33
51
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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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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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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"""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 old_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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def revision_graph(revision, revision_source):
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"""Produce a graph of the ancestry of the specified revision.
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Return root, ancestors map, descendants map
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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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descendants[revision] = {}
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while len(lines) > 0:
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if line == NULL_REVISION:
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rev = revision_source.get_revision(line)
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parents = list(rev.parent_ids)
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if len(parents) == 0:
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parents = [NULL_REVISION]
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except bzrlib.errors.NoSuchRevision:
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if parents is not None:
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for parent in parents:
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if parent not in ancestors:
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new_lines.add(parent)
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if parent not in descendants:
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descendants[parent] = {}
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descendants[parent][line] = 1
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if parents is not None:
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ancestors[line] = set(parents)
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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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root, ancestors, descendants = revision_graph(revision_a, revision_source)
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root_b, ancestors_b, descendants_b = revision_graph(revision_b,
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raise bzrlib.errors.NoCommonRoot(revision_a, revision_b)
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for node, node_anc in ancestors_b.iteritems():
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if node in ancestors:
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ancestors[node] = set()
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ancestors[node].update(node_anc)
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for node, node_dec in descendants_b.iteritems():
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if node not in descendants:
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descendants[node] = {}
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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 = \
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combined_graph(revision_a, revision_b, revision_source)
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except bzrlib.errors.NoCommonRoot:
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raise bzrlib.errors.NoCommonAncestor(revision_a, revision_b)
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distances = node_distances (descendants, ancestors, root)
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farthest = select_farthest(distances, common)
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if farthest is None or farthest == NULL_REVISION:
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raise bzrlib.errors.NoCommonAncestor(revision_a, revision_b)
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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):
280
object.__init__(self)
281
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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root, ancestors, descendants = revision_graph(rev_id, rev_source)
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if len(descendants) == 0:
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raise NoSuchRevision(rev_source, rev_id)
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if ancestor_id not in descendants:
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rev_source.get_revision(ancestor_id)
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raise bzrlib.errors.NotAncestor(rev_id, ancestor_id)
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root_descendants = all_descendants(descendants, ancestor_id)
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root_descendants.add(ancestor_id)
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if rev_id not in root_descendants:
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raise bzrlib.errors.NotAncestor(rev_id, ancestor_id)
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distances = node_distances(descendants, ancestors, ancestor_id,
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root_descendants=root_descendants)
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def best_ancestor(rev_id):
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for anc_id in ancestors[rev_id]:
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distance = distances[anc_id]
321
if revision_history is not None and anc_id in revision_history:
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elif best is None or distance > best[1]:
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best = (anc_id, distance)
329
while next != ancestor_id:
331
next = best_ancestor(next)
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from bzrlib.xml import Element, SubElement
76
root = Element('revision',
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committer = self.committer,
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timestamp = '%.9f' % self.timestamp,
79
revision_id = self.revision_id,
80
inventory_id = self.inventory_id,
81
inventory_sha1 = self.inventory_sha1,
84
root.set('timezone', str(self.timezone))
87
msg = SubElement(root, 'message')
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msg.text = self.message
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pelts = SubElement(root, 'parents')
93
pelts.tail = pelts.text = '\n'
94
for rr in self.parents:
95
assert isinstance(rr, RevisionReference)
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p = SubElement(pelts, 'revision_ref')
99
p.set('revision_id', rr.revision_id)
101
p.set('revision_sha1', rr.revision_sha1)
106
def from_element(cls, elt):
107
return unpack_revision(elt)
109
from_element = classmethod(from_element)
113
def unpack_revision(elt):
114
"""Convert XML element into Revision object."""
115
# <changeset> is deprecated...
116
from bzrlib.errors import BzrError
118
if elt.tag not in ('revision', 'changeset'):
119
raise BzrError("unexpected tag in revision file: %r" % elt)
121
rev = Revision(committer = elt.get('committer'),
122
timestamp = float(elt.get('timestamp')),
123
revision_id = elt.get('revision_id'),
124
inventory_id = elt.get('inventory_id'),
125
inventory_sha1 = elt.get('inventory_sha1')
128
precursor = elt.get('precursor')
129
precursor_sha1 = elt.get('precursor_sha1')
131
pelts = elt.find('parents')
135
assert p.tag == 'revision_ref', \
136
"bad parent node tag %r" % p.tag
137
rev_ref = RevisionReference(p.get('revision_id'),
138
p.get('revision_sha1'))
139
rev.parents.append(rev_ref)
143
prec_parent = rev.parents[0].revision_id
144
assert prec_parent == precursor
146
# revisions written prior to 0.0.5 have a single precursor
147
# give as an attribute
148
rev_ref = RevisionReference(precursor, precursor_sha1)
149
rev.parents.append(rev_ref)
151
v = elt.get('timezone')
152
rev.timezone = v and int(v)
154
rev.message = elt.findtext('message') # text of <message>
159
REVISION_ID_RE = None
161
def validate_revision_id(rid):
162
"""Check rid is syntactically valid for a revision id."""
163
global REVISION_ID_RE
164
if not REVISION_ID_RE:
166
REVISION_ID_RE = re.compile('[\w.-]+@[\w.-]+--?\d+--?[0-9a-f]+\Z')
168
if not REVISION_ID_RE.match(rid):
169
raise ValueError("malformed revision-id %r" % rid)