Hi,
pkrishna wrote:
> I have versions of both C and D. If I run the above query, I may get the
> previous versions of both C and D since both may match the criteria of
> having a frozenUuid < than the node UUID and have the same version.
what's the reason you are using the 'less than' operator for the frozenUuid?
> VersionHistory vh = node.getVersionHistory();
> VersionIterator vi = vh.getAllVersions();
> vi.skip(1); // this is the root
this might be dangerous. the specification does not mandate a specific order of
versions returned by getAllVersions().
> while (vi.hasNext()) {
>
> System.out.println("-----------------------------------------");
> Version v = vi.nextVersion();
> NodeIterator ni = v.getNodes();
> while (ni.hasNext()) {
>
> Node nv = ni.nextNode();
> Property p = nv.getProperty("ecr:version");
> long val = p.getLong();
> System.out.println("Version: " + val);
> if (val == version.longValue()) { // interested in
> matching the version
> return nv;
> }
> }
>
> }
>
> The above code execute the version loop ony once and this version reflects
> the current version. I expected the loop to execute atleast twice since I
> have a version 1 and version 2 and I was looking for version 1 but couldn't
> find it.
>
> Is there a way to find the version I am looking for? am I missing something?
did you double check that you actually created two versions?
regards
marcel
