问题描述
我在 Python 3 中有一个脚本，我正在尝试使用 tkinter 为它制作一个 GUI.
I have a script in Python 3 and I'm trying to make a GUI for it using tkinter.
这是一个完整的工作代码示例:
Here is a complete example of working code:
#!/usr/bin/python
# coding: utf-8
import pickle
import openpyxl
from tkinter import *
import threading
import queue
class Worker():
def __init__(self):
self.one_name_list = []
self.dic = {}
self.root = Tk()
self.root.title("GUI Python")
self.root.geometry("820x350")
self.thread_queue = queue.Queue()
self.btn1 = Button(text="start counting", width = '20', height = '1', background = "#555", foreground = "#ccc", command=self.start_working)
self.btn1.grid(row=0, column=0, columnspan=2, ipadx=10, ipady=6, padx=5, pady=5, sticky=N)
self.btn2 = Button(text="stop counting", width = '20', height = '1', background = "#555", foreground = "#ccc", command=self.stop_running)
self.btn2.grid(row=1, column=0, columnspan=2, ipadx=10, ipady=6, padx=5, pady=5, sticky=N)
self.btn5 = Button(text="clear window", width = '10', height = '1', background = "#555", foreground = "#ccc", command=self.tex_clear)
self.btn5.grid(row=3, column=0, columnspan=2, ipadx=10, ipady=6, padx=5, pady=5, sticky=N)
self.tex = Text(self.root, width = 72, height = 20, font="Verdana 10", wrap=WORD)
self.tex.grid(row=0, column=2, rowspan=4, ipadx=10, ipady=6, padx=5, pady=5)
self.S = Scrollbar(self.root, orient="vertical", command=self.tex.yview)
self.S.grid(row=0, column=4, rowspan=4, ipady=143, pady=5, sticky=W)
self.tex.config(yscrollcommand=self.S.set)
self.root.after(100, self.listen_for_result)
self.root.mainloop()
def read_from_pickle_file(self, filename):
""" Reads python object from pickle file. """
# with open(filename, 'rb') as handle:
# obj = pickle.load(handle)
self.thread_queue.put('Got list file.
')
return True
def get_boxes(self, xlsx_filename, txt_filename=None):
pass # does some job
self.thread_queue.put('Got boxes list.
')
def tex_clear(self):
self.tex.delete('1.0', END)
self.tex.see("end")
def stop_running(self):
pass # stops somehow
def _print(self, text):
self.tex.insert(END, text)
self.tex.see("end")
def start_working(self):
t = threading.Thread(target=self.start_working_2)
t.start()
def start_working_2(self):
self.one_name_list = self.read_from_pickle_file('1.pickle')
self.root.after(100, self.listen_for_result)
self.boxes_list = self.get_boxes('1.xlsx')
self.root.after(100, self.listen_for_result)
self.thread_queue.put('Getting files
')
self.root.after(100, self.listen_for_result)
def listen_for_result(self):
""" Check if there is something in the queue. """
try:
self.res = self.thread_queue.get(0)
self._print(self.res)
except queue.Empty:
self.root.after(100, self.listen_for_result)
if __name__ == '__main__':
se = Worker()
您可以运行它并查看工作窗口.
You can run it and see the working window.
我有几个问题.
这个 GUI 的想法是有 3 个按钮 - 开始运行、停止运行和清除文本窗口.文本窗口 - 应该替代控制台 - 所有消息都应该在文本窗口中打印，而不是控制台.
The idea of this GUI - is that there are 3 buttons - start running, stop running, and clear text window. Text window - should be a substitute for a console - all messages should be printed in text window, instead of console.
现在我使用队列来打印消息.但我想我用错了——因为每次放东西后我都需要手动检查队列.
For now I'm using queue to print messages. But I guess I'm using it in a wrong way - because I need manually to check the queue every time after I put something there.
所以，问题:
有没有办法自动检查队列一直 - 并立即将进入队列的所有内容打印到文本窗口，无论它来自哪个线程?(每次放东西后我都可以忍受检查队列，但是会有几个函数无法预测他们将发送东西多少次到队列中 - 所以我将无法检查队列是否未知次数.)
Is there a way to check queue autamatically all the time - and instantly print to text window everything which gets to the queue, no matter from which thread it came there? (I can put up with checking queue every time after I put something there, but there will be couple of functions where it is impossible to predict how many times they will send something to queue - so I will not be able to check queue for unknown number of times.)
如果您回答第一个问题，我将考虑已回答的问题.
其他问题是可选的.
I will consider the question answered if you answer the first question.
Other questions are optional.
我是否正确启动了 GUI?它应该在 __init__() 还是其他地方?
Am I starting the GUI correctly? Should it be in the __init__() or somewhere else?
如何隐藏控制台窗口?(尝试重命名为 .pyw - 控制台和 GUI 均未出现.尝试将 self.root.withdraw() 放在 self.root = Tk() 之后 - 结果:控制台出现了，GUI - 没有.)
how to hide the console window? (tried renaming to .pyw - neither console, nor GUI showed up. Tried putting self.root.withdraw() after self.root = Tk() - the result: console showed up, GUI - not.)
这段代码中是否有任何笨拙或愚蠢的地方(GUI、线程、队列)?我使用几本手册编写了这段代码，所以我可能会误解其中的部分或全部，并以错误的方式进行操作.
Is there any clumsy or stupid places in this code (GUI, threading, queue)? I wrote this code using several manuals, so I could be misunderstanding some or all of them and do it in a wrong way.
推荐答案
发现很简单:在这个函数中
It figured out to be quite simple: in this function
def listen_for_result(self):
""" Check if there is something in the queue. """
try:
self.res = self.thread_queue.get(0)
self._print(self.res)
except queue.Empty:
self.root.after(100, self.listen_for_result)
应该再添加一个对自身的调用 - 即使在成功打印之后也是如此.在此之后 - 我可以从任何地方将文本发送到队列，并且在将文本发送到队列后立即打印它而无需调用此函数.
should be added one more call to itself - even after successfull printing. After this - I can send text to queue from anywhere and it will be printed without calling this function right after sending text to queue.
def listen_for_result(self):
""" Check if there is something in the queue. """
try:
self.res = self.thread_queue.get(0)
self._print(self.res)
self.root.after(100, self.listen_for_result)
except queue.Empty:
self.root.after(100, self.listen_for_result)
那么现在
self.thread_queue.put('Getting files
')
可以在任何线程中使用.而不是
can be used in any thread. Instead of
self.thread_queue.put('Getting files
')
self.root.after(100, self.listen_for_result)
像以前一样双线.
这篇关于将队列与 tkinter(和线程)一起使用的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持编程学习网！