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A plugin for displaying what revisions are in 'other' but not in local.
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def show_missing(br_local, br_remote, verbose=False, quiet=False):
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"""Show the revisions which exist in br_remote, that
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do not exist in br_local.
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from bzrlib.log import show_one_log
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local_history = br_local.revision_history()
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remote_history = br_remote.revision_history()
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if local_history == remote_history:
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print 'Trees are identical.'
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if local_history[:len(remote_history)] == remote_history:
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# Local is not missing anything from remote, so consider it
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print 'Local tree has all of remote revisions (remote is missing local)'
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# Check for divergence
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common_idx = min(len(local_history), len(remote_history)) - 1
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if common_idx >= 0 and local_history[common_idx] != remote_history[common_idx]:
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print 'Trees have diverged'
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local_rev_set = set(local_history)
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# Find the last common revision between the two trees
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for revno, (local_rev, remote_rev) in enumerate(zip(local_history, remote_history)):
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if local_rev != remote_rev:
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for rno, rev_id in enumerate(remote_history[revno:]):
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# This assumes that you can have a revision in the
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# local history, which does not have the same ancestry
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# as the remote ancestry.
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# This may or may not be possible.
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# In the future this should also checked for merged revisions.
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if rev_id not in local_rev_set:
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missing_remote.append((rno+revno+1, rev_id))
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print 'Missing %d revisions' % len(missing_remote)
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from bzrlib.diff import compare_trees
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from bzrlib.tree import EmptyTree
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for revno, rev_id in missing_remote:
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rev = br_remote.get_revision(rev_id)
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parent_rev_id = rev.parent_ids[0]
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if last_rev_id == parent_rev_id:
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parent_tree = last_tree
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parent_tree = br_remote.revision_tree(parent_rev_id)
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revision_tree = br_remote.revision_tree(rev_id)
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last_tree = revision_tree
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delta = compare_trees(revision_tree, parent_tree)
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show_one_log(revno, rev, delta, verbose, sys.stdout, 'original')
added
removed
bzrlib/missing.py
