#🔒 How to use Image.open to open the actual image file, and not a temp file?

7 messages · Page 1 of 1 (latest)

brave vapor
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Hi all, new to python and spent the last few days learning some basics (but kind of rushing through the fundimentals; i have some coding bg and mainly wanted to dive into working on this. not good practice, but its what i did)

My end goal is to write a program that will open a random PNG file from the liss of all files in a folder. i have written that successfully, but what i need is for it to open the actual file for editing. Below is a simplified version of my code that simply opens a specified image in a given older, and writes the filepath in the console.

Currently, as per how the actual module works, it scans the chosen file and opens it as a temporary image, which is deleted as soon as its opened, making it essentially a fresh image that would need to be saved/"re"saved as a new file, with the original left untouched no matter what. How would i go about opening the original file, such that the paint window that opens can be edited and then Ctrl + S would save those changes over the original specified file?

from PIL import Image       # Initializer for images.

FileName = "3"              # File to open.
FilePath = "C:/Users/ [Username] /Desktop/RandomImageTest/" + FileName + ".png"  # Filepath for image to open.

print(FilePath)             # Writes the filename in console.
Image.open(File).show()     # Opens file.```
long oracleBOT
#

@brave vapor

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obsidian holly
#

you're opening it in memory so you can modify it. after your modifications, you save it back to disk

tardy stone
#

@brave vapor ^^

long oracleBOT
#
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