PyQt连接到KeyPressEvent [英] PyQt Connect to KeyPressEvent

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本文介绍了PyQt连接到KeyPressEvent的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

某些小部件将允许我执行以下操作:

Certain widgets will allow me to do:

self.widget.clicked.connect(on_click)

但要这样做:

self.widget.keyPressEvent.connect(on_key)

将无法说该对象没有属性"connect".

will fail saying that the object has no attribute 'connect'.

我知道子类化小部件并重新实现keyPressEvent方法将使我能够响应事件.但是,此后如何从用户上下文.connect()进入键盘事件,或者换句话说,又如何呢?

I know that sub-classing the widget and re-implementing the keyPressEvent method will allow me to respond to the event. But how can I .connect() to the keyboard event thereafter, or otherwise said, from a user context?

推荐答案

创建自定义信号,并从重新实现的事件处理程序中发出该信号:

Create a custom signal, and emit it from your reimplemented event handler:

class MyWidget(QtGui.QWidget):
    keyPressed = QtCore.pyqtSignal(int)

    def keyPressEvent(self, event):
        super(MyWidget, self).keyPressEvent(event)
        self.keyPressed.emit(event.key())
...

def on_key(key):
    # test for a specific key
    if key == QtCore.Qt.Key_Return:
        print('return key pressed')
    else:
        print('key pressed: %i' % key)

self.widget.keyPressed.connect(on_key)

(注意:为了保持事件的现有处理,需要调用基类实现).

(NB: calling the base-class implementation is required in order to keep the existing handling of events).

这篇关于PyQt连接到KeyPressEvent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!

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