Issue

In case an opener is used into open() function, for creating a file with certain permission for instance, ruff PTH123 suggests to use pathlib.Path.open() which does not implement the opener.

import os
dir_fd = os.open('somedir', os.O_RDONLY)
def opener(path, flags):
    return os.open(path, flags, dir_fd=dir_fd)

with open('spamspam.txt', 'w', opener=opener) as f:
    print('This will be written to somedir/spamspam.txt', file=f)

• ruff version range: v0.0.276 - v0.0.291

Reference

• ruff PTH123
• python open()
• python pathlib.Path.open()

Solution

• Ruff can stop reporting it in case an opener is defined
• Using # noqa: PTH123
• …