCoreApplication::exec: The event loop is already running is actually because you're trying to start app.run() twice. Qt will notice there's already an instance running and throw this exception. When instead, what you want to do is just swap the icon in the already running instance.
Your main problem here is actually the mix of libraries to solve one task if you ask me.
Rather two tasks, but using Qt5 for the graphical part is fine tho.
The way you use Pyglet is wrong from the get go.
Pyglet is intended to be a highly powerful and effective graphics library where you build a graphics engine ontop of it. For instance if you're making a game or a video-player or something.
The way you use win32api is also wrong because you're using it in a graphical window that only checks the value when a key is pressed inside that window.
Now, if you move your win32api code into a Thread (a QtThread to be precise) you can check the state no matter if you pressed your key inside your graphical window or not.
import sys
import win32api
import win32con
from PyQt5 import QtCore, QtGui, QtWidgets
from threading import Thread, enumerate
from time import sleep
class SystemTrayIcon(QtWidgets.QSystemTrayIcon):
def __init__(self, icon, parent=None):
QtWidgets.QSystemTrayIcon.__init__(self, icon, parent)
menu = QtWidgets.QMenu(parent)
exitAction = menu.addAction("Exit")
exitAction.setShortcut('Ctrl+Q')
exitAction.setStatusTip('Exit application')
exitAction.triggered.connect(QtWidgets.qApp.quit)
self.setContextMenu(menu)
class KeyCheck(QtCore.QThread):
def __init__(self, mainWindow):
QtCore.QThread.__init__(self)
self.mainWindow = mainWindow
def run(self):
main = None
for t in enumerate():
if t.name == 'MainThread':
main = t
break
while main and main.isAlive():
x = win32api.GetAsyncKeyState(win32con.VK_NUMLOCK)
## Now, GetAsyncKeyState returns three values,
## 0 == No change since last time
## -3000 / 1 == State changed
##
## Either you use the positive and negative values to figure out which state you're at.
## Or you just swap it, but if you just swap it you need to get the startup-state correct.
if x == 1:
self.mainWindow.swap()
elif x < 0:
self.mainWindow.swap()
sleep(0.25)
class GUI():
def __init__(self):
self.app = QtWidgets.QApplication(sys.argv)
self.state = True
w = QtWidgets.QWidget()
self.modes = {
True : SystemTrayIcon(QtGui.QIcon('on.png'), w),
False : SystemTrayIcon(QtGui.QIcon('off.png'), w)
}
self.refresh()
keyChecker = KeyCheck(self)
keyChecker.start()
sys.exit(self.app.exec_())
def swap(self, state=None):
if state is not None:
self.state = state
else:
if self.state:
self.state = False
else:
self.state = True
self.refresh()
def refresh(self):
for mode in self.modes:
if self.state == mode:
self.modes[mode].show()
else:
self.modes[mode].hide()
GUI()
Note that I don't do Qt programming often (every 4 years or so).
So this code is buggy at it's best. You have to press Ctrl+C + Press "Exit" in your menu for this to stop.
I honestly don't want to put more time and effort in learning how to manage threads in Qt or how to exit the application properly, it's not my area of expertis. But this will give you a crude working example of how you can swap the icon in the lower corner instead of trying to re-instanciate the main() loop that you did.
from PyQt4.QtCore import *
from PyQt4.QtGui import *
import sys, os, time
class SetName(QWidget):
def __init__(self):
QWidget.__init__(self)
self.show()
toplayout = QVBoxLayout()
self.setWindowTitle('Personal Info')
self.form_layout = QFormLayout()
self.setLayout(self.form_layout)
self.line_edit_param = QLineEdit(self)
self.line_edit_param.setPlaceholderText("Write Here")
self.form_layout.addRow('Write Name', self.line_edit_param)
toplayout.addLayout(self.form_layout)
self.setFocus()
class LearnApp(QDialog):
def __init__(self):
super(QDialog, self).__init__()
self.setWindowTitle("LearnApp")
self.active = False
close_button = QPushButton("Close")
close_button.clicked.connect(self.close)
self.check_button = QPushButton("Check")
self.check_button.clicked.connect(self.set_data)
self.tr = QTextEdit()
self.tr.setReadOnly(True)
# layout
layout = QHBoxLayout()
#layout.addWidget(self.button3)
sub_layout = QVBoxLayout()
sub_layout.addWidget(self.check_button)
sub_layout.addWidget(close_button)
layout.addLayout(sub_layout)
layout.addWidget(self.tr)
self.setLayout(layout)
self.setFocus()
def set_data(self):
print "in set_data"
SetName()
app = QApplication(sys.argv)
dialog = LearnApp()
dialog.show()
app.exec_()
This is the code I'm trying. If edit it with toplayout = QGridLayout(), program works fine but with toplayout = QVBoxLayout(), it gives message QLayout::addChildLayout: layout "" already has a parentand just flashes the new window. What could be the problem? How should I tackle this? I wanna use QVBoxLayout instead of QGridLayout
Firstly, the new window disappears straight away because you don't store a reference to it. You need to store a reference to the instance in your LearnApp class, or parent it to another Qt object outside of set_data() if you want it to stick around.
The error message regarding the layouts is not occurring because of your choice of layouts, but because you are calling
self.setLayout(self.form_layout)
and then
toplayout.addLayout(self.form_layout)
The first call assigns the layout to the instance of SetName, but in doing so also makes the instance the parent of self.form_layout. The second call is trying to add the same layout to toplayout and set it as the parent, but Qt sees that self.form_layout already has a parent (i.e. is being used elsewhere). This is what the error message is trying to tell you.
I suspect that instead of self.setLayout(self.form_layout), you intended to write something like
self.setLayout(toplayout)
I want to run my app in different tabs. Rightnow, it is running in main window. I have done some search work on how to create tabs. I found this to be useful, but not sufficient to meet my requirements
Create TAB and create textboxes that take data in the TAB-page
I want to have a feature of adding a new tab (like new tab in chrome)
Below is my code sample. I described what i require in the comments.
from PyQt4 import Qt, QtCore, QtGui
import sys
class Map(QtGui.QMainWindow):
def __init__(self,parentQExampleScrollArea=None,parentQWidget = None):
super(Map,self).__init__()
self.initUI()
#Initialize the UI
def initUI(self):
#Initilizing Menus toolbars
#This must be maintained in all tabbed panes
filename = ""
#Filename is obtained through open file button in file menu
self.filename = filename
def paintEvent(self, e):
qp = QtGui.QPainter()
qp.begin(self)
self.drawPoints(qp,self.filename)
qp.end()
def drawPoints(self, qp,FILENAME=""):
#Read contents in file
#Get the necessary coordinates
#A separate class for storing the info of all the coordinates
#Run a for loop for all the coordinates in the list
#Actually, object is created here and image of that object is set
# as a square using the coordinates
qp.setBrush(QtGui.QColor(255, 0, 20, 200))
qp.drawRect(20,20,75,75)
qp.drawRect(100,20,75,75)
self.update()
#There are many functions to handle keyboard and mouse events
def main():
#How do I modify so that I can have multiple tabs
#And show images of the coordinates in files
#Basically I want to have the feature of opening many files
# and displaying them in UI
#Basically a feature to add a new tab
#like that of in eclipse netbeans sublime etc
app = QtGui.QApplication(sys.argv)
myQExampleScrollArea = Map()
myQExampleScrollArea.show()
sys.exit(app.exec_())
if __name__ == '__main__':
main()
Thanks in advance.. :)
It's simply to use method int QTabWidget.addTab (self, QWidget widget, QString) to create widget in tab. In each tab, I suggest use QtGui.QWidget more than QtGui.QMainWindow;
Example;
import sys
from PyQt4 import QtGui
class QCustomWidget (QtGui.QWidget):
# Your widget to implement
# Put your override method here
def paintEvent (self, eventQPaintEvent):
currentQPainter = QtGui.QPainter()
currentQPainter.begin(self)
currentQPainter.setBrush(QtGui.QColor(255, 0, 20, 200))
currentQPainter.drawRect(20, 20, 75, 75)
currentQPainter.drawRect(100, 20, 75, 75)
self.update()
currentQPainter.end()
class QCustomTabWidget (QtGui.QTabWidget):
def __init__ (self, parent = None):
super(QCustomTabWidget, self).__init__(parent)
self.addTab(QtGui.QPushButton('Test'), 'Tab 1')
self.addTab(QCustomWidget(), 'Tab 2')
myQApplication = QtGui.QApplication([])
myQCustomTabWidget = QCustomTabWidget()
myQCustomTabWidget.show()
sys.exit(myQApplication.exec_())
So handle with more tab, It's bad to create many line call int QTabWidget.addTab (self, QWidget widget, QString). Anyway, All widget has add in QTabWidget is can reference in ifself. So, your can control element in it by call QWidget QTabWidget.widget (self, int index).
Example to call widget in tab widget;
import os
import sys
from PyQt4 import QtCore, QtGui
class QCustomLabel (QtGui.QLabel):
def __init__(self, imagePath, parentQWidget = None):
super(QCustomLabel, self).__init__(parentQWidget)
self.setPixmap(QtGui.QPixmap(imagePath))
class QCustomWidget (QtGui.QWidget):
def __init__ (self, parentQWidget = None):
super(QCustomWidget, self).__init__(parentQWidget)
self.addQPustButton = QtGui.QPushButton('Open image')
self.addQPustButton.setMaximumWidth(120)
self.addQPustButton.released.connect(self.openImage)
self.workSpaceQTabWidget = QtGui.QTabWidget()
self.workSpaceQTabWidget.setTabsClosable(True)
self.workSpaceQTabWidget.tabCloseRequested.connect(self.closeImage)
allQVBoxLayout = QtGui.QVBoxLayout()
allQVBoxLayout.addWidget(self.addQPustButton)
allQVBoxLayout.addWidget(self.workSpaceQTabWidget)
self.setLayout(allQVBoxLayout)
def openImage (self):
path = QtGui.QFileDialog.getOpenFileName(self, 'Open image')
if not path.isEmpty():
self.workSpaceQTabWidget.addTab(QCustomLabel(path), QtCore.QString(os.path.basename(str(path))))
def closeImage (self, currentIndex):
currentQWidget = self.workSpaceQTabWidget.widget(currentIndex)
currentQWidget.deleteLater()
self.workSpaceQTabWidget.removeTab(currentIndex)
myQApplication = QtGui.QApplication([])
myQCustomWidget = QCustomWidget()
myQCustomWidget.show()
sys.exit(myQApplication.exec_())
When I minimize the application window on Windows XP and restore it later, the dock will be hidden. This has to do with view menu which has toggles to set visibility and of course is connected by signals.
I hope this will save someone a few hours of debugging.
Here is a full functional example with both wrong and right code:
# -*- coding: utf-8 -*-
import sys
from PyQt4 import QtCore, QtGui
class Ui_QMainWindow(QtGui.QMainWindow):
def __init__(self):
QtGui.QMainWindow.__init__(self)
self.resize(200, 200)
self.menubar = QtGui.QMenuBar(self)
self.menubar.setGeometry(QtCore.QRect(0, 0, 800, 27))
self.menuMenu = QtGui.QMenu(self.menubar)
self.setMenuBar(self.menubar)
self.dock = QtGui.QDockWidget(self)
self.dock.setObjectName("dock")
self.dockContents = QtGui.QWidget()
self.dockContents.setObjectName("dockContents")
self.dock.setWidget(self.dockContents)
self.addDockWidget(QtCore.Qt.DockWidgetArea(4), self.dock)
self.action = QtGui.QAction(self)
self.action.setCheckable(True)
self.action.setChecked(True)
self.action.setObjectName("action")
self.menuMenu.addAction(self.action)
self.menubar.addAction(self.menuMenu.menuAction())
self.setWindowTitle("Example of dock remaining minimized")
self.menuMenu.setTitle("Menu")
self.dock.setWindowTitle("I'm a dock")
self.action.setText("Dock visibility")
if True:
# This is NOT working on Windows XP.
# Minimize the window and restore again, the dock is gone.
# Other than that it works.
QtCore.QObject.connect(self.action,
QtCore.SIGNAL("toggled(bool)"),
self.dock.setVisible)
QtCore.QObject.connect(self.dock,
QtCore.SIGNAL("visibilityChanged(bool)"),
self.action.setChecked)
else:
# This DOES work, but boy it looks nasty, writing useless
# per dock is not nice.
QtCore.QObject.connect(self.action,
QtCore.SIGNAL("triggered()"),
self.toggle_dock)
QtCore.QObject.connect(self.dock,
QtCore.SIGNAL("visibilityChanged(bool)"),
self.action.setChecked)
def toggle_dock(self):
self.dock.setVisible(not self.dock.isVisible())
def main():
app = QtGui.QApplication(sys.argv)
ui = Ui_QMainWindow()
ui.show()
sys.exit(app.exec_())
if __name__ == "__main__":
main()
There is a much simpler way to do this, using QDock.toggleViewAction. This function returns a ready-made action that handles the checked state automatically.
So your code would become simply:
self.action = self.dock.toggleViewAction()
self.action.setObjectName("action")
self.menuMenu.addAction(self.action)
self.menubar.addAction(self.menuMenu.menuAction())
self.setWindowTitle("Example of dock remaining minimized")
self.menuMenu.setTitle("Menu")
self.dock.setWindowTitle("I'm a dock")
self.action.setText("Dock visibility")
and you can then get rid of all the signal handling.
I'm trying to open a window (QWidget) when clicking on a button. My problem is that the second window doesn't show up when I click on the button no matter what I've tried. The two windows are created using QTDesigner.
Here is a little snippet explaining what I'm trying to do:
from PyQt4.QtGui import *
from PyQt4.QtCore import *
from win1 import *
from win2 import *
import sys
class win1(QWidget, Ui_Win1):
def __init__(self, parent = None):
self.parent = parent
QWidget.__init__(self)
self.setupUi(parent)
self.connect(self.pushButton, SIGNAL("clicked()"), self.on_btn_clicked)
def on_btn_clicked(self):
self.child = win2(self.parent)
self.child.show()
class win2(QWidget, Ui_Win2):
def __init__(self, parent = None):
QWidget.__init__(self)
self.setupUi(parent)
def main(args):
app = QApplication(args)
win = QWidget()
a = win1(win)
win.show()
result = app.exec_()
if __name__=="__main__":
main(sys.argv)
What am I missing here ?
Thanks.
Not sure, but two random thoughts:
If you add a print statement to on_btn_clicked, do you see anything when you click on the button? This would diagnose whether it's an event triggering issue
Does it work if you change the setupUI(parent) commands to setupUI(self)?