pynput 侦听器在实现时崩溃代码

Question

我对 python 很陌生，我不知道为什么，但是当我专门在 with Listener(on_press=press) as listener: listener.join() 当它崩溃时它不会给我一个错误消息
这是我的全部代码：

import tkinter as tk
from tkinter.constants import LEFT, N, NE, NW, W
from PIL import Image, ImageTk
import tkinter.ttk as ttk
import time
from pynput.keyboard import Key, Controller
from pynput.keyboard import Listener
self = tk.Tk()
self.title('Key spammer')
self.iconbitmap("D:\Vs code repos\Key spammer\keyboard-icon.ico")

#Set the geometry of frame
self.geometry("350x550")
self.resizable(False, False)

#Button exit function
def exit_prog():
    exit()

def press(key):
    print(key)

#press interval
self.grid_rowconfigure(20, weight=1)
self.grid_columnconfigure(20, weight=1)

labelframe = ttk.Labelframe(self,text= "Press interval")
labelframe.grid(row=1, column=0, padx= 25, sticky=N)

class Lotfi(ttk.Entry):
    def __init__(self, master=None, **kwargs):
        self.var = tk.StringVar()
        ttk.Entry.__init__(self, master, textvariable=self.var, **kwargs)
        self.old_value = ''
        self.var.trace('w', self.check)
        self.get, self.set = self.var.get, self.var.set

    def check(self, *args):
        if self.get().isdigit(): 
            # the current value is only digits; allow this
            self.old_value = self.get()
        else:
            # there's non-digit characters in the input; reject this 
            self.set(self.old_value)

#Entry interval
hoursentry = Lotfi(labelframe, width= 10)
hoursentry.grid(row=1, column=1, padx= 5, pady= 10)
hoursentry.insert(0, '0')
hourslabel = ttk.Label(labelframe, text= "hours")
hourslabel.grid(row=1, column=2)
minutesentry = Lotfi(labelframe, width= 10)
minutesentry.grid(row=1, column=3)
minutesentry.insert(0, '0')
minuteslabel = ttk.Label(labelframe, text= "mins")
minuteslabel.grid(row=1, column=4)
secondsentry = Lotfi(labelframe, width= 10)
secondsentry.grid(row=1, column=5)
secondsentry.insert(0, '0')
secondeslabel = ttk.Label(labelframe, text= "secs")
secondeslabel.grid(row=1, column=6)

labelframekey = ttk.Labelframe(self,text= "Key options")
labelframekey.grid(row=2, column=0, padx= 25, pady= 25, sticky=NW)
#labelframekey.pack(side= LEFT)

keypress = ttk.Label(labelframekey, text= "Key pressed", font= 10)
keypress.grid(row=2, column=0, sticky= N)

with Listener(on_press=press) as listener:
    listener.join()    
class Keyreg(ttk.Entry):
    def __init__(self, master=None, **kwargs):
        self.var = tk.StringVar()
        ttk.Entry.__init__(self, master, textvariable=self.var, **kwargs)
        self.old_value = ''
        self.var.trace('w', self.check)
        self.get, self.set = self.var.get, self.var.set

    def check(self, *args):
        if press: 
            self.set(self.old_value)

#Keyreg = Keyreg(labelframekey, width= 10)
#keyreg.grid(row=2, column=1, sticky= N)
'''
#Creation of Option buttons
button = ttk.Button(self, text = 'Exit', command = exit_prog, width = 25)
button.grid(row=2, column=1,padx= 5, ipadx=15, ipady=10)
button2 = ttk.Button(self, text = 'Exit', command = exit_prog, width = 25)
button2.grid(row=2, column=2, ipadx=15, ipady=10)
button3 = ttk.Button(self, text = 'Exit', command = exit_prog, width = 25)
button3.grid(row=3, column=1, ipadx=15, ipady=10)
button4 = ttk.Button(self, text = 'Exit', command = exit_prog, width = 25)
button4.grid(row=3, column=2, ipadx=15, ipady=10)
'''
#Top menu for keybinds

tk.mainloop()

我试图自己找出问题，但是当我将它导入另一个选项卡并使用多种不同的方法时，它仍然从未在这些测试中崩溃

请帮帮我

Answer 1

您没有显示完整的错误消息，所以我不知道您的真正问题是什么，但我发现了一个问题。

代码

with Listener(on_press=press) as listener:
    listener.join()    

运行等待listener 结束的代码并停止其余代码

如果你想和tkinter同时运行它，那么你应该运行为

with Listener(on_press=press) as listener:

    tk.mainloop()

    listener.join()    

或更类似于threading（由listener使用）

listener = Listener(on_press=press)
listener.start()

tk.mainloop()

listener.join()