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# along with this program; if not, write to the Free Software
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# Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA 02111-1307 USA
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# TODO: Some kind of command-line display of revision properties:
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# perhaps show them in log -v and allow them as options to the commit command.
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import bzrlib.errors
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from bzrlib.graph import node_distances, select_farthest, all_descendants
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from bzrlib.osutils import contains_whitespace
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NULL_REVISION="null:"
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class RevisionReference(object):
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Reference to a stored revision.
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Includes the revision_id and revision_sha1.
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def __init__(self, revision_id, revision_sha1=None):
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if revision_id == None \
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or isinstance(revision_id, basestring):
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self.revision_id = revision_id
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raise ValueError('bad revision_id %r' % revision_id)
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if revision_sha1 != None:
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if isinstance(revision_sha1, basestring) \
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and len(revision_sha1) == 40:
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self.revision_sha1 = revision_sha1
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raise ValueError('bad revision_sha1 %r' % revision_sha1)
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class Revision(object):
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"""Single revision on a branch.
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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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Dictionary of revision properties. These are attached to the
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revision as extra metadata. The name must be a single
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word; the value can be an arbitrary string.
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List of parent revisions, each is a RevisionReference.
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def __init__(self, revision_id, properties=None, **args):
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self.revision_id = revision_id
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self.properties = properties or {}
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self._check_properties()
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self.parent_sha1s = []
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def __init__(self, **args):
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self.__dict__.update(args)
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def __repr__(self):
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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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# FIXME: rbc 20050930 parent_ids are not being compared
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self.inventory_sha1 == other.inventory_sha1
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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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and self.properties == other.properties)
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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 _check_properties(self):
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"""Verify that all revision properties are OK.
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for name, value in self.properties.iteritems():
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if not isinstance(name, basestring) or contains_whitespace(name):
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raise ValueError("invalid property name %r" % name)
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if not isinstance(value, basestring):
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raise ValueError("invalid property value %r for %r" %
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def is_ancestor(revision_id, candidate_id, branch):
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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, revision_source):
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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.repository.get_ancestry(revision_id)
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if candidate_id is None:
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for ancestor_id, distance in iter_ancestors(revision_id, revision_source):
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if ancestor_id == candidate_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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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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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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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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return a_closest[0]
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elif b_closest[0] in a_closest:
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TODO: Produce graphs with the NULL revision as root, so that we can find
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a common even when trees are not branches don't represent a single line
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RBC: 20051024: note that when we have two partial histories, this may not
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be possible. But if we are willing to pretend :)... sure.
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if parents is not None:
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ancestors[line] = set(parents)
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lines = new_lines
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# The history for revision becomes inaccessible without
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# actually hitting a no-parents revision. This then
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# makes these asserts below trigger. So, if root is None
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# determine the actual root by walking the accessible tree
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# and then stash NULL_REVISION at the end.
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descendants[root] = {}
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# for every revision, check we can access at least
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# one parent, if we cant, add NULL_REVISION and
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for rev in ancestors:
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if len(ancestors[rev]) == 0:
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raise RuntimeError('unreachable code ?!')
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for parent in ancestors[rev]:
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if parent in ancestors:
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descendants[root][rev] = 1
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ancestors[rev].add(root)
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ancestors[root] = set()
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assert root not in descendants[root]
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assert root not in ancestors[root]
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return root, ancestors, descendants
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def combined_graph(revision_a, revision_b, revision_source):
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"""Produce a combined ancestry graph.
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Return graph root, ancestors map, descendants map, set of common nodes"""
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descendants[node].update(node_dec)
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return root, ancestors, descendants, common
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def common_ancestor(revision_a, revision_b, revision_source):
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root, ancestors, descendants, common = \
added
removed
bzrlib/revision.py
