Question

以帖子here中的示例为例，我创建了一个切换框架应用程序。当我将StringVar用作修改为以下代码的“框架”页面内的变量时，UI不显示，该代码在后台运行

from tkinter import *
from tkinter import font  as tkfont
import time
import threading

class RMS_APP(Tk):
    def __init__(self, *args, **kwargs):
        Tk.__init__(self, *args, **kwargs)
        self.title_font = tkfont.Font(family='Helvetica', size=18, weight="bold", slant="italic")
        self.title("Restaurant Management System")
        self.master = Frame(self)
        self.master.grid(row=0,column=0,sticky='nwes')
        #self.geometry("1600x800+0+0")

        #self.master.value = StringVar()
        self.master.grid_rowconfigure(0, weight=1)
        self.master.grid_columnconfigure(0, weight=1)

        self.frames = {}
        for F in (LogIn,):#,Kitchen):
            page_name = F.__name__
            frame = F(parent=self.master, controller=self)
            self.frames[page_name] = frame

            # put all of the pages in the same location;
            # the one on the top of the stacking order
            # will be the one that is visible.
            frame.grid(row=0, column=0, sticky="nsew")

        self.show_frame("LogIn")

    def show_frame(self, page_name,arg = None):
        '''Show a frame for the given page name'''
        frame = self.frames[page_name]


        if arg:
            #frame.OnClick(arg)
            #self.label1.grid_forget()
            frame.label1['text'] = arg
            pass
        frame.tkraise()

class LogIn(Frame):
    def __init__(self, parent, controller):
        Frame.__init__(self, parent)
        self.controller = controller

        self.localtime=time.asctime(time.localtime(time.time()))
        label = Label(self, text="Restaurant Management System", font=controller.title_font)
        label.grid(row=0,column=0,sticky='news',columnspan=3,padx = 500,pady=(150,10))
        lblinfo = Label(self, font=controller.title_font,text=self.localtime,fg="black",anchor=W)
        lblinfo.grid(row=1,column=0,columnspan=3,padx=500,pady=(0,10))
        button1 = Button(self, text="Table",height = 10, width = 15,
                            command=lambda: controller.show_frame("Table"))
        button2 = Button(self, text="Kitchen",height = 10, width = 15,
                            command=lambda: controller.show_frame("Kitchen"))
        button3 = Button(self, text="Bill Table",height = 10, width = 15,
                            command=lambda: controller.show_frame("Kitchen"))
        button1.grid(row=2,column=0,sticky='news',padx=(250,10),pady = (100,10))
        button2.grid(row=2,column=1,sticky='news',padx=(0,10),pady = (100,10))
        button3.grid(row=2,column=2,sticky='news',pady = (100,10))
if __name__ == "__main__":
    app = RMS_APP()
    app.mainloop()

当我取消注释self.master.value行时，代码无法启动。通过注释它可以启动代码，并显示LogIn页面。还有其他需要检查的内容

Answer 1

我试图通过减少代码来查找问题，但找不到任何东西。如果我取消对StringVar行的注释，则此代码将不会显示窗口：

from tkinter import *

class RMS_APP(Tk):
    def __init__(self):
        Tk.__init__(self)
        self.master = Frame(self)

        self.my_integer = 4     # Works!
        #self.my_stringvar = StringVar()    # Works not!

RMS_APP().mainloop()

但是，如果我从Frame继承我的课程，那么它可以正常工作：

class RMS_APP(Frame):

我不知道为什么会这样...

更新 Tk和Frame都已经具有属性“ master”。和RMS_APP.master =='。'当您更改此设置时，事情显然会中断。

在切换框架程序中使用StringVar挂起

1 个答案: