Python 如何正确地在QThread循环中发出信号
PyQT QThread似乎没有在应该发出信号的时候发出信号。我想用PyQt5构建代码的一部分,它将与服务器通信。我希望它在单独的线程中,因为我主要操作图像,所以发送它可能需要一些时间。我希望能够发送它,当它收到来自服务器的响应时,它应该发出一个信号,允许发送下一组数据 下面是代码缩减到最小值 导入系统 导入时间Python 如何正确地在QThread循环中发出信号,python,python-3.x,qt,qt5,pyqt5,Python,Python 3.x,Qt,Qt5,Pyqt5,PyQT QThread似乎没有在应该发出信号的时候发出信号。我想用PyQt5构建代码的一部分,它将与服务器通信。我希望它在单独的线程中,因为我主要操作图像,所以发送它可能需要一些时间。我希望能够发送它,当它收到来自服务器的响应时,它应该发出一个信号,允许发送下一组数据 下面是代码缩减到最小值 导入系统 导入时间 from PyQt5.Qt import QThread, QWidget, QApplication from PyQt5.QtCore import pyqtSignal de
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()的代码的输出:
(它卡住了)
使用loop2()的代码的输出:
在我看来,从
receive_data()
中发出的信号不知何故被卡住了,并且没有将数据发送到QtClientManager类。如何做到这一点,使其不会被阻止?在loop1
中的while
循环阻止了Qt
事件循环,阻止了排队信号的传递。在loop1
中的while
循环阻止了Qt
事件循环,阻止了排队信号的传递。
adding data b'sssss'
receiving_datab'some_message_received'
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'