lfs: don't add extension to hgrc after clone or share (BC)
Now that repository loading in core supports automatically loading
the lfs extension when the "lfs" requirement is present, we no
longer need to update the .hg/hgrc of newly-created repos to load
the lfs extension!
I'm marking this as BC because it is a change in behavior. But users
should not notice unless they create an LFS repo with new Mercurial
and then attempt to use it with an old versions that doesn't support
automatic extension loading.
Differential Revision: https://phab.mercurial-scm.org/D4712
# pushkey.py - dispatching for pushing and pulling keys
#
# Copyright 2010 Matt Mackall <mpm@selenic.com>
#
# This software may be used and distributed according to the terms of the
# GNU General Public License version 2 or any later version.
from __future__ import absolute_import
from . import (
bookmarks,
encoding,
obsolete,
phases,
)
def _nslist(repo):
n = {}
for k in _namespaces:
n[k] = ""
if not obsolete.isenabled(repo, obsolete.exchangeopt):
n.pop('obsolete')
return n
_namespaces = {"namespaces": (lambda *x: False, _nslist),
"bookmarks": (bookmarks.pushbookmark, bookmarks.listbookmarks),
"phases": (phases.pushphase, phases.listphases),
"obsolete": (obsolete.pushmarker, obsolete.listmarkers),
}
def register(namespace, pushkey, listkeys):
_namespaces[namespace] = (pushkey, listkeys)
def _get(namespace):
return _namespaces.get(namespace, (lambda *x: False, lambda *x: {}))
def push(repo, namespace, key, old, new):
'''should succeed iff value was old'''
pk = _get(namespace)[0]
return pk(repo, key, old, new)
def list(repo, namespace):
'''return a dict'''
lk = _get(namespace)[1]
return lk(repo)
encode = encoding.fromlocal
decode = encoding.tolocal
def encodekeys(keys):
"""encode the content of a pushkey namespace for exchange over the wire"""
return '\n'.join(['%s\t%s' % (encode(k), encode(v)) for k, v in keys])
def decodekeys(data):
"""decode the content of a pushkey namespace from exchange over the wire"""
result = {}
for l in data.splitlines():
k, v = l.split('\t')
result[decode(k)] = decode(v)
return result