似乎无法在简单的 PyQt 程序中修复 PyCharm 警告 [英] can't seem to fix PyCharm warnings in a simple PyQt program

问题描述

以下程序(我的真实代码摘录)

The following program (an extract of my real code)

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 QMessageBox.warning 调用相关的以下三个警告(例如，在运行 Code/Inspect Code... 时)

triggers the following three warnings in PyCharm (eg when running Code/Inspect Code... ) related to the QMessageBox.warning call

使用不同类的实例按类调用方法；传递 PyQt4.QtGui.QPushButton.QPushButton 而不是PyQt4.QtGui.QMessageBox.QMessageBox.这是故意的吗?

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?

调用参数不正确；参数QString_1"未填充

Incorrect call arguments; Parameter 'QString_1' unfilled

类型检查器；预期类型'QMessageBox'，改为'QPushButton'

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

和一个与 PyQt connect 调用相关的警告

and one warning related to the PyQt connect call

在函数"中找不到引用连接"

Cannot find reference 'connect' in 'function'

知道如何解决/避免这些警告吗?

Any idea how I can work-around/avoid these warnings?

推荐答案

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.所以就这样传递它:

I came across the same problem. You're passing the function a QPushButton, in your case the 'w'. The function is expecting a QMessageBox. So just pass it one like this:

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()

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