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#!/usr/bin/python
#
# Copyright(c) 2004-2010, Gentoo Foundation
#
# Licensed under the GNU General Public License, v2
#
# $Header$
"""Provides common methods on Gentoo GLEP 53 keywords.
http://www.gentoo.org/proj/en/glep/glep-0053.html
"""
__all__ = (
'Keyword',
'compare_strs'
)
# =======
# Imports
# =======
# =======
# Classes
# =======
class Keyword(object):
"""Provides common methods on a GLEP 53 keyword."""
def __init__(self, keyword):
self.keyword = keyword
def __str__(self):
return self.keyword
def __repr__(self):
return "<Keyword {0.keyword!r}>".format(self)
# =========
# Functions
# =========
def compare_strs(kw1, kw2):
"""Similar to the builtin cmp, but for keyword strings. Usually called
as: keyword_list.sort(keyword.compare_strs)
An alternative is to use the Keyword descriptor directly:
>>> kwds = sorted(Keyword(x) for x in keyword_list)
@see: >>> help(cmp)
"""
kw1_arch, sep, kw1_os = kw1.partition('-')
kw2_arch, sep, kw2_os = kw2.partition('-')
if kw1_arch != kw2_arch:
if kw1_os != kw2_os:
return cmp(kw1_os, kw2_os)
return cmp(kw1_arch, kw2_arch)
return cmp(kw1_os, kw2_os)
def reduce_keywords(keywords):
"""Reduce a list of keywords to a unique set of stable keywords.
Example usage:
>>> reduce_keywords(['~amd64', 'x86', '~x86'])
set(['amd64', 'x86'])
@type keywords: array
@rtype: set
"""
return set(x.lstrip('~') for x in keywords)
abs_keywords = reduce_keywords
# FIXME: this is unclear
# dj, how about 'deduce_keyword'
# I was trying to avoid a 2nd use of determine_keyword name (in analyse.lib)
# but that one is a little different and not suitable for this task.
def determine_keyword(arch, accepted, keywords):
"""Determine a keyword from matching a dep's KEYWORDS
list against the ARCH & ACCEPT_KEYWORDS provided.
@type arch: string
@param arch: portage.settings["ARCH"]
@type accepted: string
@param accepted: portage.settings["ACCEPT_KEYWORDS"]
@type keywords: string
@param keywords: the pkg ebuilds keywords
"""
if not keywords:
return ''
keys = keywords.split()
if arch in keys:
return arch
keyworded = "~" + arch
if keyworded in keys:
return keyworded
match = list(set(accepted.split(" ")).intersection(keys))
if len(match) > 1:
if arch in match:
return arch
if keyworded in match:
return keyworded
return 'unknown'
if match:
return match[0]
return 'unknown'
|
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