CODEWITHSUNDEEP
pythontkinterpathfilenames
A. Blackmagic
A. Blackmagic

Reputation: 233

askopenfilename is not defined? - Making a browse button in tkinter

The purpose of my code is to create a GUI that has 4 buttons. 2 of them are to open a "browse" window, allowing the user to select a file from a directory. The third button is to allow the user to choose a directory for the final document to be outputted to. The fourth button applies my python code to both files, creating the outputted document.

In attempting to create the "browse" buttons, through many posts here on stackoverflow and on the internet, most solutions include the "askopenfilename" module that is often imported from tkFileDialog. However no matter how I word it, or whatever variations of tkinter modules that i import, I consistently receive the same error messages of "no module name tkfileDialog" or "askopenfilename is not defined".

Am I doing something wrong with my code? Is this a common error found in tkinter with python 3.6? How would one go about creating a browse button that finds a file and adds its path?

Please let me know!

Thanks.

Below is my code:

import os
#from tkFileDialog import *
from tkinter import filedialog
from Tkinter import *
from tkfileDialog import askopenfilename
content = 'apple'
file_path = 'squarebot'


#FUNCTIONS
def browsefunc(): #browse button to search for files
    filename = askopenfilename()
    infile = open(filename, 'r')
    content = infile.read()
    pathadd = os.path.dirname(filename)+filename

    pathlabel.delete(0, END)
    pathlabel.insert(0, pathadd)

    return content
def open_file(): #also browse button to search for files - im trying various things to get this to work!
    global content
    global file_path

    #filename = filedialog.askopenfilename(filetypes = (typeName {.txt},))
    filename = askopenfilename()                                       
    infile = open(filename, 'r')
    content = infile.read()
    file_path = os.path.dirname(filename)
    entry.delete(0, END)
    entry.insert(0, file_path)
    return content

def process_file(content): #process conversion code
    print(content)

def directoryname():
    directoryname = filedialog.askdirectory() # pick a folder


#GUI

root = Tk()
root.title('DCLF Converter')
root.geometry("598x600")

mf = Frame(root)
mf.pack()


f1 = Frame(mf, width=600, height=250) #DC file
f1.pack(fill=X)

f2 = Frame(mf, width=600, height=250) #LF file
f2.pack(fill=X)

f3 = Frame(mf, width=600, height=250) #destination folder
f3.pack(fill=X)

f4 = Frame(mf, width=600, height=250) #convert button
f4.pack()

file_path = StringVar


Label(f1,text="Select Your DC File (Only txt files)").grid(row=0, column=0, sticky='e') #DC button
entry = Entry(f1, width=50, textvariable=file_path)
entry.grid(row=0,column=1,padx=2,pady=2,sticky='we',columnspan=25)

Label(f2,text="Select Your LF File (Only csv files)").grid(row=0, column=0, sticky='e') #LF button
entry = Entry(f2, width=50, textvariable=file_path)
entry.grid(row=0,column=1,padx=2,pady=2,sticky='we',columnspan=25)

Label(f3,text="Select Your Destination Folder").grid(row=0, column=0, sticky='e') #destination folder button
entry = Entry(f3, width=50, textvariable=directoryname)
entry.grid(row=0,column=1,padx=2,pady=2,sticky='we',columnspan=25)

Button(f1, text="Browse", command=browsefunc).grid(row=0, column=27, sticky='ew', padx=8, pady=4)#DC button
Button(f2, text="Browse", command=browsefunc).grid(row=0, column=27, sticky='ew', padx=8, pady=4)#LF button
Button(f3, text="Browse", command=browsefunc).grid(row=0, column=27, sticky='ew', padx=8, pady=4)#destination folder button
Button(f4, text="RECONCILE NOW", width=32, command=lambda: process_file(content)).grid(sticky='ew', padx=10, pady=10)#convert button


root.mainloop()

P.S If you have found any other errors in my code please let me know. I am just starting with tkinter, and as such this may be attributed to something completely unrelated!

Much Appreciated

Upvotes: 1

Views: 7661

Answers (4)

martineau
martineau

Reputation: 123393

This is what I use in my code so it will work with the Tkinter module in both Python 2 and 3:

try:
    import Tkinter as tk
    import ttk
    from tkFileDialog import askopenfilename
    import tkMessageBox
    import tkSimpleDialog
    from tkSimpleDialog import Dialog
except ModuleNotFoundError:   # Python 3
    import tkinter as tk
    from tkinter import ttk
    from tkinter.filedialog import askopenfilename
    import tkinter.messagebox as tkMessageBox
    import tkinter.simpledialog as tkSimpleDialog
    from tkinter.simpledialog import Dialog

You asked to be notified any of other errors and I noticed the way you're using askopenfilename doesn't look right. Specifically, the filetypes keyword argument should be a sequence of two-element tuples containing file type names and patterns that will select what appears in the file listing. So for text files you would use:

filename = askopenfilename(filetypes=[('text files', '*.txt')])

I usually also include a generic pattern to allow easy access to files with other extensions thusly:

filename = askopenfilename(filetypes=[('text files', '*.txt'), ("all files", "*")])

Either way, it's important to remember to check the value returned because it might be the empty string it the user didn't select anything.

Upvotes: 6

MorganTonz
MorganTonz

Reputation: 1

You can try this:

from tkinter.filedialog import askopenfilename

Upvotes: -2

v.coder
v.coder

Reputation: 1932

The module name is misnamed.

Since the python version is 3.6 you need to use filedialog library. The includes should look something like this:

import os
from tkinter import *
import tkinter.filedialog

or

import os
from tkinter import *
from tkinter import filedialog

Upvotes: 2

A. Blackmagic
A. Blackmagic

Reputation: 233

The problem was actually that I needed to append askopenfilename() to filedialog as mentioned by Roars in a now deleted comment!(it looks like this --> filedialog.askopenfilename().

Upvotes: 2

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