python - PyQt4 : inheriting QThread

Question

我有一个QWidget，它使用QThread调用一些python代码。然后，该代码还执行一些逻辑并调用另一个类。在此类中，我想将信号传递到我的 QWidget，以便打印“Hello World”。目前的代码给出了错误 sm instance has no attribute 'sayHello'，我知道 - 我只是非常不知道如何使 run 方法调用不同的方法类以便所有信号正常工作。

widget.py

from prog import TaskThread
from PyQt4 import QtCore, QtGui
class MyWidget(QtGui.QWidget):
  def __init__(self):
    self.btn = QtGui.QPushButton('Run!', self)
    self.btn.clicked.connect(self.onStart)
    self.myLongTask = TaskThread()
    self.myLongTask.sayHello.connect(self.sayHi)

  def onStart(self):
    self.myLongTask.start()

  def sayHi(self):
    print "hello world"

prog.py

from PyQt4 import QtCore 
from sm import sc
class TaskThread(QtCore.QThread):
  sayHello = QtCore.pyqtSignal()
  def run(self):
    sm.sc()

sm.py

class sc():
  def __init__(self):
    for i in range(0,50):
      print i
      if i == 5: self.sayHello.emit()

Answer 1

我必须稍微改变你的代码才能让它在我的机器上运行，但是手动将信号实例传递给 sc 类肯定会引发信号。输出可能与 i 循环的打印交错。

widget.py

from prog import TaskThread
import sys
from PyQt4 import QtCore, QtGui

class flexemWidget(QtGui.QWidget):
  def __init__(self):
    super(flexemWidget, self).__init__()
    self.btn = QtGui.QPushButton('Run!', self)
    self.btn.clicked.connect(self.onStart)
    self.myLongTask = TaskThread()
    self.myLongTask.sayHello.connect(self.sayHi)
    self.show()

  def onStart(self):
    self.myLongTask.start()

  def sayHi(self):
    print "hello world" 

app = QtGui.QApplication(sys.argv)
ex = flexemWidget()
sys.exit(app.exec_())

prog.py

from PyQt4 import QtCore 
from sm import sc
class TaskThread(QtCore.QThread):
  sayHello = QtCore.pyqtSignal()
  def run(self):
    sc(self.sayHello)

sc.py

from PyQt4 import QtCore
class sc():
  def __init__(self, signal):
    for i in range(0,50):
      print i
      if i == 5: signal.emit()