Question

我用Tkinter在Python（2.7）中开发了一个简单的应用程序。但我的状态栏只是有点工作。这是精简代码：

from Tkinter import *
import os
import sys

def fixFiles():
    inputFilePath= input_dir.get()

    #Build a list of files in a directory
    fileList = os.listdir(inputFilePath)

    #Loop through those files, open the file, do something, close the file
    for filename in fileList:
        infile = open(inputfilepath + "/" + filename,'r')

        #Update the status with the filename
        status_string = 'Status: Working on file: ' + str(filename)
        status.set(status_string)

        for line in infile:
            #Do some stuff here
        infile.close()  

class App:
    def __init__(self, master):
            i = 0
        status.set("Status: Press 'Fix Files!'")
        statuslabel = Label(master, textvariable=status, relief = RIDGE, width = 65, pady = 5, anchor=W)
        bFixFiles = Button(root, text='Fix Files!', command = fixFiles)
        bQuit = Button(root, text='Quit', command = root.destroy)

        statuslabel.grid(row=i, column = 0, columnspan = 2)
        bFixFiles.grid(row=i, column=2, sticky=E)
        bQuit.grid(row=i, column=3, sticky=W)

root = Tk()
root.title("FIX Files")
input_dir = StringVar()
status = StringVar()
choice = IntVar()
app = App(root)
root.mainloop()

目前发生的情况是状态栏显示“状态：按'修复文件！'”，直到程序循环完成文件，此时显示“Status：Working on file：XXXXX.txt”（其中是程序打开和关闭的最后一个文件的名称。

我希望每次程序打开新文件时都会使用文件名更新状态栏。任何帮助表示赞赏！

Answer 1

愚蠢的方式是使用root.update_idletasks()：

#Update the status with the filename
status_string = 'Status: Working on file: ' + str(filename)
status.set(status_string)
root.update_idletasks()

值得赞扬的是，它很简单，但它确实不起作用 - 虽然statuslabel得到更新，但Quit按钮会冻结，直到fixFiles完成。这不是非常友好的GUI。 Here are some more reasons为什么update和update_idletasks被视为有害。

那么我们应该如何在不冻结GUI的情况下运行长时间运行的任务呢？

关键是让你的回调功能快速结束。而不是长时间运行for-loop，而是创建一个贯穿for-loop内部的函数。希望这很快就会结束，用户不会觉得GUI被冻结了。

然后，要替换for-loop，您可以使用root.after调用多次调用快速运行的函数。

from Tkinter import *
import tkFileDialog
import os
import sys
import time


def startFixFiles():
    inputFilePath = tkFileDialog.askdirectory()
    # inputFilePath= input_dir.get()

    # Build a list of files in a directory
    fileList = os.listdir(inputFilePath)
    def fixFiles():
        try:
            filename = fileList.pop()
        except IndexError:
            return
        try:
            with open(os.path.join(inputFilePath, filename), 'r') as infile:
                # Update the status with the filename
                status_string = 'Status: Working on file: ' + str(filename)
                status.set(status_string)
                for line in infile:
                    # Do some stuff here
                    pass
        except IOError:
            # You might get here if file is unreadable, you don't have read permission,
            # or the file might be a directory...
            pass
        root.after(250, fixFiles)
    root.after(10, fixFiles)

class App:
    def __init__(self, master):
        i = 0
        status.set("Status: Press 'Fix Files!'")
        statuslabel = Label(
            master, textvariable=status, relief=RIDGE, width=65,
            pady=5, anchor=W)
        bFixFiles = Button(root, text='Fix Files!', command=startFixFiles)
        bQuit = Button(root, text='Quit', command=root.destroy)

        statuslabel.grid(row=i, column=0, columnspan=2)
        bFixFiles.grid(row=i, column=2, sticky=E)
        bQuit.grid(row=i, column=3, sticky=W)

root = Tk()
root.title("FIX Files")
input_dir = StringVar()
status = StringVar()
choice = IntVar()
app = App(root)
root.mainloop()

上面提出了一个问题，如果我们长时间运行的任务没有循环，我们该怎么办？或者即使一个人通过循环需要很长时间？

这是一种在单独的进程（或线程）中运行长时间运行的任务的方法，并让它通过队列传递信息，主进程可以定期轮询（使用root.after）来更新GUI状态栏。我认为这种设计更容易适用于这个问题，因为它不需要你拆分for-loop。

请注意，所有与Tkinter GUI相关的函数调用必须来自单个线程。这就是为什么长时间运行的进程只是通过队列发送字符串而不是试图直接调用status.set。

import Tkinter as tk
import multiprocessing as mp
import tkFileDialog
import os
import Queue

sentinel = None

def long_running_worker(inputFilePath, outqueue):
    # Build a list of files in a directory
    fileList = os.listdir(inputFilePath)  
    for filename in fileList:
        try:
            with open(os.path.join(inputFilePath, filename), 'r') as infile:
                # Update the status with the filename
                status_string = 'Status: Working on file: ' + str(filename)
                outqueue.put(status_string)
                for line in infile:
                    # Do some stuff here
                    pass
        except IOError:
            # You might get here if file is unreadable, you don't have read permission,
            # or the file might be a directory...
            pass
    # Put the sentinel in the queue to tell update_status to end
    outqueue.put(sentinel)

class App(object):
    def __init__(self, master):
        self.status = tk.StringVar()
        self.status.set("Status: Press 'Fix Files!'")
        self.statuslabel = tk.Label(
            master, textvariable=self.status, relief=tk.RIDGE, width=65,
            pady=5, anchor='w')
        bFixFiles = tk.Button(root, text='Fix Files!', command=self.startFixFiles)
        bQuit = tk.Button(root, text='Quit', command=root.destroy)
        self.statuslabel.grid(row=1, column=0, columnspan=2)
        bFixFiles.grid(row=0, column=0, sticky='e')
        bQuit.grid(row=0, column=1, sticky='e')

    def update_status(self, outqueue):
        try:
            status_string = outqueue.get_nowait()
            if status_string is not sentinel:
                self.status.set(status_string)
                root.after(250, self.update_status, outqueue)
            else:
                # By not calling root.after here, we allow update_status to truly end
                pass
        except Queue.Empty:
            root.after(250, self.update_status, outqueue)

    def startFixFiles(self):
        inputFilePath = tkFileDialog.askdirectory()
        # Start long running process
        outqueue = mp.Queue()    
        proc = mp.Process(target=long_running_worker, args=(inputFilePath, outqueue))
        proc.daemon = True
        proc.start()
        # Start a function to check a queue for GUI-related updates
        root.after(250, self.update_status, outqueue)

root = tk.Tk()
root.title("FIX Files")
app = App(root)
root.mainloop()

Python / Tkinter状态栏未正确更新

1 个答案: