ZODB.FileStorage -- copy? save as?

Mitch Chapman Mitch.Chapman at bioreason.com
Wed Feb 6 14:50:10 EST 2002

Jeremy Hylton wrote:
> Mitch Chapman <Mitch.Chapman at bioreason.com> wrote in message news:<3C606CCE.ACA1472D at bioreason.com>...
> > I'd like to be able to do something like "save as" with
> > ZODB.FileStorage.FileStorage or a derivative thereof.  I'm having a
> > hard time even identifying alternatives, mostly because I don't yet
> > understand the FileStorage implementation.
> I hope you don't need to understand the implementation in order to use
> it.  I'll grant that there's a lot of folklore about how to manage a
> FileStorage that isn't documented anywhere.

That was my hope, too.  The problem is, I'm dealing with very large 
object graphs, and am trying to find a way to implement "Save As..." 
without tying up the application for several minutes at a time.  
So I was looking for a way to do a (fast) file copy rather than 
a (slow) copy-and-save of a large Python object graph.

> [...] It's more 
> likely that you'd want to copy objects at the ZODB level; load an
> object graph out of one storage and create copies of all the objects
> in a new storage.

Yes, that's basically it:  I want to load an object graph out of 
one storage, make changes to the graph, commit the changes to
a new storage, and henceforth use the new storage.  Ideally I'd
like to be able to do this *without* having to copy all of the
objects in the graph or -- worse yet -- having to create new copies
of an associated host of "wrapper" objects.

(Never mind what those wrapper objects suggest about the
application architecture :)

Just to be sure I understand, are you suggesting something like
oldGraph = oldConnection.root()["graph"]
newGraph = Graph()
newConnection.root()["graph"] = newGraph

# Create copies of all objects from oldGraph, in newGraph:

# Don't commit any pending changes to the old store.

# *Do* commit all changes to the new store.

> Hope this was helpful.

Yes, it was.  Obviously, I still have questions (feel free to 
answer :), but you've already helped a lot.  Thanks.


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