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python - 如何在 tkinter 中使用 Entry.insert() 框架？

问题描述

我正在尝试制作一个程序来帮助我在学校的工作，我在这方面取得了很大进展，但我现在搞砸了，我需要罚款如何使文件的路径显示在 Entry 小部件上我尝试了很多这里的东西，我可能没有想到它背后的想法，请帮助我

import tkinter as tk 
from tkinter.filedialog import askopenfilename
from tkinter.messagebox import showerror
from tkinter import ttk
import xlrd

class SchoolProjict(tk.Tk):
    def __init__(self, *args, **kwargs):
        tk.Tk.__init__(self, *args, **kwargs)
        self.app_data = {"name": tk.StringVar(),
                         }
        container = tk.Frame(self)
        container.pack(side = "top", fill = "both", expand = True)
        container.grid_rowconfigure(0, weight = 1)
        container.grid_columnconfigure(0, weight = 1)
        self.frames = {}
        for F in (StartPage,  SetingPage):
            frame = F(container, self)
            self.frames[F] = frame
            frame.grid(row = 0, column = 0, sticky = "nsew")
        self.show_frame(StartPage)

    def show_frame(self, cont):
        frame = self.frames[cont]
        frame.tkraise()

    def get_page(self, classname):
        for page in self.frames.values():
            if str(page.__class__.__name__) == classname:
                return page
        return None

def printingstuff(var1):
    print (var1)

def printontherthing(page_class):
    print(page_class)


class StartPage(tk.Frame):
    def __init__(self, parent, controller):
        self.controller = controller
        tk.Frame.__init__(self, parent)
        lablel = ttk.Label(self, text = "Main Page")
        lablel.pack(pady = 10, padx = 10)
        button2 = ttk.Button(self, text = "Siting", command = lambda: controller.show_frame(SetingPage))
        button2.pack()

class SetingPage(tk.Frame):
    def __init__(self, parent, controller):
        tk.Frame.__init__(self, parent)
        self.controller = controller
        lablel = tk.Label(self, text = "Siting Page")
        lablel.grid(row = 0, column = 0)
        text1 = tk.Entry(self)  #<== i want to show the path of the file i am going to open Here after i select it from openfile 
        text1.grid(row = 2, column = 0)
        text1.focus()
        button1 = tk.Button(self, text = "print text1", command = lambda: printingstuff(text1.get()))
        button1.grid(row = 3, column = 0)
        button2 = tk.Button(self, text="open file", command= self.load_file, width=10)
        button2.grid(row = 3, column = 1)
        button4 = tk.Button(self, text = "Main Page", command = lambda: controller.show_frame(StartPage))
        button4.grid(row = 4, column = 1)

    def load_file(self):
        fname = askopenfilename(filetypes=(("Excel file", "*.xls"),
                                           ("HTML files", "*.html;*.htm"),
                                           ("All files", "*.*") ))
        if fname:
            try:
                # print(fname)
                value = str(fname)
                page_var = self.controller.get_page("SetingPage")
                page_var.text1.insert(0, value)
                return

            except:                    
                showerror("Open Source File", "Failed to read file\n'%s'" % fname)
            return

app = SchoolProjict()
app.mainloop()

它给了我一个错误，我运行它并选择文件 这是我的错误

标签： pythontkintertkinter-entrytkinter-layout

解决方案

您需要通过添加来使您text成为该类的属性。SetingPageself

class SetingPage(tk.Frame):
    def __init__(self, parent, controller):
        ...
        self.text1 = tk.Entry(self)  #<== i want to show the path of the file i am going to open Here after i select it from openfile
        self.text1.grid(row = 2, column = 0)
        self.text1.focus()
        button1 = tk.Button(self, text = "print text1", command = lambda: printingstuff(self.text1.get()))
        ...

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