如何对按钮单击pyqt5做出反应 [英] How to react to a button click in pyqt5

问题描述

因此，我是python编程的新手。我已经开始在pyqt5中实现UI，并且有一个按钮，我想在用户单击它时做出反应。

So, I am new to python programming. I have started to implement a UI in pyqt5 and there I have a button and I want to react when the user clicks it.

根据此链接我应该只写 btn.clicked.connect（self.buton_pressed）但是我收到消息在函数中找不到引用连接。 （周围的代码在问题的末尾）。

According to this Link I should simply write btn.clicked.connect(self.buton_pressed) however I get the message "Cannot find reference connect in function". (The surrounding code is at the end of the question)

所以我在Google上搜索了一下，发现所有内容只能以这种方式工作。我不明白为什么不这样做。我发现了这个 Stackoverflow问题，该问题还描述了如何执行此操作的旧方法。在进行了一些糊涂之后，我发现它在pyqt5或其他软件包中不再受支持。

So I googled a bit and all I found is that it should just work that way. I just don't get why it does not. I found this Stackoverflow question which also describes the old variant of how to do it. That did not work either, after some googeling I found out that it is no longer supported in pyqt5 or in some other package.

我尝试与该函数连接的函数event：

The function where I try to connec to the event:

def __add_button(self, text: str, layout: QLayout):
    btn = QPushButton(text, self)
    layout.addWidget(btn)
    btn.clicked.connect(self.button_pressed)
    # TODO: fix this.
    return btn

生成GUI的代码和调用的函数，在 __ init __ 函数

The code where the GUI is generated and the function called, in the __init__ function

lblhm = QLabel("Hauptmessung", self)
layout.addWidget(lblhm)

self.__hm_b = self.__add_button("Messung öffnen", layout)
self.__hm_config_b = self.__add_button("Configruation öffnen", layout)

lblzm = QLabel("Zusatzmessung", self)
layout.addWidget(lblzm)

self.__zm_b = self.__add_button("Messung öffnen", layout)
self.__zm_config_b = self.__add_button("Configuration öffnen", layout)

The button_pressed函数尚未实现，但是应该打开一个openFile对话框来选择文件。

The button_pressed function is not yet implemented, but it is supposed to open a openFile dialog for file selection.

根据这篇文章我可以在返回函数后直接连接，但随后我必须写它4次不是很好。

According to this post i could just connect after returning the function, but then i would have to write it 4 times which is not very nice. Isn't the signal bound to the object not to the variable?

感谢您的帮助：）

推荐答案

由于您没有向我们提供工作示例，因此很难理解您的问题，即，一个可以使代码按原样运行的和平方式。像这样：

It's hard to understand your problem since you don't provide us with a working example, i.e. a peace of code one can run "as is". Something like this:

from PyQt4 import QtCore, QtGui

class MyWindow(QtGui.QWidget):
    def __init__(self):
        super().__init__()

        layout = QtGui.QVBoxLayout()
        self.setLayout(layout)

        lblhm = QtGui.QLabel("Hauptmessung", self)
        layout.addWidget(lblhm)

        self.__hm_b = self.__add_button("Messung öffnen", layout)
        self.__hm_config_b = self.__add_button("Configruation öffnen", layout)

        lblzm = QtGui.QLabel("Zusatzmessung", self)
        layout.addWidget(lblzm)

        self.__zm_b = self.__add_button("Messung öffnen", layout)
        self.__zm_config_b = self.__add_button("Configuration öffnen", layout)

    def button_pressed(self):
        print('Button pressed')

    def __add_button(self, text: str, layout: QtGui.QLayout):
        btn = QtGui.QPushButton(text, self)
        layout.addWidget(btn)
        btn.clicked.connect(self.button_pressed)
        return btn



if __name__== '__main__':
    import sys
    app = QtGui.QApplication(sys.argv)
    wnd = MyWindow()
    wnd.show()
    sys.exit(app.exec_())

在 PyQt4 。它与 PyQt5 一起使用吗？

这篇关于如何对按钮单击pyqt5做出反应的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！