将额外的参数传递给 PyQt 插槽而不丢失默认信号参数 [英] Pass extra arguments to PyQt slot without losing default signal arguments

问题描述

PyQt 按钮事件可以以正常方式连接到函数，以便函数接收默认信号参数(在本例中为按钮选中状态):

A PyQt button event can be connected in the normal way to a function so that the function receives the default signal arguments (in this case the button checked state):

def connections(self):
    my_button.clicked.connect(self.on_button)

def on_button(self, checked):
    print checked   # prints "True"

或者，可以使用 lambda 覆盖默认信号参数:

Or, the default signal arguments can be overridden using lambda:

def connections(self):
    my_button.clicked.connect(lambda: self.on_button('hi'))

def on_button(self, message):
    print message   # prints "hi"

有没有一种很好的方法来保留两个信号参数，以便它可以直接被下面的函数接收?

Is there a nice way to keep both signal arguments so it can be directly received by a function like below?

def on_button(self, checked, message):
    print checked, message   # prints "True, hi"

推荐答案

你的 lambda 可以接受一个参数:

Your lambda could take an argument:

def connections(self):
    my_button.clicked.connect(lambda checked: self.on_button(checked, 'hi'))

def on_button(self, checked, message):
    print checked, message   # prints "True, hi"

或者你可以使用 functools.partial:

Or you could use functools.partial:

# imports the functools module
import functools 

def connections(self):
    my_button.clicked.connect(functools.partial(self.on_button, 'hi'))

def on_button(self, message, checked):
    print checked, message   # prints "True, hi"

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