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tkinter等待按钮按下.invoke（）

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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，文件浏览器窗口会在单击openButton之前打开 . 在从按钮调用值之前，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该按钮 . 但是，我觉得程序没有重复检查这个布尔值 . 也许有一个特定的功能，必须添加playFile？

python button tkinter invoke

1 回答

  • 0

    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执行操作

    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

    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()
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