Finds and returns the index of child XML node in the given parent's children array - Java XML

Java examples for XML:XML Node Child

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Description

Finds and returns the index of child XML node in the given parent's children array

Demo Code

/*/* w  ww.jav a 2s  .co m*/

   Licensed to the Apache Software Foundation (ASF) under one or more
   contributor license agreements.  See the NOTICE file distributed with
   this work for additional information regarding copyright ownership.
   The ASF licenses this file to You under the Apache License, Version 2.0
   (the "License"); you may not use this file except in compliance with
   the License.  You may obtain a copy of the License at

       http://www.apache.org/licenses/LICENSE-2.0

   Unless required by applicable law or agreed to in writing, software
   distributed under the License is distributed on an "AS IS" BASIS,
   WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
   See the License for the specific language governing permissions and
   limitations under the License.

 */
//package com.java2s;

import org.w3c.dom.Node;
import org.w3c.dom.NodeList;

public class Main {
    /**
     * Finds and returns the index of child node in the given parent's children
     * array
     *
     * @param child
     *            The child node
     * @param parent
     *            The parent node
     * @return the index
     */
    public static int getChildIndex(Node child, Node parent) {
        if (child == null || child.getParentNode() != parent
                || child.getParentNode() == null) {
            return -1;
        }
        return getChildIndex(child);
    }

    /**
     * Finds and returns the index of child node in its parent's children array
     *
     * @param child
     *            The child node
     * @return the index in children array
     */
    public static int getChildIndex(Node child) {
        NodeList children = child.getParentNode().getChildNodes();
        for (int i = 0; i < children.getLength(); i++) {
            Node currentChild = children.item(i);
            if (currentChild == child) {
                return i;
            }
        }
        return -1;
    }
}

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