keep version in release
This commit is contained in:
parent
6dfa80b646
commit
8bfbaef598
3 changed files with 21 additions and 13 deletions
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@ -1,3 +1,4 @@
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build
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build
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dist
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dist
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oxutils.egg-info
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oxutils.egg-info
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ox/__version.py
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@ -1,7 +1,11 @@
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# -*- coding: utf-8 -*-
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# -*- coding: utf-8 -*-
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# vi:si:et:sw=4:sts=4:ts=4
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# vi:si:et:sw=4:sts=4:ts=4
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# GPL 2011
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# GPL 2011
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__version__ = '2.1.1'
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try:
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from . import __version
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__version__ = __version.VERSION
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except:
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__version__ = '2.1.x'
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from . import cache
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from . import cache
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from . import js
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from . import js
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27
setup.py
27
setup.py
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@ -6,29 +6,32 @@ try:
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except:
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except:
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from distutils.core import setup
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from distutils.core import setup
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def get_bzr_version():
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def get_version():
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import os
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import os
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import re
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import re
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info = os.path.join(os.path.dirname(__file__), '.bzr/branch/last-revision')
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info = os.path.join(os.path.dirname(__file__), '.bzr/branch/last-revision')
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changelog = os.path.join(os.path.dirname(__file__), 'debian/changelog')
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__version = os.path.join(os.path.dirname(__file__), 'ox/__version.py')
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if os.path.exists(info):
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if os.path.exists(info):
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f = open(info)
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f = open(info)
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rev = int(f.read().split()[0])
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rev = int(f.read().split()[0])
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f.close()
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f.close()
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if rev:
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if rev:
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return u'%s' % rev
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version = u"2.1.%s"%rev
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elif os.path.exists(changelog):
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with open(__version, 'w') as fd:
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f = open(changelog)
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fd.write('VERSION="%s"'%version)
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head = f.read().strip().split('\n')[0]
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return version
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f.close()
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elif os.path.exists(__version):
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rev = re.compile('\d+\.\d+\.(\d+)').findall(head)
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with open(__version) as fd:
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if rev:
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data = fd.read().strip().split('\n')[0]
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return u"%s" % rev[0]
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version = re.compile('VERSION="(.*)"').findal(data)
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return u'unknown'
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if version:
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version = version[0]
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return version
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return u'2.1.x'
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setup(
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setup(
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name="ox",
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name="ox",
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version="2.1.%s" % get_bzr_version() ,
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version=get_version() ,
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description="python-ox - the web in a dict",
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description="python-ox - the web in a dict",
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author="0x2620",
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author="0x2620",
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author_email="0x2620@0x2620.org",
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author_email="0x2620@0x2620.org",
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