How to open branch object equivalent of -r ancestor::parent

[Aaron Bentley]

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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 =
> mod_branch.Branch.open_containing(branch.get_parent())[0]
> 
> 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(),
                            old_branch.last_revision())

That should give you the revision you need to diff against.

If you really need to specify a branch, not a revision, then I guess
you can make a memory branch, stacked on either of the other branches,
and call Branch.set_last_revision_info(1, lca) on it.  But I recommend
avoiding this if at all possible.

Aaron
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