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，在单击openButton之前将打开文件浏览器窗口。在调用按钮中的值之前，如何使程序等待按钮按下？

我尝试使用带有True / False的全局变量来查找是否已单击按钮。但是，我觉得程序没有重复检查此布尔值。也许有一个必须添加playFile的特定功能？

Answer 1

您的代码

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

从不使用。

您需要做的是在弹出消息框后采取行动

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

成为

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用.invoke（）等待按钮按下

1 个答案: