pycharm - 似乎无法在简单的 PyQt 程序中修复 PyCharm 警告

Question

以下程序(我的真实代码的摘录)

from PyQt4 import QtGui
import sys
from PyQt4.QtGui import QMessageBox


def main():
    app = QtGui.QApplication([])

    w = QtGui.QPushButton("Test")

    def on_pressed():
        print("Pressed")
        QMessageBox.warning(w, 'Info', 'Button Pressed')

    w.pressed.connect(on_pressed)
    w.show()

    sys.exit(app.exec_())


if __name__ == '__main__':
    main()

在 PyCharm 中触发与 Code/Inspect Code... 相关的以下三个警告(例如，在运行 QMessageBox.warning 时)称呼

Calling a method by class using an instance of a different class; Passing PyQt4.QtGui.QPushButton.QPushButton instead of PyQt4.QtGui.QMessageBox.QMessageBox. Is this intentional?

Incorrect call arguments; Parameter 'QString_1' unfilled

Type checker; Expected type 'QMessageBox', got 'QPushButton' instead

和一个与 PyQt 相关的警告 connect称呼

Cannot find reference 'connect' in 'function'

知道如何解决/避免这些警告吗？

Answer 1

我遇到了同样的问题。您将函数传递给 QPushButton，在您的情况下为“w”。该函数需要一个 QMessageBox。所以只需像这样传递一个:

from PyQt4 import QtGui
import sys
from PyQt4.QtGui import QMessageBox


def main():
    app = QtGui.QApplication([])

    w = QtGui.QPushButton("Test")
    dialog = QtGui.QMessageBox()

    def on_pressed():
        print("Pressed")
        QMessageBox.warning(dialog, 'Info', 'Button Pressed', 'Okay')

    w.pressed.connect(on_pressed)
    w.show()

    sys.exit(app.exec_())


if __name__ == '__main__':
    main()