Question

我试图输入（text1）以获得我打开显示的文件路径我的意思是我打开了一个文件，它向我显示了条目（文本框）上的文件路径 ps：对不起，这个错误的例子和英语

''''python

acomber@mail:~/Documents/projects/modem/serial/gdbplay$ ls
main.cpp  main.o  Makefile  serial
acomber@mail:~/Documents/projects/modem/serial/gdbplay$ make serial
g++ -Wall -Werror -ggdb3 -std=c++17 -pedantic  -c main.cpp
g++ -o serial -Wall -Werror -ggdb3 -std=c++17 -pedantic  main.o -L/usr/lib -lstdc++fs
acomber@mail:~/Documents/projects/modem/serial/gdbplay$ sudo ./serial
[sudo] password for acomber: 
AT
No. found modems: 1
                   /dev/ttyACM0->
                                 Operation took 8643 milliseconds
                                                                 acomber@mail:~/Documents/projects/modem/serial/gdbplay$

这是我的怪胎课程

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

此操作将测试条目中的打印内容

class SchoolProjict(tk.Tk):
    def __init__(self, *args, **kwargs):
        tk.Tk.__init__(self, *args, **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)
        self.frames = {}
        for F in (StartPage, PageOne, 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()

这是打开文件，我希望它更改条目以显示文件路径

def printingstuff(var1):

    print (var1)

这是程序的框架

def load_file():
        fname = askopenfilename(filetypes=(("Excel file", "*.xls"),
                                           ("HTML files", "*.html;*.htm"),
                                           ("All files", "*.*") ))
        if fname:
            try:
                print(fname)

                return

            except:                     # <- naked except is a bad idea
                showerror("Open Source File", "Failed to read file\n'%s'" % fname)
            return

主循环事物

class StartPage(tk.Frame):
    def __init__(self, parent, controller):
        tk.Frame.__init__(self, parent)
        lablel = tk.Label(self, text = "Start Page")
        lablel.pack(pady = 10, padx = 10)
        button1 = tk.Button(self, text = "Main Menu", command = lambda: controller.show_frame(PageOne))
        button1.pack()
        button2 = tk.Button(self, text = "Siting", command = lambda: controller.show_frame(SetingPage))
        button2.pack()

class PageOne(tk.Frame):
    def __init__(self, parent, controller):
        tk.Frame.__init__(self, parent)
        lablel = tk.Label(self, text = "Main Menu")
        lablel.pack(pady = 10, padx = 10)
        button1 = tk.Button(self, text = "Start Page", command = lambda: controller.show_frame(StartPage))
        button1.pack()
        button2 = tk.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)
        lablel = tk.Label(self, text = "Siting Page")
        lablel.pack(pady = 10, padx = 10)
        text1 = ttk.Entry(self)
        text1.pack()
        text1.focus()
        button1 = tk.Button(self, text = "Print from Entry", command = lambda: printingstuff(text1.get()))
        button1.pack()
        button2 = tk.Button(self, text="open File", command= load_file, width=10)
        button2.pack()
        button3 = tk.Button(self, text = "Main Menu", command = lambda: controller.show_frame(PageOne))
        button3.pack()
        button4 = tk.Button(self, text = "Start Page", command = lambda: controller.show_frame(StartPage))
        button4.pack()

'''' 抱歉，如果没有任何意义

Answer 1

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()))
        ...

我如何在tkinter中将文件的路径发布到entry（）中

1 个答案: