Working with Python and Tkinter, I have been trying to find out the way to show the file_path beside the Browse Button but unable to do so.

Here is my code:

  import os
  from tkFileDialog import askopenfilename
  from Tkinter import *

  content = ''
  file_path = ''

  #~~~~ FUNCTIONS~~~~

  def open_file():
    global content
    global file_path

    filename = askopenfilename()
    infile = open(filename, 'r')
    content = infile.read()
    file_path = os.path.dirname(filename)
    return content

  def process_file(content):
    print content

  #~~~~~~~~~~~~~~~~~~~

  #~~~~~~ GUI ~~~~~~~~

  root = Tk()
  root.title('Urdu Mehfil Ginti Converter')
  root.geometry("598x120+250+100")

  mf = Frame(root)
  mf.pack()

  f1 = Frame(mf, width=600, height=250)
  f1.pack(fill=X)
  f2 = Frame(mf, width=600, height=250)
  f2.pack()

  file_path = StringVar

  Label(f1,text="Select Your File (Only txt files)").grid(row=0, column=0, sticky='e')
  Entry(f1, width=50, textvariable=file_path).grid(row=0,column=1,padx=2,pady=2,sticky='we',columnspan=25)
  Button(f1, text="Browse", command=open_file).grid(row=0, column=27, sticky='ew', padx=8, pady=4)
  Button(f2, text="Process Now", width=32, command=lambda: process_file(content)).grid(sticky='ew', padx=10, pady=10)

  root.mainloop()

  #~~~~~~~~~~~~~~~~~~~

Kindly guide me as how I can show the file path along with the "Browse Button" button after the user selects the file as shown in this image.

Thanks in advance!

First, change this line:

    Entry(f1, width=50, textvariable=file_path).grid(row=0,column=1,padx=2,pady=2,sticky='we',columnspan=25)

to this:

entry = Entry(f1, width=50, textvariable=file_path)
entry.grid(row=0,column=1,padx=2,pady=2,sticky='we',columnspan=25)

Then, in the open_file() function, add these two lines, just before the return:

entry.delete(0, END)
entry.insert(0, file_path)

Explanation: First, we give the entry a name, so that it can be modified. Then, in the open_file() function we clear it and add the text for the file-path.

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