mercurial/pushkey.py

rebase: fix warning about ignoring tool option on rebase continue (issue4698)

Before this patch rebase --continue with specified --tool option outputs
warnings "tool option will be ignored". It is false statement because
in case of any merge conflicts it uses specified tool to resolve it.

This patch makes this warning appears only when user specified --tool
when running rebase --abort , in this case tool doesn't have any
sense

# pushkey.py - dispatching for pushing and pulling keys
#
# Copyright 2010 Matt Mackall <[e-mail]>
#
# 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