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@ -12,6 +12,7 @@ import json
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import locale
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import operator
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import os
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import pathlib
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import random
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import re
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import shutil
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@ -1481,27 +1482,39 @@ it checks if the length does not exceed the OS limit. However currently
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any error is simply raised, independent of the cause.
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Without this, download would fail only after the entire file is downloaded."""
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cwd = os.getcwd()
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# An improvement idea:
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# by default, retry (exec yt-dlp itself) by
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# -o "%(id)s.%(ext)s" --write-info-json,
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# but respect the directory from --output of the original call.
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with tempfile.TemporaryDirectory() as d:
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os.chdir(d)
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try:
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with open(filename, 'w') as f:
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f.close()
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# To make sure it's confined under tmpdir
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tmpfn = '.' + os.sep + os.path.splitdrive(filename)[1]
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parentDirStr = os.sep + '..' + os.sep
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# This may contain '../' so remove them.
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while parentDirStr in tmpfn:
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tmpfn = tmpfn.replace(parentDirStr, os.sep)
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tmpfn = os.path.join(d, tmpfn)
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pathlib.Path(os.path.dirname(tmpfn)).mkdir(parents=True, exist_ok=True)
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open(tmpfn, 'w').close()
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except OSError as e:
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if (os.name == 'nt' and e.errno == 206) or (e.errno == errno.ENAMETOOLONG):
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# The first condition is for windows,
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# and the second for unix-ish systems.
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# An improvement idea:
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# by default, retry (exec yt-dlp itself) by
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# -o "%(id)s.%(ext)s" --write-info-json,
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# but respect the directory from --output of the original call.
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if (os.name == 'nt' and e.errno == 206) or (os.name != 'nt' and e.errno == errno.ENAMETOOLONG):
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# For Win, 206 means filename length exceeds MAX_PATH.
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e.filename = filename
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self.to_screen('''[Notice] The file name to be saved is too long, exceeding the OS limit.
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[Notice] Consider options --trim-filenames or -o (--output).''')
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raise
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finally:
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os.chdir(cwd)
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elif os.name == 'nt' and e.errno == 22:
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# Even when MAX_PATH is disabled, 255 chars is the limit, resulting in 22.
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# https://github.com/python/cpython/issues/126929#issuecomment-2483684861
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e.filename = filename
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self.to_screen(f'''[Notice] Attempt to create file {filename} resulted in Errno 22.
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This is often caused e.g. by too long filename or forbidden characters.''')
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raise
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def prepare_filename(self, info_dict, dir_type='', *, outtmpl=None, warn=False):
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"""Generate the output filename"""
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