Question

PyQT QThread似乎没有发出信号。我想用PyQt5构建我的代码的一部分，它将与服务器通信。我希望它在单独的线程中，因为我主要在图像上操作，所以发送它可能需要一些时间。我希望能够发送它，当它从服务器收到响应时，它应该发出一个信号，允许发送下一组数据。

以下代码减少到最低限度。

导入系统进口时间

string sEmail = string.Empty;
foreach (ListItem item in myListItem.Items)
{
    if (item.Selected)
    {
        sEmail += item.Value + "; ";
    }
}

带有loop1（）的代码的输出：

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_())

（它被卡住了）

带有loop2（）的代码的输出：

adding data b'sssss'
receiving_datab'some_message_received'

在我看来，从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'内部发出的信号在某种程度上被卡住了，而不是向QtClientManager类发送数据。怎么做才能阻止？

如何正确地在QThread循环中发出信号

0 个答案: