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

[Alexander Belchenko]

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.

[Aaron Bentley-4]

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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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[Alexander Belchenko]

Aaron Bentley пишет:

> -----BEGIN PGP SIGNED MESSAGE-----
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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.

Yes, you're right. I don't need a branch object actually, for
obtaining delta between 2 tree states I need tree object corresponding
to that revision. Thank you for your example! Having revision id I
should be able to obtain revision tree to compare with.

>
> 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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>

[Alexander Belchenko]

Aaron Bentley пишет:

> -----BEGIN PGP SIGNED MESSAGE-----
> Hash: SHA1
>
> 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.

I have lock error which I'm not quite understand. here is the code:

def get_parent_submission_info(branch):
     """Get the status information for a branch versus its submit branch.

     :return: delta
     """
     new_tree = branch.basis_tree()
     old_branch = mod_branch.Branch.open_containing(branch.get_parent())[0]
     # 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_branch.last_revision())
     old_tree = old_branch.repository.revision_tree(lca)
     return changes_between_trees(new_tree, old_tree)[0]

I've got error message instead of expected delta between 2 trees:

_KnitGraphIndex(CombinedGraphIndex(<bzrlib.btree_index.BTreeGraphIndex
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

[Aaron Bentley-4]

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On 12-03-29 12:12 PM, Alexander Belchenko wrote:
> I've got error message instead of expected delta between 2 trees:
>
> _KnitGraphIndex(CombinedGraphIndex(<bzrlib.btree_index.BTreeGraphIndex
>
>
object at 0x02EEEF10>, <bzrlib.btree_index.BTreeGraphIndex object at
> 0x02EEEE30>)) is not locked
>
> What did I do wrong?

You need to take read (or write) locks on both repositories.  One way
to do this is to take locks on both branches.

Aaron

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[Alexander Belchenko]

Aaron Bentley пишет:

> -----BEGIN PGP SIGNED MESSAGE-----
> Hash: SHA1
>
> On 12-03-29 12:12 PM, Alexander Belchenko wrote:
>> I've got error message instead of expected delta between 2 trees:
>>
>> _KnitGraphIndex(CombinedGraphIndex(<bzrlib.btree_index.BTreeGraphIndex
>>
>>
> object at 0x02EEEF10>, <bzrlib.btree_index.BTreeGraphIndex object at
>> 0x02EEEE30>)) is not locked
>>
>> What did I do wrong?
>
> You need to take read (or write) locks on both repositories.  One way
> to do this is to take locks on both branches.

Thank you! That helps.

>
> Aaron
>
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