Is there simple way how to replace (ordered) existing node in JCR 2.0?

Question

Is there some simple way how to replace existing node with another node in JCR 2.0?

Due to the ordering of nodes, currently I am doing these steps:

step 1: Find sibling node which is right after existing node i want to replace:

        if (preserveOrdering) {
        NodeIterator iter = parent.getNodes();
        boolean found = false;
        while (iter.hasNext()) {
            if (tempNode.equals(iter.nextNode())) {
                found = true;
                if (iter.hasNext()) {
                    tempNodeSibling = iter.nextNode();
                    break;
                }
            }
        }
        assert found;
    }

step 2: delete existing node:

tempNode.remove();

step 3: Create new node (I am doing clone, but probably node.addNode() method can be used, new node is appended to the end of the child node list):

workspace.clone(workspace.getName(), existingNodePath, tempNodePath, false);

step 4: Move new node before his old tempNode sibling (remebered in the first step)

parent.orderBefore(tempNodeName, tempNodeSibling.getName());

All these steps looks to me quite cumbersome. But I cannot find in JCR API better way. Mainly because there is only one method orderBefore() working with ordering.

Do you think there is some different/more simple approach for solving this problem?

Randall Hauch Randall Hauch · Accepted Answer · 2014-01-10T14:20:05

Unfortunately, with JCR 2.0 new nodes are always added at the end and using javax.jcr.Node.reorder(...) is the only way to change the position of a child node within the parent's list of children. It is inconvenient to say the least, but I suspect adding such methods would have added too much complexity to an already complicated API.

Is there simple way how to replace (ordered) existing node in JCR 2.0?

1 Answers