Question

我有一个线程应用程序，我有一个网络线程。 UI部分将callback传递给此线程。线程是普通 python线程 - 它是否 QThread。

是否可以在此线程中发出PyQT Slot？

Answer 1

不，不可能像这样从python线程发出PyQt信号。

但是，可能的解决方案是使用共享的其他对象两个线程，进行必要的操作，最终发出一个线程安全的 PyQt信号。

这是“SafeConnector”类的实现，使用了一对连接的套接字和一个Queue在两个线程之间交换数据，并使用QSocketNotifier返回Qt循环。 QObject用于可以发出适当的Qt信号：

from PyQt4 import Qt, QtCore, QtGui
import threading
import socket
import Queue
import time

# Object of this class has to be shared between
# the two threads (Python and Qt one).
# Qt thread calls 'connect',   
# Python thread calls 'emit'.
# The slot corresponding to the emitted signal
# will be called in Qt's thread.
class SafeConnector:
    def __init__(self):
        self._rsock, self._wsock = socket.socketpair()
        self._queue = Queue.Queue()
        self._qt_object = QtCore.QObject()
        self._notifier = QtCore.QSocketNotifier(self._rsock.fileno(),
                                                QtCore.QSocketNotifier.Read)
        self._notifier.activated.connect(self._recv)

    def connect(self, signal, receiver):
        QtCore.QObject.connect(self._qt_object, signal, receiver)

    # should be called by Python thread
    def emit(self, signal, args):
        self._queue.put((signal, args))
        self._wsock.send('!')

    # happens in Qt's main thread
    def _recv(self):
        self._rsock.recv(1)
        signal, args = self._queue.get()
        self._qt_object.emit(signal, args)

class PythonThread(threading.Thread):
    def __init__(self, connector, *args, **kwargs):
        threading.Thread.__init__(self, *args, **kwargs)
        self.connector = connector
        self.daemon = True

    def emit_signal(self):
        self.connector.emit(QtCore.SIGNAL("test"), str(time.time()))

    def run(self):
        while True:
            time.sleep(1)
            self.emit_signal()

if __name__ == '__main__':
    app = QtGui.QApplication([])
    mainwin = QtGui.QMainWindow()
    label = QtGui.QLabel(mainwin)
    mainwin.setCentralWidget(label)

    connector = SafeConnector()
    python_thread = PythonThread(connector)
    connector.connect(QtCore.SIGNAL("test"), label.setText)
    python_thread.start()

    mainwin.show()
    app.exec_()

在标准python线程中发出信号

1 个答案: