在java中使用regexp来修改xml

Question

I'm trying to change an xml by using regular expressions in java, but I can't find the right way. I have an xml like this (simplified):

<ROOT>
   <NODE ord="1" />
   <NODE ord="3,2" />
</ROOT>

The xml actually shows a sentence with its nodes, chunks ... in two languages and has more attributes. Each sentence it's loaded in two RichTextAreas (one for the source sentence, and the other for the translated one).

What I need to do is add a style attribute to every node that has an specific value in its ord attribute (this style attribute will show correspondences between two languages, like Google Translate does when you mouse over a word). I know this could be done using DOM (getting all the NODE nodes and then seeing the ord attribute one by one), but I am looking for the fastest way to do the change as it is going to execute in the client side of my GWT app.

When that ord attribute has a single value (like in the first node) it is easy to do just taking the xml as a string and using the replaceAll() function . The problem is when the attribute has composed values (like in the second node).

For example, how could I do to add that attribute if the value I'm looking for is 2? I believe this could be done using regular expressions, but I can't find out how. Any hint or help would be appreciated (even if it doesn't use regexp and replaceAll function).

Thanks in advance.

Answer 1

String resultString = subjectString.replaceAll("<NODE ord=\"([^\"]*\\b2\\b[^\"]*)\" />", "<NODE ord=\"$1\" style=\"whatever\"/>");

will find any <NODE> tag that has a single ord attribute with a value of "2" (or "1,2" or "2,3" or "1,2,3" but not "12") and adds a style attribute.

This is quick and dirty, and rightfully advised against by many here, but for a one-off quick job it should be OK.

Explanation:

<NODE ord="  # Match <NODE ord:" verbatim
(            # Match and capture...
 [^"]*       #  any number of characters except "
 \b2\b       #  "2" as a whole word (surrounded by non-alphanumerics)
 [^"]*       #  any number of characters except "
)            # End of capturing group
" />         # Match " /> verbatim

Answer 2

XPath can do this for you. You could select:

/ROOT/NODE[contains(concat(',', @ord, ','), ',2,')]

Since you intend to use GWT on the client, you could give gwtxslt a try. With it you could specify an XSLT stylesheet to do the transformation (ie adding the attribute) for you:

XsltProcessor processor = new XsltProcessor();
processor.importStyleSheet(styleSheetText);
processor.importSource(sourceText);
processor.setParameter("ord", "2");
processor.setParameter("style", "whatever");
String resultString = processor.transform();
// do something with resultString

where styleSheetText could be an XSLT document along the lines of

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <xsl:param name="ord"   select="''" />
  <xsl:param name="style" select="''" />

  <xsl:template match="node()|@*">
    <xsl:copy>
      <xsl:apply-templates select="node()|@*" />
    </xsl:copy>
  </xsl:template>

  <xsl:template match="NODE">
    <xsl:copy>
      <xsl:apply-templates select="@*" />
      <xsl:if test="contains(concat(',', @ord, ','), concat(',', $ord, ','))">
        <xsl:attribute name="style">
          <xsl:value-of select="$style" />
        </xsl:attribute>
      </xsl:if>
      <xsl:apply-templates select="node()" />
    </xsl:copy>
  </xsl:template>
</xsl:stylesheet>

Note that I use concat() to prevent partial matches in the comma-separated list that the attribute value of @ord actually is.

Answer 3

I'm trying to change an xml by using regular expressions in java, but I can't find the right way.

That's because there isn't a right way. Regular expressions are not the right way to manipulate XML. That's because XML is not a regular grammar (which is a technical term in computer science, not a generalized insult.)

Answer 4

It might sound like overkill, but I'd consider using the standard DOM parsers to read the fragment, modify it using setAttribute() calls, and then write it out again. I know you said that efficiency is important, but how long does this really take? Testing shows 60ms on my ageing 2GHz pentium.

This approach will be more robust against comments, things split across lines etc. It is also much more likely to give you well-formed XML. Also things like your requirement of only doing it if certain values are present will become trivial.

public class AddStyleExample {

    public static void main(final String[] args) {
        String input = "<ROOT> <NODE ord=\"1\" /> <NODE ord=\"3,2\" /> </ROOT>";
        try {
            final DocumentBuilderFactory factory = DocumentBuilderFactory
                    .newInstance();
            factory.setValidating(false);
            factory.setNamespaceAware(false);
            DocumentBuilder builder;

            builder = factory.newDocumentBuilder();

            final Document doc = builder.parse(new InputSource(
                    new StringReader(input)));

            NodeList tags = doc.getElementsByTagName("NODE");
            for (int i = 0; i < tags.getLength(); i++) {
                Element node = (Element) tags.item(i);
                node.setAttribute("style", "example value");
            }
            StringWriter writer = new StringWriter();
            final StreamResult result = new StreamResult(writer);
            final Transformer t = TransformerFactory.newInstance()
                    .newTransformer();
            t.setOutputProperty(OutputKeys.INDENT, "yes");
            t.setOutputProperty(OutputKeys.OMIT_XML_DECLARATION, "yes");
            t.transform(new DOMSource(doc), result);
            System.out.println(writer.toString());

        } catch (ParserConfigurationException e) {
            e.printStackTrace();
        } catch (TransformerException e) {
            e.printStackTrace();
        } catch (SAXException e) {
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        }

    }
}