如何正确地在Qthread循环中发出信号

Question

pyqt qthread似乎在应该时不会发出信号。我想用PYQT5构建代码的一部分，该代码可以与服务器通信。我希望它在单独的线程中，因为我主要在图像上操作，因此可能需要一些时间发送它。我希望能够发送它，当它从服务器接收响应时，它应该发出一个信号，允许发送下一个数据。

这是将代码减少到最低。

导入系统 导入时间

from PyQt5.Qt import QThread, QWidget, QApplication
from PyQt5.QtCore import pyqtSignal

def ask_server():
    return b'some_message_received'
def send_to_server(data):
    pass
class QTClient(QThread):
    result_signal = pyqtSignal(bytes)
    def __init__(self, *args, **kwargs):
        QThread.__init__(self)
        self._data = None
        self._result = ' '
    def set_data(self, data):
        self._data = data
    def set_result(self, result):
        self._result = result
    def receive_data(self):
        result = ask_server()
        print('receiving_data{}'.format(result))
        self.result_signal.emit(result)
        self._result = result
    def send_data(self, data):
        send_to_server(data)
        self._data = None
    def run(self):
        broadcast = True
        while broadcast:
            if self._data is not None:
                self.send_data(self._data)
                self.receive_data()
            if self._result is None:
                self.receive_data()

class QtClientManager(QWidget):
    data = pyqtSignal(bytes)
    def __init__(self, *args, **kwargs):
        QWidget.__init__(self, *args, **kwargs)
        self.slam_client = QTClient()
        self.data[bytes].connect(self.slam_client.set_data)
        self._result = ''
        self.slam_client.result_signal[bytes].connect(self.get_result)
        self.thread_slam = QThread()
        self.slam_client.moveToThread(self.thread_slam)
        self.slam_client.start()
    def add_image(self, data):
        print('adding data {}'.format(data))
        self.data.emit(data)
    def get_result(self, result):
        print('getting result data {}'.format(result))
        self._result = result
def loop1():
    for i in range(2):
        qqq.add_image(b'sssss')
        while qqq._result != b'some_message_received':
            time.sleep(1)

def loop2():
    for i in range(2):
        qqq.add_image(b'sssss')
        #while qqq._result != b'some_message_received':
        time.sleep(1)
if __name__ == "__main__":
    app = QApplication([])
    qqq = QtClientManager()
    loop1()
    sys.exit(app.exec_())

使用loop1（）的代码输出：

adding data b'sssss'
receiving_datab'some_message_received'

（它被卡住了）

loop2（）代码的输出：

adding data b'sssss'
receiving_datab'some_message_received'
adding data b'sssss'
receiving_datab'some_message_received'
getting result data b'some_message_received'
getting result data b'some_message_received'

在我看来，从 receive_data()中发出的信号以某种方式卡住，而不是将数据发送到qtclientmanager类。如何这样做，以免阻止？

Answer 1

我在代码中也有类似的问题，然后发现这是由于连接函数多次注册，这又导致了对emit函数的多次调用