How to open branch object equivalent of -r ancestor::parent
bialix at ukr.net
Thu Mar 29 16:12:34 UTC 2012
Aaron Bentley пишет:
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> On 12-03-16 11:41 AM, Alexander Belchenko wrote:
>> To fix the bug https://bugs.launchpad.net/bzr-explorer/+bug/956268
>> I need to open ancestor related to parent branch instead of tip of
>> parent branch itself.
>> Current code does today:
>> old_branch =
>> Which is equivalent of -r branch::parent
>> I need this to be changed to be equivalent -r ancestor::parent
>> Any suggestions or code snippets will help.
> What you ask is technically possible, but the difference between -r
> branch::parent and -r ancestor::parent is the *revision* they select,
> not the branch they open.
> We'll need the revision anyhow, so let's start there. To get the
> revision used by ancestor::parent, you'll need to
> # Ensure you have both branches open
> old_branch = mod_branch.Branch.open(branch.get_parent())
> # Get a graph, so you can find the common ancestor (LCA)
> graph = old_branch.repository.get_graph(branch.repository)
> # Find the LCA of the revisions
> lca = graph.find_unique_lca(branch.last_revision(),
> That should give you the revision you need to diff against.
I have lock error which I'm not quite understand. here is the code:
"""Get the status information for a branch versus its submit branch.
new_tree = branch.basis_tree()
old_branch = mod_branch.Branch.open_containing(branch.get_parent())
# Thanks to Aaron for code snippet
# Get a graph, so we can find the common ancestor (LCA)
graph = old_branch.repository.get_graph(branch.repository)
# Find the LCA of the revisions
lca = graph.find_unique_lca(branch.last_revision(),
old_tree = old_branch.repository.revision_tree(lca)
return changes_between_trees(new_tree, old_tree)
I've got error message instead of expected delta between 2 trees:
object at 0x02EEEF10>, <bzrlib.btree_index.BTreeGraphIndex object at
0x02EEEE30>)) is not locked
What did I do wrong?
All the dude wanted was his rug back
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