问如何将文件的路径发布到tkinter中的entry()中
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Stack Overflow用户

提问于 2019-05-26 17:24:22

回答 1查看 49关注 0票数 0

我正在尝试创建(text1)条目来获取我打开的文件路径以显示我的意思当我打开一个文件时它会在条目(文本框)上显示文件路径ps:很抱歉我的例子和英语都不好

‘python

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

这是我的fream课程

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()

此选项用于测试从条目打印

def printingstuff(var1):

    print (var1)

这是为了打开一个文件，我想让它更改条目以显示文件路径

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

以下是程序的框架

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()

主循环的事情

app = SchoolProjict()
app.mainloop()

‘如果这没有任何意义，我很抱歉

python

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回答 1

Stack Overflow用户

回答已采纳

发布于 2019-06-04 01:09:14

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()))
        ...