pyqtSlot装饰器为什么会导致“ TypeError：connect（）失败”？ [英] Why does pyqtSlot decorator cause "TypeError: connect() failed"?

问题描述

我有这个带有PyQt5的Python 3.5.1程序，以及一个从QtCreator ui文件创建的GUI，其中pyqtSlot装饰器导致 TypeError：textChanged（QString）和edited（）之间的connect（）失败。

I have this Python 3.5.1 program with PyQt5 and a GUI created from a QtCreator ui file where the pyqtSlot decorator causes "TypeError: connect() failed between textChanged(QString) and edited()".

在重现该问题的示例代码中，我有2个自定义类：MainApp和LineEditHandler。 MainApp实例化主GUI（从文件 mainwindow.ui），并且LineEditHandler处理QLineEdit对象。 LineEditHandler存在的原因是将与QLineEdit对象主要相关的自定义方法集中在一个类中。它的构造函数需要QLineEdit对象和MainApp实例（在需要时访问其他对象/属性）。

In the sample code to reproduce the problem I have 2 custom classes: MainApp and LineEditHandler. MainApp instantiates the main GUI (from the file "mainwindow.ui") and LineEditHandler handles a QLineEdit object. LineEditHandler reason to exist is to concentrate the custom methods that relate mostly to the QLineEdit object in a class. Its constructor needs the QLineEdit object and the MainApp instance (to access other objects/properties when needed).

在MainApp中，我将QLineEdit的textChanged信号连接到LineEditHandler.edited （）。如果我不使用pyqtSlot（）装饰LineEditHandler.edited（），则一切正常。如果我对方法使用@pyqtSlot（），则代码运行将失败，并显示 TypeError：textChanged（QString）和edited（）之间的connect（）失败。我在这里做什么错了？

In MainApp I connect the textChanged signal of the QLineEdit to LineEditHandler.edited(). If I don't decorate LineEditHandler.edited() with pyqtSlot() everything works fine. If I do use @pyqtSlot() for the method, the code run will fail with "TypeError: connect() failed between textChanged(QString) and edited()". What am I doing something wrong here?

您可以在以下位置获取mainwindow.ui文件： https://drive.google.com/file/d/0B70NMOBg3HZtUktqYVduVEJBN2M/view

You can get the mainwindow.ui file at: https://drive.google.com/file/d/0B70NMOBg3HZtUktqYVduVEJBN2M/view

是生成问题的示例代码：

And this is the sample code to generate the problem:

import sys
from PyQt5 import uic
from PyQt5.QtWidgets import *
from PyQt5.QtCore import pyqtSlot


Ui_MainWindow, QtBaseClass = uic.loadUiType("mainwindow.ui")


class MainApp(QMainWindow, Ui_MainWindow):

    def __init__(self):
        # noinspection PyArgumentList
        QMainWindow.__init__(self)
        Ui_MainWindow.__init__(self)
        self.setupUi(self)

        # Instantiate the QLineEdit handler.
        self._line_edit_handler = LineEditHandler(self, self.lineEdit)
        # Let the QLineEdit handler deal with the QLineEdit textChanged signal.
        self.lineEdit.textChanged.connect(self._line_edit_handler.edited)


class LineEditHandler:

    def __init__(self, main_window, line_edit_obj):
        self._line_edit = line_edit_obj
        self._main_window = main_window

    # FIXME The pyqtSlot decorator causes "TypeError: connect() failed between
    # FIXME textChanged(QString) and edited()"
    @pyqtSlot(name="edited")
    def edited(self):
        # Copy the entry box text to the label box below.
        self._main_window.label.setText(self._line_edit.text())


def main():
    app = QApplication(sys.argv)
    window = MainApp()
    window.show()
    sys.exit(app.exec_())


if __name__ == "__main__":
    main()

推荐答案

为什么想要使用 @pyqtSlot ？

失败的原因是 LineEditHandler 不是 QObject 。 @pyqtSlot 所做的基本上是创建一个真正的Qt插槽，而不是在内部使用代理对象（这是默认行为，没有 @pyqtSlot ）。

The reason it fails is that LineEditHandler is not a QObject. What @pyqtSlot does is basically creating a real Qt slot instead of internally using a proxy object (which is the default behavior without @pyqtSlot).

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