Python Scripting Help Needed Gimp 3

Python Scripting Help Needed Gimp 3 - Printable Version

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Python Scripting Help Needed Gimp 3 - silenuznowan - 03-29-2025

I have some simple batch scripts for version 2 that I am trying to convert to work with version 3, which I really like working with.

Anyway one simple script was this:

Code:

import os

import gimpfu

def convert(filename):

    img = pdb.gimp_file_load(filename, filename)

    new_name = filename.rsplit(".",1)[0] + ".png"

    layer = pdb.gimp_image_merge_visible_layers(img, 1)

    pdb.gimp_file_save(img, layer, new_name, new_name)

    pdb.gimp_image_delete(img)

    savepath = ('${DEST}' + '/' + "$1" )

    if os.path.exists(savepath):

        print('the folder exists')

    else:

        os.makedirs(savepath)

    os.rename(new_name, savepath + "/" + new_name)

from glob import glob

for filename in glob("*.xcf"):

    print(filename)

    convert(filename)

pdb.gimp_quit(1)

I'm now trying to do the equivalent with Gimp 3 but am unable to even load the file.

In Gimp 2 you could load a file with the one line:

Code:

img = pdb.gimp_file_load(filename, filename)

So now this is replaced with:

Code:

def load_file(filename):

    procedure = Gimp.get_pdb().lookup_procedure('gimp-file-load');

    config = procedure.create_config();

    config.set_property('run-mode', Gimp.RunMode.NONINTERACTIVE);    

    config.set_property('file', filename);

    result = procedure.run(config);

    success = result.index(0);

    image = result.index(1)

    return image;

def convert(filename):

    img = load_file(filename)

Only this results in the following error

Quote:TypeError: could not convert 'interior-starship-doors.xcf' to type 'GFile' when setting property 'GimpProcedureConfig-gimp-file-load.file'

So it looks like instead of expecting a filename for file the new python api expects an object of type GFile to be passed, and I was wondering how I do this?

Thanks in advance.

Update: I managed to figure things out using Gio so my new load_file function looks like this:

Code:

def load_file(filename):

    file = Gio.File.new_for_path(filename)

    procedure = Gimp.get_pdb().lookup_procedure('gimp-file-load');

    config = procedure.create_config();

    config.set_property('run-mode', Gimp.RunMode.NONINTERACTIVE);

    config.set_property('file', file);

    result = procedure.run(config);

    success = result.index(0);

    image = result.index(1)

    return image;

While this works, I'm wondering if I'm also supposed to dispose of the Gio.File object, and if so how do I do that?

Thanks in advance. Also if there's any interest I can post the whole version of the new script.

RE: Python Scripting Help Needed Gimp 3 - Ofnuts - 03-29-2025

The Gio.File is a rather small object (only holds the name, and no state or file dara...) and will disappear when your script ends, so no need to worry.