tkinter 等待使用 .invoke() 按下按钮

Question

import time
from pygame import mixer
from tkinter import filedialog, Tk, BOTH
from tkinter.ttk import Frame, Button
from tkinter import *


def playFile(filePath, interval = 5, playTime = 60):
    playCount = int(playTime//interval)
    for play in range(0, playCount):
        mixer.init()
        mixer.music.load(filePath)
        mixer.music.play()
        time.sleep(interval*60)

global clicked
clicked = False


def findFile():
    global clicked
    clicked = True
    fileLocation = filedialog.askopenfilename(initialdir = "C:/", title = "Select file", filetypes = (("mp3 files","*.mp3"), ("m4a files", ".m4a"), ("all files","*.*")))
    return fileLocation


file = ''


class Example(Frame):

    def __init__(self):
        super().__init__()   

        self.initUI()


    def initUI(self):

        self.master.title("Interval Player")
        self.pack(fill=BOTH, expand = 1)

        openButton = Button(self, text = "Open", command=findFile)
        openButton.place(x=0, y=0)

def main():

    root = Tk()
    root.geometry("250x150+300+300")
    if clicked == True:
            file = str(openButton.invoke())
            playFile(file)      
    app = Example()   
    root.mainloop()  

if __name__ == '__main__':
    main()

当我选择调用按钮的结果时，文件资源管理器 window 在单击 openButton 之前打开。如何让程序在从按钮调用值之前等待按钮按下？

我试过使用全局变量 True/False 来确定按钮是否被点击。但是，我觉得程序并没有反复检查这个布尔值。也许有一个特定的功能必须添加播放文件？

Answer 1

你的代码

if clicked == True:
            file = str(openButton.invoke())
            playFile(file)

从未使用过。

您需要做的是在消息框弹出后进行操作

因此

def findFile():
    global clicked
    clicked = True
    fileLocation = filedialog.askopenfilename(initialdir = "C:/", title = "Select file", filetypes = (("mp3 files","*.mp3"), ("m4a files", ".m4a"), ("all files","*.*")))
    return fileLocation

变成

def findFile():
    global clicked
    clicked = True
    fileLocation = filedialog.askopenfilename(initialdir = "C:/", title = "Select file", filetypes = (("mp3 files","*.mp3"), ("m4a files", ".m4a"), ("all files","*.*")))
    if fileLocation:
          playFile(fileLocation)
    return fileLocation

完整的工作示例

import time
from pygame import mixer
from tkinter import filedialog, Tk, BOTH
from tkinter.ttk import Frame, Button
from tkinter import *


def playFile(filePath, interval = 5, playTime = 60):
    playCount = int(playTime//interval)
    for play in range(0, playCount):
        mixer.init()
        mixer.music.load(filePath)
        mixer.music.play()
        time.sleep(interval*60)

global clicked
clicked = False


def findFile():
    global clicked
    clicked = True
    fileLocation = filedialog.askopenfilename(initialdir = "C:/", title = "Select file", filetypes = (("mp3 files","*.mp3"), ("m4a files", ".m4a"), ("all files","*.*")))
    playFile(fileLocation)
    return fileLocation


file = ''


class Example(Frame):

    def __init__(self):
        super().__init__()

        self.initUI()


    def initUI(self):

        self.master.title("Interval Player")
        self.pack(fill=BOTH, expand = 1)

        openButton = Button(self, text = "Open", command=findFile)
        openButton.place(x=0, y=0)

def main():

    root = Tk()
    root.geometry("250x150+300+300")
    app = Example()
    root.mainloop()

if __name__ == '__main__':
    main()

tkinter 等待使用 .invoke() 按下按钮

tkinter wait for button press with .invoke()

python

tkinter

button

invoke