CODEWITHSUNDEEP
pythontkinter
PyNoob enthusiast
PyNoob enthusiast

Reputation: 1

How do I locate my file/image in tkinter without C: User?

How do I make a code that locates the image inside a folder without using the direct location with C:User. So that whenever I send the GUI to someone, they can run it without any errors. Is there any other codes that I can use?

photo = tk.PhotoImage(file="C:/Users/Me/Downloads/xxxx/Pics/xxxx.png")
label = tk.Label(window, image=photo)
label.place(x=0, y=0, relwidth=1, relheight=1)

Upvotes: 0

Views: 361

Answers (2)

frippe
frippe

Reputation: 1393

Reading a bit between the lines, it sounds like:

  1. You want to distribute your code by sending a directory or archive to people so that they can run it on their machine
  2. You have resources inside this package that you'd like to reference

You can't really assume anything about where the user will place this directory on their machine, so you need to work with relative paths.

This line:

photo = tk.PhotoImage(file=os.path.join(downloadsDirectory,"xxxx","Pics","xxxx.png"))

seems to suggest your script is placed in folder xxxx of the user's Downloads directory, which means the relative path of the resource is Pics/xxxx.png.

Based on these assumptions, you could do something like:

from pathlib import Path

script_dir = Path(__file__).parent.absolute()
resource_path = script_dir / "Pics" / "xxxx.png"

photo = tk.PhotoImage(file=str(resource_path))  # use resource_path.as_posix() instead if you want forward slashes on all platforms instead of native path format
label = tk.Label(window, image=photo)
label.place(x=0, y=0, relwidth=1, relheight=1)

Upvotes: 1

user17505586
user17505586

Reputation:

Use pathlibe instead, if you want to access user home, it will also work in LINUX OR MacOS as its platform independent

import os
from pathlib import Path

downloadsDirectory = os.path.join(Path.home(),'Downloads')
photo = tk.PhotoImage(file=os.path.join(downloadsDirectory,"xxxx","Pics","xxxx.png"))
label = tk.Label(window, image=photo)
label.place(x=0, y=0, relwidth=1, relheight=1)

Upvotes: 0

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