How to Show File Path with Browse Button in Python / Tkinter

Question

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!

laurencevs · Accepted Answer · 2014-08-07 16:51:13Z

4

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.

answered Aug 7, 2014 at 16:51

laurencevs

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2 Comments

Muhammad Ahmed Over a year ago

Thank you very much @laurencevs! I spend hours to get this functional but I was unable. Thanks a lot.

laurencevs Over a year ago

@pleasureinblues You can show others that this answer was useful by clicking on the little tick on its left.

Baris Demiray · Accepted Answer · 2016-08-10 16:29:33Z

0

Here is a diff that fixes instead the file_path, i.e. the StringVar() usage:

--- old.py      2016-08-10 18:22:16.203016340 +0200
+++ new.py      2016-08-10 18:24:59.115328029 +0200
@@ -4,7 +4,6 @@


 content = ''
-file_path = ''


 #~~~~ FUNCTIONS~~~~
@@ -16,7 +15,7 @@
   filename = askopenfilename()
   infile = open(filename, 'r')
   content = infile.read()
-  file_path = os.path.dirname(filename)
+  file_path.set(os.path.dirname(filename))
   return content

 def process_file(content):
@@ -40,7 +39,7 @@
 f2 = Frame(mf, width=600, height=250)
 f2.pack()

-file_path = StringVar
+file_path = StringVar(root)


 Label(f1,text="Select Your File (Only txt files)").grid(row=0, column=0, sticky='e')

answered Aug 10, 2016 at 16:29

Baris Demiray

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