PyQt sending parameter to slot when connecting to a signal

问题

I have a taskbar menu that when clicked is connected to a slot that gets the trigger event. Now the problem is that I want to know which menu item was clicked, but I don't know how to send that information to the function connected to. Here is the used to connect the action to the function:

QtCore.QObject.connect(menuAction, 'triggered()', menuClickedFunc)

I know that some events return a value, but triggered() doesn't. So how do I make this happen? Do I have to make my own signal?

回答1:

Use a lambda

Here's an example from the PyQt book:

self.connect(button3, SIGNAL("clicked()"),
    lambda who="Three": self.anyButton(who))

By the way, you can also use functools.partial, but I find the lambda method simpler and clearer.

回答2:

As already mentioned here you can use the lambda function to pass extra arguments to the method you want to execute.

In this example you can pass a string obj to the function AddControl() invoked when the button is pressed.

# Create the build button with its caption
self.build_button = QPushButton('&Build Greeting', self)
# Connect the button's clicked signal to AddControl
self.build_button.clicked.connect(lambda: self.AddControl('fooData'))
def AddControl(self, name):
    print name

Source: snip2code - Using Lambda Function To Pass Extra Argument in PyQt4

回答3:

use functools.partial

otherwise you will find you cannot pass arguments dynamically when script is running, if you use lambda.

回答4:

In general, you should have each menu item connected to a different slot, and have each slot handle the functionality only for it's own menu item. For example, if you have menu items like "save", "close", "open", you ought to make a separate slot for each, not try to have a single slot with a case statement in it.

If you don't want to do it that way, you could use the QObject::sender() function to get a pointer to the sender (ie: the object that emitted the signal). I'd like to hear a bit more about what you're trying to accomplish, though.

回答5:

I'd also like to add that you can use the sender method if you just need to find out what widget sent the signal. For example:

def menuClickedFunc(self):
    # The sender object:
    sender = self.sender()
    # The sender object's name:
    senderName = sender.objectName()
    print senderName

来源：https://stackoverflow.com/questions/940555/pyqt-sending-parameter-to-slot-when-connecting-to-a-signal

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python

qt4

pyqt