Tkinter widget after方法

2024-03-29 09:51:37 发布

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我还没有找到一条线索来回答我的问题,所以:

我的图形用户界面由3个独立的“窗口”(框架?)as类:TextInput、TextOutput和Statistic。出于性能原因,我只运行统计.mainloop()但其他班级也出现了。在

我想要我的TextInput类做的是 迭代一个包含字符串的列表,并将它们插入到textwidget中”自我参考文本". 在

class TextInput(tk.Frame):

LARGE_FONT = ("Arial Bold ", 18)
SMALL_FONT = ("Arial", 16)
BGC = '#CDCDC1'
FG = ['#000000', '#f44242']

def __init__(self, parent, controller):
    tk.Frame.__init__(self, parent)
    # print('415 TextInput instance: ', type(self))
    self.controller = controller
    self.reference = [] # some strings in here 
    self.createText()

def createText(self):
    self.ref_text = tk.Text(self, height=30, width=100, font=TextInput.SMALL_FONT)
    self.ref_text.config(background=TextInput.BGC)
    self.ref_text.grid(column=0, columnspan=4, row=1, padx=5, sticky="W")

def display_ref(self, line):
    print('line: ', line)
    self.ref_text.insert('end', line)

def read_ref(self):
    for line in self.reference:
        self.ref_text.insert('end', line)
        self.ref_text.after(1500, self.read_ref)

after()方法插入“的所有字符串”自我参照“而不是预期的FOR循环。而且,整个TextInput应用程序似乎倾斜(太多递归?) 在另一个版本里我试着打电话给你

^{pr2}$

1500毫秒后,再次将所有文本放入小部件中

我做错什么了? 我只跑是不是有问题

Statistik.mainloop()

在底部而不是文本输入.mainloop(). 谢谢你的帮助

至于最小的例子:

import tkinter as tk

class Interface(tk.Tk):
def __init__(self, name, page,  *kwargs):
    tk.Tk.__init__(self, name, *kwargs)
    container = tk.Frame(self)
    container.pack(side="top", fill="both", expand=True)

    container.grid_rowconfigure(0, weight=1)
    container.grid_columnconfigure(0, weight=1)
    container.master.title(name)

    self.frames = {}
    self.windows = {}
    self.windows[name] = page

    self.window = page(container, self)
    self.frames[name] = self.window
    self.window.grid(row=0, column=0, sticky='nsew')

    self.show_window(name)

def show_window(self, cont):
    window = self.frames[cont]
    window.tkraise()

class TextInput(tk.Frame):

def __init__(self, parent, controller):
    tk.Frame.__init__(self, parent)
    self.reference = ['a is a sentence', 'b follows a', 'c closes the session']
    self.createText()
    self.read = True
    self.read_ref()

def stop_read(self):
    self.read = False

def createText(self):
    self.ref_text = tk.Text(self, height=30, width=80,)
    self.ref_text.grid(column=0, row=1, columnspan=3, padx=5, sticky="W")

def display_ref(self, line):
    print('line: ', line)
    self.ref_text.insert('end', line)

def read_ref(self):
    '''
    the goal is to loop through self.reference line by line
    with a 1500 ms pause inbetween
    '''
    for line in self.reference:
        if self.read:
            self.ref_text.insert('end', line + '\n')
            self.ref_text.after(1500, self.read_ref)
        else:
            return

class Statistik(tk.Frame):
def __init__(self, parent, controller):
    tk.Frame.__init__(self, parent)
'''does some other stuff'''

textinput_instance = Interface('TextInput', TextInput)
statistik_instance = Interface('Statistik', Statistik)

statistik_instance.mainloop()

Tags: textnameselfrefreadinitcontainerdef
1条回答
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1楼 · 发布于 2024-03-29 09:51:37

如果您的目标是在列表中循环,以1500毫秒的间隔显示每一行,那么最简单的方法就是让一个函数执行一次迭代,然后使用after重复它自己

可能是这样的:

def read_ref(self, lines):
    # remove one item from the list
    line = lines.pop(0)

    # insert it
    self.ref_text.insert("end", line + "\n")

    # run again in 1500ms if there's still more work to do
    if lines:
        self.after(1500, self.read_ref, lines)

然后,只调用此函数一次以启动该过程:

^{pr2}$

如果要停止它,可以检查函数中的标志:

def read_ref(self):
    ...
    if self.reference and not self.stop:
        self.after(1500, self.read_ref)

上面的代码缓慢地从self.reference中删除项。如果不希望发生这种情况,请在启动时传递self.reference的副本,以便函数将从原始数据的副本中删除项。在

self.read_ref(self, self.reference[:])

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