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# Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA 02111-1307 USA
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from xml import XMLMixin
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from cElementTree import Element, ElementTree, SubElement
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from elementtree.ElementTree import Element, ElementTree, SubElement
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class Revision(XMLMixin):
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class Revision(object):
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"""Single revision on a branch.
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Revisions may know their revision_hash, but only once they've been
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written out. This is not stored because you cannot write the hash
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into the file it describes.
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:todo: Perhaps make predecessor be a child element, not an attribute?
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After bzr 0.0.5 revisions are allowed to have multiple parents.
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List of parent revision_ids
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def __init__(self, **args):
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self.inventory_id = None
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self.revision_id = None
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self.__dict__.update(args)
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self.parent_sha1s = []
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def __repr__(self):
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return "<Revision id %s>" % self.revision_id
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root = Element('revision',
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committer = self.committer,
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timestamp = '%.9f' % self.timestamp,
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revision_id = self.revision_id,
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inventory_id = self.inventory_id,
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timezone = str(self.timezone))
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root.set('precursor', self.precursor)
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msg = SubElement(root, 'message')
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msg.text = self.message
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def from_element(cls, elt):
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# <changeset> is deprecated...
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if elt.tag not in ('revision', 'changeset'):
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bailout("unexpected tag in revision file: %r" % elt)
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cs = cls(committer = elt.get('committer'),
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timestamp = float(elt.get('timestamp')),
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precursor = elt.get('precursor'),
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revision_id = elt.get('revision_id'),
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inventory_id = elt.get('inventory_id'))
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v = elt.get('timezone')
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cs.timezone = v and int(v)
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cs.message = elt.findtext('message') # text of <message>
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from_element = classmethod(from_element)
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return "<Revision id %s>" % self.revision_id
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def __eq__(self, other):
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if not isinstance(other, Revision):
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return (self.inventory_id == other.inventory_id
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and self.inventory_sha1 == other.inventory_sha1
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and self.revision_id == other.revision_id
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and self.timestamp == other.timestamp
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and self.message == other.message
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and self.timezone == other.timezone
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and self.committer == other.committer)
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def __ne__(self, other):
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return not self.__eq__(other)
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def validate_revision_id(rid):
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"""Check rid is syntactically valid for a revision id."""
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if not REVISION_ID_RE:
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REVISION_ID_RE = re.compile('[\w.-]+@[\w.-]+--?\d+--?[0-9a-f]+\Z')
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if not REVISION_ID_RE.match(rid):
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raise ValueError("malformed revision-id %r" % rid)
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def is_ancestor(revision_id, candidate_id, branch):
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"""Return true if candidate_id is an ancestor of revision_id.
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A false negative will be returned if any intermediate descendent of
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candidate_id is not present in any of the revision_sources.
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revisions_source is an object supporting a get_revision operation that
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behaves like Branch's.
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return candidate_id in branch.get_ancestry(revision_id)
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def iter_ancestors(revision_id, revision_source, only_present=False):
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ancestors = (revision_id,)
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while len(ancestors) > 0:
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for ancestor in ancestors:
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yield ancestor, distance
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revision = revision_source.get_revision(ancestor)
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except bzrlib.errors.NoSuchRevision, e:
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if e.revision == revision_id:
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yield ancestor, distance
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new_ancestors.extend(revision.parent_ids)
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ancestors = new_ancestors
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def find_present_ancestors(revision_id, revision_source):
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"""Return the ancestors of a revision present in a branch.
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It's possible that a branch won't have the complete ancestry of
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one of its revisions.
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anc_iter = enumerate(iter_ancestors(revision_id, revision_source,
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for anc_order, (anc_id, anc_distance) in anc_iter:
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if not found_ancestors.has_key(anc_id):
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found_ancestors[anc_id] = (anc_order, anc_distance)
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return found_ancestors
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def __get_closest(intersection):
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for entry in intersection:
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if entry[0] == intersection[0][0]:
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matches.append(entry[2])
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def common_ancestor(revision_a, revision_b, revision_source):
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"""Find the ancestor common to both revisions that is closest to both.
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from bzrlib.trace import mutter
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a_ancestors = find_present_ancestors(revision_a, revision_source)
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b_ancestors = find_present_ancestors(revision_b, revision_source)
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# a_order is used as a tie-breaker when two equally-good bases are found
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for revision, (a_order, a_distance) in a_ancestors.iteritems():
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if b_ancestors.has_key(revision):
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a_intersection.append((a_distance, a_order, revision))
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b_intersection.append((b_ancestors[revision][1], a_order, revision))
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mutter("a intersection: %r" % a_intersection)
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mutter("b intersection: %r" % b_intersection)
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a_closest = __get_closest(a_intersection)
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if len(a_closest) == 0:
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b_closest = __get_closest(b_intersection)
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assert len(b_closest) != 0
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mutter ("a_closest %r" % a_closest)
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mutter ("b_closest %r" % b_closest)
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if a_closest[0] in b_closest:
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elif b_closest[0] in a_closest:
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raise bzrlib.errors.AmbiguousBase((a_closest[0], b_closest[0]))
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class MultipleRevisionSources(object):
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"""Proxy that looks in multiple branches for revisions."""
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def __init__(self, *args):
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object.__init__(self)
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assert len(args) != 0
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self._revision_sources = args
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def get_revision(self, revision_id):
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for source in self._revision_sources:
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return source.get_revision(revision_id)
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except bzrlib.errors.NoSuchRevision, e:
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def get_intervening_revisions(ancestor_id, rev_id, rev_source,
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revision_history=None):
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"""Find the longest line of descent from maybe_ancestor to revision.
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Revision history is followed where possible.
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If ancestor_id == rev_id, list will be empty.
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Otherwise, rev_id will be the last entry. ancestor_id will never appear.
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If ancestor_id is not an ancestor, NotAncestor will be thrown
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[rev_source.get_revision(r) for r in (ancestor_id, rev_id)]
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if ancestor_id == rev_id:
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def historical_lines(line):
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"""Return a tuple of historical/non_historical lines, for sorting.
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The non_historical count is negative, since non_historical lines are
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for revision in line:
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if revision in revision_history:
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return good_count, bad_count
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successful_lines = []
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while len(active) > 0:
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for parent in rev_source.get_revision(line[-1]).parent_ids:
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if parent == ancestor_id:
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successful_lines.append(line_copy)
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line_copy.append(parent)
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new_active.append(line_copy)
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if len(successful_lines) == 0:
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raise bzrlib.errors.NotAncestor(rev_id, ancestor_id)
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for line in successful_lines:
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if revision_history is not None:
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by_historical_lines = []
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for line in successful_lines:
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count = historical_lines(line)
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by_historical_lines.append((count, line))
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by_historical_lines.sort()
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if by_historical_lines[-1][0][0] > 0:
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return by_historical_lines[-1][1]
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assert len(successful_lines)
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successful_lines.sort(cmp, len)
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return successful_lines[-1]
added
removed
bzrlib/revision.py
