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/*
* This file is part of the LibreOffice project.
*
* This Source Code Form is subject to the terms of the Mozilla Public
* License, v. 2.0. If a copy of the MPL was not distributed with this
* file, You can obtain one at http://mozilla.org/MPL/2.0/.
*
* This file incorporates work covered by the following license notice:
*
* 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 .
*/
package org.openoffice.xmerge.util;
import org.w3c.dom.Document;
import org.w3c.dom.Element;
import org.w3c.dom.Node;
import org.w3c.dom.Text;
import org.openoffice.xmerge.converter.xml.OfficeConstants;
import java.util.ArrayList;
/**
* Class providing utility methods for OpenOffice plugins.
*
* @version 1.1
*/
public class OfficeUtil implements OfficeConstants {
/**
* <p>Method to replace whitespace character within text with appropriate
* OpenOffice tags.</p>
*
* @param text The text to parse for whitespace.
*
* @return <code>Node</code> array containing OpenOffice XML nodes
* representing the text.
*/
public static Node[] parseText(String text, Document parentDoc) {
ArrayList<Node> nodeVec = new ArrayList<Node>();
/*
* Break up the text from the text run into Open
* Office text runs. There may be more runs in OO because
* runs of 2 or more spaces map to nodes.
*/
while ((text.indexOf(" ") != -1) || (text.indexOf("\t") != 1)) {
/*
* Find the indices of tabs and multiple spaces, and
* figure out which of them occurs first in the string.
*/
int spaceIndex = text.indexOf(" ");
int tabIndex = text.indexOf("\t");
if ((spaceIndex == -1) && (tabIndex == -1))
break; // DJP This should not be necessary. What is wrong
// with the while() stmt up above?
int closerIndex; // Index of the first of these
if (spaceIndex == -1)
closerIndex = tabIndex;
else if (tabIndex == -1)
closerIndex = spaceIndex;
else
closerIndex = (spaceIndex > tabIndex) ? tabIndex : spaceIndex;
/*
* If there is any text prior to the first occurrence of a
* tab or spaces, create a text node from it, then chop it
* off the string we're working with.
*/
if (closerIndex > 0) {
String beginningText = text.substring(0, closerIndex);
Text textNode = parentDoc.createTextNode(beginningText);
nodeVec.add(textNode);
}
text = text.substring(closerIndex);
/*
* Handle either tab character or space sequence by creating
* an element for it, and then chopping out the text that
* represented it in "text".
*/
if (closerIndex == tabIndex) {
Element tabNode = parentDoc.createElement(TAG_TAB_STOP);
nodeVec.add(tabNode);
text = text.substring(1); // tab is always a single character
} else {
// Compute length of space sequence.
int nrSpaces = 2;
while ((nrSpaces < text.length())
&& text.substring(nrSpaces, nrSpaces + 1).equals(" "))
nrSpaces++;
Element spaceNode = parentDoc.createElement(TAG_SPACE);
spaceNode.setAttribute(ATTRIBUTE_SPACE_COUNT,
new Integer(nrSpaces).toString());
nodeVec.add(spaceNode);
text = text.substring(nrSpaces);
}
}
/*
* No more tabs or space sequences. If there's any remaining
* text create a text node for it.
*/
if (text.length() > 0) {
Text textNode = parentDoc.createTextNode(text);
nodeVec.add(textNode);
}
// Now create and populate an array to return the nodes in.
Node nodes[] = new Node[nodeVec.size()];
for (int i = 0; i < nodeVec.size(); i++)
nodes[i] = nodeVec.get(i);
return nodes;
}
}
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