I need help on how to make a button after I choose a folder in dropdown list.
For Example: I have 3 folders name "Folder 1","Folder 2" & "Folder 3". Inside "Folder 1", I have 5 excel(.xlsx) files. So I need help on how to read and display the data in 1 excel(.xlsx) file.
My current situation: I choose "Folder 1" in the dropdown menu. The next thing that I need is a button which can open the "Folder 1" and display the other list of 5 excel(.xlsx) files. And then, I can choose one of the excel(.xlsx) file and display the data inside the gui.
Here is my code.... Help me :'(
import os
import tkinter as tk
from tkinter import ttk
#import tkinter as tk
from tkinter import filedialog, messagebox, ttk
folder = r'C:\Users\test folder'
filelist = [fname for fname in os.listdir(folder)]
master = tk.Tk()
master.geometry('1200x800')
master.title('Select a file')
optmenu = ttk.Combobox(master, values=filelist, state='readonly')
optmenu.pack(fill='x')
master.mainloop()
You cannot just select and read a file's contents from tkinter. You have to write some other scripts for that reading part.
What the selection of filename does from tkinter combo box, is nothing but, get the particular file name as a string type.
However, in Python it's pretty straight forward to read a .xlsx file.
You can use Pandas module for that.
I have written the code for you to read the file, (but you have to install pandas)
from functools import partial
import os
import tkinter as tk
from tkinter import ttk
#import tkinter as tk
from tkinter import filedialog, messagebox, ttk
import pandas
def get_selected_file_name(file_menu):
filename = file_menu.get()
print("file selected:", filename)
reader = pandas.read_excel(filename) # code to read excel file
# now you can use the `reader` object to get the file data
folder = os.path.realpath('./test_folder')
filelist = [fname for fname in os.listdir(folder)]
master = tk.Tk()
master.geometry('1200x800')
master.title('Select a file')
optmenu = ttk.Combobox(master, values=filelist, state='readonly')
optmenu.pack(fill='x')
button_select = tk.Button(master, text="Read File",
width=20,
height=7,
compound=tk.CENTER,
command=partial(get_selected_file_name, optmenu))
button_select.pack(side=tk.RIGHT)
master.mainloop()
The window should look somewhat like this:
I'd explore using the filedialog module in tkinter.
import tkinter as tk
from tkinter import filedialog
def load_file():
f_in = filedialog.askopenfilename( filetypes = [ ( 'Python', '*.py' ) ]) # Change to appropriate extension.
if len( f_in ) > 0:
with open( f_in, 'r' ) as file:
filedata = file.read()
print( filedata ) # printed to the terminal for simplicity.
# process it as required.
root = tk.Tk()
tk.Button( root, text = 'Find File', command = load_file ).grid()
root.mainloop()
askopenfilename allows a user to navigate the folder tree to find the correct file. Basic documentation
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
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.
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().
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
You can try this:
from tkinter.filedialog import askopenfilename
Right now, I have a GUI program that allows you to change parameters and stuff like that. I want to make it so you can choose a picture instead of only having one for the whole thing.
I have this:
import tkinter as tk
from tkinter import filedialog
root = tk.Tk()
root.withdraw()
file_path = filedialog.askopenfilename()
When I try to use this, it will just say that it cannot import filedialog.
EDIT:
Okay, so I just fixed that error by using:
import tkFileDialog as filedialog
Now I just need help making the file I choose be the one that appears on the canvas. Right now, I have this:
__dir__ = os.path.dirname(os.path.abspath(__file__))
filename = os.path.join(__dir__, root.filename)
img = PIL.Image.open(filename)
shrek= img.resize((100,100))
root = Tk() # create main window; must be done before using ImageTk
root.filename = filedialog.askopenfilename(initialdir = "E:/Images",title = "choose your file",filetypes = (("jpeg files","*.jpg"),("all files","*.*")))
print (root.filename)
I am trying to make the file that I choose from the explorer replace the current file.
This is for Python 2.x import tkFileDialog as filedialog
I guess you use Python 2.x..
I want to launch an "Open File" dialog in Tkinter in Python 2.7.
My code starts with:
from Tkinter import Frame, Tk, BOTH, Text, Menu, END
import tkFileDialog as tkfd
import fileinput
root = Tk()
global strTab
strTab = ""
def openTab(event):
r = tkfd.askopenfilename()
strTab = unicodedata.normalize('NFKD', r).encode('ascii','ignore')
Later in the code I have:
btnLoadTab = Button(root,
text="Load Tab",
width=30,height=5,
bg="white",fg="black")
btnLoadTab.bind("<Button-1>", openTab)
btnLoadTab.pack()
root.mainloop()
When I press the button an "Open File" dialog is shown, but when I select a file it closes and the button remains "clicked".
If I later call to strTab outside of openTab, it remains equal to "".
You can find workable example here: http://www.python-course.eu/tkinter_dialogs.php
trying to make a GUI with an 'open file' button. When I run the code shown below, the open file dialog opens straight away, and not when I press the button. Why? Is there a simple way to fix this that doesn't involve using classes? (I don't currently know anything about classes and am working on a time-pressured project)
from tkinter import *
interface = Tk()
def openfile():
return filedialog.askopenfilename()
button = ttk.Button(interface, text = "Open", command = openfile())
button.grid(column = 1, row = 1)
interface.mainloop()
The code is passing the return value of the openfile function call, not the function itself. Pass the function itself by removing trailing () which cause a call.
from tkinter import *
from tkinter import ttk
from tkinter import filedialog
interface = Tk()
def openfile():
return filedialog.askopenfilename()
button = ttk.Button(interface, text="Open", command=openfile) # <------
button.grid(column=1, row=1)
interface.mainloop()