mirror of https://github.com/borgbackup/borg.git
66 lines
2.3 KiB
Python
66 lines
2.3 KiB
Python
import os
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import re
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def translate(pat, match_end=r"\Z"):
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"""Translate a shell-style pattern to a regular expression.
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The pattern may include ``**<sep>`` (<sep> stands for the platform-specific path separator; "/" on POSIX systems) for
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matching zero or more directory levels and "*" for matching zero or more arbitrary characters with the exception of
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any path separator. Wrap meta-characters in brackets for a literal match (i.e. "[?]" to match the literal character
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"?").
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Using match_end=regex one can give a regular expression that is used to match after the regex that is generated from
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the pattern. The default is to match the end of the string.
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This function is derived from the "fnmatch" module distributed with the Python standard library.
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Copyright (C) 2001-2016 Python Software Foundation. All rights reserved.
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TODO: support {alt1,alt2} shell-style alternatives
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"""
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sep = os.path.sep
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n = len(pat)
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i = 0
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res = ""
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while i < n:
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c = pat[i]
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i += 1
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if c == "*":
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if i + 1 < n and pat[i] == "*" and pat[i + 1] == sep:
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# **/ == wildcard for 0+ full (relative) directory names with trailing slashes; the forward slash stands
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# for the platform-specific path separator
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res += fr"(?:[^\{sep}]*\{sep})*"
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i += 2
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else:
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# * == wildcard for name parts (does not cross path separator)
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res += r"[^\%s]*" % sep
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elif c == "?":
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# ? == any single character excluding path separator
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res += r"[^\%s]" % sep
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elif c == "[":
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j = i
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if j < n and pat[j] == "!":
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j += 1
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if j < n and pat[j] == "]":
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j += 1
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while j < n and pat[j] != "]":
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j += 1
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if j >= n:
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res += "\\["
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else:
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stuff = pat[i:j].replace("\\", "\\\\")
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i = j + 1
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if stuff[0] == "!":
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stuff = "^" + stuff[1:]
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elif stuff[0] == "^":
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stuff = "\\" + stuff
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res += "[%s]" % stuff
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else:
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res += re.escape(c)
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return "(?ms)" + res + match_end
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