如何在pyqt5中将函数输出设置为信号

Question

我想显示在标签小部件中显示的登录用户 os.getlogin()。

到目前为止我需要设置一个自定义信号并将其连接到标签

这是我的（无法运行的）代码，希望您能指导我如何正确编写它

import sys
import os

from PyQt5 import QtWidgets as qtw
from PyQt5 import QtCore as qtc
from PyQt5 import QtGui as qtg



class UserTest(qtw.QWidget):
    def __init__(self, *args, **kwargs):
        super().__init__(*args, **kwargs)
        # your code will go here

        user_in_label = qtw.QLabel()

        layout = qtw.QVBoxLayout()

        layout.addWidget(user_in_label)

        # funktionen

        self.trigger = qtc.pyqtSignal(str)


        # fire signal to slot
        self.set_signal.connect(user_in_label.setText)


        # your code ends here
        self.show()

    def set_signal(self):
        active_user = os.getlogin()
        self.trigger.emit(active_user)

if __name__ == '__main__':
    app = qtw.QApplication(sys.argv)
    w = UserTest()
    sys.exit(app.exec_())

self.set_signal.connect(user_in_label.setText) 这一行缺少一个对象，我需要为此设置一个新的class吗，好像是胡说八道

Answer 1

试一试：

import sys
import os
from PyQt5 import QtWidgets as qtw
from PyQt5 import QtCore as qtc
from PyQt5 import QtGui as qtg


class UserTest(qtw.QWidget):

    trigger = qtc.pyqtSignal(str)                                             # +++

    def __init__(self, *args, **kwargs):
        super().__init__(*args, **kwargs)

        user_in_label = qtw.QLabel()

        layout = qtw.QVBoxLayout(self)                                         # + self
        layout.addWidget(user_in_label)
        layout.addWidget(qtw.QPushButton('Click me', clicked=self.set_signal)) # +++

        self.trigger.connect(user_in_label.setText)                            # +++

    def set_signal(self):
        active_user = os.getlogin()
        self.trigger.emit(active_user)


if __name__ == '__main__':
    app = qtw.QApplication(sys.argv)
    w = UserTest()
    w.show()
    sys.exit(app.exec_())

更新

can you explain why you relocate trigger = qtc.pyqtSignal(str) above init ? and any suggestions how to set active user automatically to the label without the push button?

trigger 必须是 class 属性

import sys
import os
from PyQt5 import QtWidgets as qtw
from PyQt5 import QtCore as qtc
from PyQt5 import QtGui as qtg


class UserTest(qtw.QWidget):

    trigger = qtc.pyqtSignal(str)                                            

    def __init__(self, *args, **kwargs):
        super().__init__(*args, **kwargs)

        user_in_label = qtw.QLabel()

        layout = qtw.QVBoxLayout(self)                                         
        layout.addWidget(user_in_label)
#        layout.addWidget(qtw.QPushButton('Click me', clicked=self.set_signal)) 

        self.trigger.connect(user_in_label.setText)                            

        self.set_signal()                                                # +++

    def set_signal(self):
        active_user = os.getlogin()
        self.trigger.emit(active_user)


if __name__ == '__main__':
    app = qtw.QApplication(sys.argv)
    w = UserTest()
    w.show()
    sys.exit(app.exec_())

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