CODEWITHSUNDEEP
pythonlinuxpyqtpyqt4
user285594
user285594

Reputation:

Python - How to show the window + print the text? Where its only printing but not showing the window

How to show the window + print that text? Where it does not show window anymore if i have my while loop on.

import sys
import datetime
import time
from PyQt4 import QtCore, QtGui

class Main(QtGui.QMainWindow):
  def __init__(self, parent=None):
    super(Main, self).__init__(parent)       
    self.b = QtGui.QPushButton("exit", self, clicked=self.close)
    self.c = QtGui.QLabel("Test", self)

if __name__ == "__main__":
  app=QtGui.QApplication(sys.argv)
  myapp=Main()
  myapp.show()     
  while True:
    time.sleep(2)
    print "Print this + Show the Window???!!!"
  sys.exit(app.exec_())

Tried: 

import sys
import datetime
import time
from PyQt4 import QtCore, QtGui

class Main(QtGui.QMainWindow):
  def __init__(self, parent=None):
    super(Main, self).__init__(parent)       
    self.b = QtGui.QPushButton("exit", self, clicked=self.close)
    self.c = QtGui.QLabel("Test", self)

  def myRun():
    while True:
      time.sleep(2)
      print "Print this + Show the Window???!!!"      

if __name__ == "__main__":
  app=QtGui.QApplication(sys.argv)
  myapp=Main()
  myapp.show()     

  thread = QtCore.QThread()
  thread.run = lambda self: myRun()
  thread.start()    
  sys.exit(app.exec_())

Output:

TypeError: () takes exactly 1 argument (0 given)

Upvotes: 1

Views: 373

Answers (2)

PenguinCoder
PenguinCoder

Reputation: 4367

Few issues: 1) You're not calling or initializing the thread correctly. 2) You need to tell your main thread to keep processing events while the other thread is running 3) Your label is hovering over the 'exit' button, and therefore you'll not be able to ever click it! 

import sys
import datetime
import time
from PyQt4 import QtCore, QtGui

class Main(QtGui.QMainWindow):
  def __init__(self, parent=None):
    super(Main, self).__init__(parent)       
    self.b = QtGui.QPushButton("exit", self, clicked=self.close)

  def myRun(self):
    while True:
      time.sleep(2)
      print "Print this + Show the Window???!!!"      

if __name__ == "__main__":
  app=QtGui.QApplication(sys.argv)
  myapp=Main()
  myapp.show()     

  thread = QtCore.QThread()
  thread.run = lambda myapp=myapp: myapp.myRun()
  thread.start()    

  app.connect(app, QtCore.SIGNAL("lastWindowClosed()"), app, QtCore.SLOT("quit()"))

  sys.exit(app.exec_())

  while thread.isAlive():
    #Make sure the rest of the GUI is responsive
    app.processEvents()

Upvotes: 2

Aaron Digulla
Aaron Digulla

Reputation: 328724

lambda self: myRun() tries to invoke the global function myRun(). Try 

lambda myapp=myapp: myapp.myRun()

instead. The odd assignment will create a default parameter because thread.run() doesn't get one.

Upvotes: 0

Related Questions