WxPython PyPubSub，使用咖喱函数不起作用

Question

我正在使用PyPubSub模块的WxPython来发送消息，我希望有一个函数订阅一个主题，其中函数有一些临时参数。不幸的是，它似乎并没有像我所期望的那样使用咖喱功能。

from functools import partial
import time

import wx
from pubsub import pub

def MakeStatusBar(top_frame):

    statusbar = top_frame.CreateStatusBar()
    statusbar.SetFieldsCount(2)
    statusbar.SetStatusWidths([400, 400])

    # pub.subscribe(listener=got_msg, topicName='game.new')  # Works, when alone

    status_func = partial(got_status_msg, statusbar)
    print(f"status_func[callable={callable(status_func)}] = {status_func}")
    status_func(some_data='Directly calling the curried-function')  # Calling the function directly - works
    pub.subscribe(listener=status_func, topicName='game.new')  # Breaks

    return statusbar

def got_msg(some_data):
    print(f"\n*** got_msg: data = {some_data}\n")

def got_status_msg(status_bar, some_data):
    print(f"\n*** got_status_msg, args = {some_data}\n")
    status_bar.SetStatusText(some_data, 0)

class topFrame(wx.Frame):
    def __init__(self, parent):
        super().__init__(parent, wx.ID_ANY, title='testing', size=(500, 500))
        self.status_bar = (MakeStatusBar(self))

class myApp(wx.App):
    def __init__(self):
        wx.App.__init__(self, redirect=True, filename='errlog.txt')  # To log error to a file
        # super().__init__(self)  # Doesn't take redirect? Weird.

        self.topFrame = topFrame(None)
        self.SetTopWindow(self.topFrame)
        self.topFrame.Show()
        print(f"in 2 seconds, will send a message to the Statusbar")
        time.sleep(2)
        print(f"sending...\n")

        try:
            pub.sendMessage(topicName='game.new', some_data='publishing a new-game message')
        except Exception as ex:
            print(f"WTF? got {str(ex)}\n")
        print(f"message has been sent.\n")

if __name__ == '__main__':
    my_app = myApp()
    my_app.MainLoop()

我得到的错误是：

Traceback (most recent call last):
  File "/Code/wxpy/hex_test/src/wx/menu_bar.py", line 11, in new_game_msg
    pub.sendMessage(topicName='game.new', some_data='this is a test message')
    raise SenderUnknownMsgDataError(self.topicNameTuple,
pubsub.core.topicargspec.SenderUnknownMsgDataError: 
Some optional args unknown in call to sendMessage('('game', 'new')', some_data): some_data

那么，我如何将一个监听器函数传递给至少有一个参数的主题呢？显然还有其他的方法(比如全局变量)，但我觉得这应该是可以用部分函数实现的。

Answer 1

嗯，盯着它看，再读一遍文档，是的，有个解决办法。它已经被放入PyPubSub中了，侦听器本身已经允许使用草率的参数。我可以直接传入我想要的StatusBar参数。

    # status_func = partial(got_status_msg, statusbar)
    # print(f"status_func[callable={callable(status_func)}] = {status_func}")
    # status_func(some_data='Directly calling the curried-function')  # Calling the function directly - works

    pub.subscribe(listener=got_status_msg, topicName='game.new', status_bar=statusbar)  # Now works

因此，显然，PyPubSub的开发人员预料到了这一用途。干得好。:)