通过单击按钮使用askopenfilename获取文件名后，如何在python的条目中显示该文件名

Question

Here is my code, so after I click the button, it will call loadFile function, and the file name will be saved in the function, after that how can I change the text of Entry(self.filedir) to that file name?

from tkinter import *
from tkinter.filedialog import askopenfilename

class Checker:
    def loadFile(self):
        self.filename = askopenfilename(filetypes=(("info", "*.xlsx"), ("all file", "*.*")))

    def __init__(self, master):
        master.title("Checker")

        self.load_button = Button(master, text="load file", command=self.loadFile)
        self.load_button.grid(row=0, column=0)
        self.filedir = Entry(master, text=" ")
        self.filedir.grid(row=0, column=1)

if __name__=='__main__':
    root = Tk()
    k = Checker(root)
    root.mainloop()

Answer 1

Add the following two lines to loadFile :

self.filedir.delete(0, "end")
self.filedir.insert(0, self.filename)

Answer 2

I have a tkinter class that creates a frame containing a label, entry field to contain the filepath and a browse button to select the file.

Try this simple example (Sorry it is python 2 not python 3 given its age).

from Tkinter import *
import tkFileDialog

class FileSelect(Frame):
    def __init__(self,master,label,opensave,filetype,**kw):
        Frame.__init__(self,master)
        self.configure(**kw)
        self.file = StringVar()
        self.opensave = opensave
        self.filetypes = filetype

        self.Label = Label(self, text=label)
        self.Label.config(width=10,anchor=E)
        self.filenamebox = Entry(self,text=self.file)
        self.filenamebox.config(width=50)
        self.btnBrowse = Button(self,text='Browse',command=self.browse_file)
        self.btnBrowse.config(width=10)
        self.Label.grid(row=0,column=0,pady=5,sticky=E)
        self.filenamebox.grid(row=0,column=1,pady=5)
        self.btnBrowse.grid(row=0,column=2,pady=5,padx=5)
    def browse_file(self):
        filename = []
        if self.opensave == "open":
            filename = tkFileDialog.askopenfilename(filetypes=self.filetypes)
        else:
            filename = tkFileDialog.asksaveasfilename(filetypes=self.filetypes)
        self.file.set(filename)
    def get_filename(self):
        return self.file.get()

def main():
    root = Tk()
    root.title("Select File Example")
    selectFile = FileSelect(root,"My File","open",[('All Files','*.*')])
    selectFile.grid()    
    root.mainloop()


if __name__ == '__main__':
    main()