Question

我的目标是在按 Enter 时向服务器发送消息。下面我发布的解决方案每次按下按键时都有效，并且与哪个按键无关。我如何得到acctualy按下的键？

from sys import argv, exit

from interface import Ui_MainWindow
from PyQt5.QtGui import QKeyEvent
from PyQt5.QtCore import Qt, pyqtSlot
from PyQt5.QtWidgets import QApplication, QMainWindow
from client import connect_to_server, send_message


class Communicator(QMainWindow):
    def __init__(self):
        super(QMainWindow, self).__init__()
        self.ui = Ui_MainWindow()
        self.ui.setupUi(self)

    @pyqtSlot()
    def on_message_box_textChanged(self):
        server_socket = connect_to_server()
        message = self.ui.message_box.toPlainText()
        send_message(message, server_socket)


def main():
    main_application = QApplication(argv)
    window = Communicator()
    window.show()
    exit(main_application.exec_())

if __name__ == '__main__':
    main()

Answer 1

假设您使用auto-connection，只需将广告位的名称更改为：

    def on_message_box_returnPressed(self):
        ...

这会使用returnPressed信号，但稍微更好的选择可能是使用editingFinished信号：

    def on_message_box_editingFinished(self):
        ...

使用显式信号连接，您的代码如下所示：

class Communicator(QMainWindow):
    def __init__(self):
        super(QMainWindow, self).__init__()
        self.ui = Ui_MainWindow()
        self.ui.setupUi(self)
        self.ui.message_box.editingFinished.connect(self.sendMessage)

    def sendMessage(self):
        server_socket = connect_to_server()
        message = self.ui.message_box.toPlainText()
        send_message(message, server_socket)

在QLineEdit中按下的处理键，哪个实例在QMainWindow中

1 个答案: