Quick and Easy File Dialog in Python

Quick and easy file dialog in Python?

Tkinter is the easiest way if you don't want to have any other dependencies.
To show only the dialog without any other GUI elements, you have to hide the root window using the withdraw method:

import tkinter as tk
from tkinter import filedialog

root = tk.Tk()
root.withdraw()

file_path = filedialog.askopenfilename()

Python 2 variant:

import Tkinter, tkFileDialog

root = Tkinter.Tk()
root.withdraw()

file_path = tkFileDialog.askopenfilename()

Choosing a file in Python with simple Dialog

How about using tkinter?

from Tkinter import Tk     # from tkinter import Tk for Python 3.x
from tkinter.filedialog import askopenfilename

Tk().withdraw() # we don't want a full GUI, so keep the root window from appearing
filename = askopenfilename() # show an "Open" dialog box and return the path to the selected file
print(filename)

Done!

How to open and select a file in Python

As mentioned in the comments, you should provide a minimal reproducible example. Since you are a new member, I am giving you this example, which can be found here. https://www.geeksforgeeks.org/loading-images-in-tkinter-using-pil/

from tkinter import *
from PIL import ImageTk, Image
from tkinter import filedialog

def open_img():
    # Select the Imagename  from a folder
    x = openfilename()

    # opens the image
    img = Image.open(x)

    # resize the image and apply a high-quality down sampling filter
    img = img.resize((250, 250), Image.ANTIALIAS)

    # PhotoImage class is used to add image to widgets, icons etc
    img = ImageTk.PhotoImage(img)

    # create a label
    panel = Label(root, image=img)

    # set the image as img
    panel.image = img
    panel.grid(row=2)

def openfilename():
    # open file dialog box to select image
    # The dialogue box has a title "Open"
    filename = filedialog.askopenfilename(title='"pen')
    return filename

# Create a window
root = Tk()

# Set Title as Image Loader
root.title("Image Loader")

# Set the resolution of window
root.geometry("550x300+300+150")

# Allow Window to be resizable
root.resizable(width=True, height=True)

# Create a button and place it into the window using grid layout
btn = Button(root, text='open image', command=open_img).grid(row=1, columnspan=4)
root.mainloop()

Python, Dash: Tkinter openfile dialog works fine only the first time but fails afterwards

In callback you

• create tkinter main window root = tkinter.Tk()
• hide it root.withdraw()

but you never destroy it when you exit file dialog.

When it runs again callback then it tries to create second main window and it makes conflict with previous (hidden) main window.

If I use root.destroy() then code works correctly.

app.callback(
    Output(component_id='selected_directory', component_property='children'),
    Input(component_id='open_excel_button', component_property='n_clicks'),
    prevent_initial_call=True
)
def open_excel_function(open_excel): 
    print ('*** 1A. Callback open_file_dialog')
    ctx = dash.callback_context
    trigger = ctx.triggered[0]['prop_id'].split('.')[0]
    print("***", trigger, "is triggered.")

    root = tkinter.Tk()
    root.withdraw()
    # root.iconbitmap(default='Extras/transparent.ico')

    if trigger == 'open_excel_button':
        file_directory = tkinter.filedialog.askopenfilename(initialdir=FILE_DIR)
        print('***', file_directory)
    else:
        file_directory = None

    root.destroy()  # <--- SOLUTION
    
    return file_directory

or even

def open_excel_function(open_excel): 
    print ('*** 1A. Callback open_file_dialog')
    ctx = dash.callback_context
    trigger = ctx.triggered[0]['prop_id'].split('.')[0]
    print("***", trigger, "is triggered.")

    if trigger == 'open_excel_button':
        root = tkinter.Tk()
        root.withdraw()
       # root.iconbitmap(default='Extras/transparent.ico')

        file_directory = tkinter.filedialog.askopenfilename(initialdir=FILE_DIR)
        print('***', file_directory)

        root.destroy()  # <--- SOLUTION
    else:
        file_directory = None
    
    return file_directory

---

EDIT:

If you want to get only dirname then maybe you should use tkinter.filedialog.askdirectory() instead of tkinter.filedialog.askopenfilename()

---

Related Topics