如何从另一个更改层次结构的XML序列化android中的xml

Question

I've got this source xml:

<source>
 <category id="1" />  
 <item1 />
 <item2 />
 <category id="2"/>
 <item1 />
 <item2 />
</source>

As you can see all items have the same hierarchy. And I need to "translate"/serialize it to another XML like this:

 <source>
   <category id="1">
     <item1  />
     <item2  />
   </category>
   <category id="2">
      <item1  />
      <item2  />
    </category>
 </source>

Where "items" are children of "category".

I'm using XmlPullParser and XmlSerializer from Android tools but I don't mind to use another if they are compatible with Android environment

Tx

Answer 1

I've found another way using XSLT: This way we use XML-only specific tools for transform without handling any data with objects.

create a transform.xsl file to handle transformation:

<?xml version="1.0" encoding="UTF-8" ?>
<xsl:stylesheet version="1.0"
    xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:output method="xml" indent="yes" encoding="UTF-8" />
    <xsl:strip-space elements="*" />
    <xsl:template match="/">
        <xsl:apply-templates />
    </xsl:template>
    <xsl:template match="source">
        <source>
            <xsl:apply-templates select="category" />
        </source>
    </xsl:template>
    <xsl:template match="category">
        <xsl:variable name="place" select="count(preceding-sibling::category)" />
        <category>
            <xsl:attribute name="id">
    <xsl:value-of select="@id" />
  </xsl:attribute>
            <xsl:apply-templates select="following-sibling::*[not(self::category)]">
                <xsl:with-param name="slot" select="$place" />
            </xsl:apply-templates>
        </category>
    </xsl:template>
    <xsl:template match="item1">
        <xsl:param name="slot" />
        <xsl:choose>
            <xsl:when test="count(preceding-sibling::category) = $slot + 1">
                <xsl:copy-of select="." />
            </xsl:when>
            <xsl:otherwise />
        </xsl:choose>
    </xsl:template>
    <xsl:template match="item2">
        <xsl:param name="slot" />
        <xsl:choose>
            <xsl:when test="count(preceding-sibling::category) = $slot + 1">
                <xsl:copy-of select="." />
            </xsl:when>
            <xsl:otherwise />
        </xsl:choose>
    </xsl:template>
</xsl:stylesheet>

And then write the code to handle transformation to a file with desired output data.xml

         AssetManager am = getAssets();
         xml = am.open("source.xml");
         xsl = am.open("transform.xsl");

        Source xmlSource = new StreamSource(xml);
        Source xsltSource = new StreamSource(xsl);

        TransformerFactory transFact = TransformerFactory.newInstance();
        Transformer trans = transFact.newTransformer(xsltSource);


        File f = new File(Environment.getExternalStorageDirectory().getAbsolutePath()+"/data.xml");
        StreamResult result = new StreamResult(f);
        trans.transform(xmlSource, result);

And its done. more info here: http://www.dpawson.co.uk/xsl/sect2/flatfile.html

Answer 2

For an easy xml format this shouldn't be that hard.

A possible way to read the first xml file with a sax parser would be:

public class MySaxHandler extends DefaultHandler {
  private List<Category> items = new LinkedList<Category>();
  private Category currentCategory;
  public void startElement(String uri, String localName, String qName, Attributes attributes) {
    if (localName.equals("category")) {
      currentCategory = new Category(attributes.getValue("id"));
      items.add(currentCategory);
    }
    if (localName.equals("item1") {
      currentCategory.setItem1(new Item1(...));
    }
    if (localName.equals("item2") {
      currentCategory.setItem2(new Item2(...));
    }
  }
}

For each <category> tag you create a new Category object. The following items will be added to the last category object. While reading the contents you create the hierarchy you need later (items are added to the appropriate category). It should be easy to transform this code to use XmlPullParser instead of sax parser. I just used sax because I am more familiar with it.

When you finished reading the first file you need to write your hierarchy to a new file.

You can do this in a way like this:

StringBuilder b = new StringBuilder();
for (int i = 0; i < categories.size(); i++) {
  b.append(categories.get(i).getXml());
}
// write content of b into file

getXml() for each category could look like:

public String getXml() {
  StringBuilder b = new StringBuilder();
  b.append("<category id=\"" + this.id + "\">");
  for (int i = 0; i < items.size(); i++) {
    b.append(items.get(i).getXml());
  }
  b.append("</category>");
  return b.toString();
}

Each item creates its own xml in its getXml() method which could be

public String getXml() {
  return "<item1 />";
}

in the easiest case.

Please note that building xml by hand is only suitable if you xml structure stays that simple. If the structure is becoming more complicated you should make use of some lightweight xml libraries that work on Android (like xstream ).