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https://issues.apache.org/jira/browse/JCR-3793?page=com.atlassian.jira.plugin.system.issuetabpanels:comment-tabpanel&focusedCommentId=14050740#comment-14050740
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Tobias Bocanegra commented on JCR-3793:
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Of course we can optimize the clients to do only the ordering if needed. but
the algorithm in NodeNameList does "restore" the order by ordering the nodes
from back to front. if the list is already in the correct order, it should be a
no-op. So I don't see that there should be a performance difference if the
client first checks the order or if Oak does it.
btw: the API could benefit of a {{Node.setChildNodeOrder(String[] names)}}
method for such cases.
> vlt: with many child nodes, NodeNameList.restoreOrder is very slow with Oak
> ---------------------------------------------------------------------------
>
> Key: JCR-3793
> URL: https://issues.apache.org/jira/browse/JCR-3793
> Project: Jackrabbit Content Repository
> Issue Type: Improvement
> Reporter: Thomas Mueller
> Attachments: JCR-3793.patch, ReorderTest.java
>
>
> The method org.apache.jackrabbit.vault.fs.api.NodeNameList.restoreOrder
> re-orders orderable child nodes by using Node.orderBefore. This is very slow
> if there are many child nodes, specially with Oak (minutes for 10'000 nodes,
> while only about 1 second for Jackrabbit 2.x).
> [~tripod], I wonder if a possible solution is to first check whether
> re-ordering is needed? For example using:
> {noformat}
> boolean isOrdered(ArrayList<String> names, Node parent)
> throws RepositoryException {
> NodeIterator it1 = parent.getNodes();
> for (Iterator<String> it2 = names.iterator(); it2.hasNext();) {
> if (!it1.hasNext() ||
> !it1.nextNode().getName().equals(it2.next())) {
> return false;
> }
> }
> return !it1.hasNext();
> }
> {noformat}
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