在Java中从XSD创建Xpath

Question

我正在使用Apache XMLSchema框架来解析和获取XSD的元素。 现在我需要将XPath字符串与每个元素相关联。 有人可以就如何做到这一点提出任何想法。 这样做的任何现有算法或框架？

例：

<?xml version="1.0" encoding="ISO-8859-1" ?>
<xs:schema xmlns:xs="http://www.w3.org/2001/XMLSchema">

<xs:simpleType name="stringtype">
  <xs:restriction base="xs:string"/>
</xs:simpleType>

<xs:simpleType name="inttype">
  <xs:restriction base="xs:positiveInteger"/>
</xs:simpleType>

<xs:simpleType name="dectype">
  <xs:restriction base="xs:decimal"/>
</xs:simpleType>

<xs:simpleType name="orderidtype">
  <xs:restriction base="xs:string">
    <xs:pattern value="[0-9]{6}"/>
  </xs:restriction>
</xs:simpleType>

<xs:complexType name="shiptotype">
  <xs:sequence>
    <xs:element name="name" type="stringtype"/>
    <xs:element name="address" type="stringtype"/>
    <xs:element name="city" type="stringtype"/>
    <xs:element name="country" type="stringtype"/>
  </xs:sequence>
</xs:complexType>

<xs:complexType name="itemtype">
  <xs:sequence>
    <xs:element name="title" type="stringtype"/>
    <xs:element name="note" type="stringtype" minOccurs="0"/>
    <xs:element name="quantity" type="inttype"/>
    <xs:element name="price" type="dectype"/>
  </xs:sequence>
</xs:complexType>

<xs:complexType name="shipordertype">
  <xs:sequence>
    <xs:element name="orderperson" type="stringtype"/>
    <xs:element name="shipto" type="shiptotype"/>
    <xs:element name="item" maxOccurs="unbounded" type="itemtype"/>
  </xs:sequence>
  <xs:attribute name="orderid" type="orderidtype" use="required"/>
</xs:complexType>

<xs:element name="shiporder" type="shipordertype"/>

</xs:schema>

XPath for

orderperson-> ./orderperson
name-> ./shipto/name

etc

Answer 1

如果模式允许递归结构，或者它包含通配符，那么这是一个非常棘手的问题，您需要更精确地指定您的需求。 对于相对简单的非递归模式，它应该更直接，但您需要提供额外的信息，例如根元素名称是什么。

Answer 2

我认为这可能会有所帮助：

import java.io.File;
import java.util.HashMap;
import java.util.Map;
import java.util.Stack;
import javax.xml.parsers.*;

import org.xml.sax.*;
import org.xml.sax.helpers.DefaultHandler;

/**
 * SAX handler that creates and prints XPath expressions for each element encountered.
 *
 * The algorithm is not infallible, if elements appear on different levels in the hierarchy.
 * Something like the following is an example:
 * - <elemA/>
 * - <elemA/>
 * - <elemB/>
 * - <elemA/>
 * - <elemC>
 * -     <elemB/>
 * - </elemC>
 *
 * will report
 *
 * //elemA[0]
 * //elemA[1]
 * //elemB[0]
 * //elemA[2]
 * //elemC[0]
 * //elemC[0]/elemB[1]       (this is wrong: should be //elemC[0]/elemB[0] )
 *
 * It also ignores namespaces, and thus treats <foo:elemA> the same as <bar:elemA>.
 */

public class SAXCreateXPath extends DefaultHandler {

    // map of all encountered tags and their running count
    private Map<String, Integer> tagCount;
    // keep track of the succession of elements
    private Stack<String> tags;

    // set to the tag name of the recently closed tag
    String lastClosedTag;

    /**
     * Construct the XPath expression
     */
    private String getCurrentXPath() {
        String str = "//";
        boolean first = true;
        for (String tag : tags) {
            if (first)
                str = str + tag;
            else
                str = str + "/" + tag;
            str += "["+tagCount.get(tag)+"]";
            first = false;
        }
        return str;
    }

    @Override
    public void startDocument() throws SAXException {
        tags = new Stack();
        tagCount = new HashMap<String, Integer>();
    }

    @Override
    public void startElement (String namespaceURI, String localName, String qName, Attributes atts)
        throws SAXException
    {
        boolean isRepeatElement = false;

        if (tagCount.get(localName) == null) {
            tagCount.put(localName, 0);
        } else {
            tagCount.put(localName, 1 + tagCount.get(localName));
        }

        if (lastClosedTag != null) {
            // an element was recently closed ...
            if (lastClosedTag.equals(localName)) {
                // ... and it's the same as the current one
                isRepeatElement = true;
            } else {
                // ... but it's different from the current one, so discard it
                tags.pop();
            }
        }

        // if it's not the same element, add the new element and zero count to list
        if (! isRepeatElement) {
            tags.push(localName);
        }

        System.out.println(getCurrentXPath());
        lastClosedTag = null;
    }

    @Override
    public void endElement (String uri, String localName, String qName) throws SAXException {
        // if two tags are closed in succession (without an intermediate opening tag),
        // then the information about the deeper nested one is discarded
        if (lastClosedTag != null) {
            tags.pop();
        }
        lastClosedTag = localName;
    }

    public static void main (String[] args) throws Exception {
        if (args.length < 1) {
            System.err.println("Usage: SAXCreateXPath <file.xml>");
            System.exit(1);
        }

        // Create a JAXP SAXParserFactory and configure it
        SAXParserFactory spf = SAXParserFactory.newInstance();
        spf.setNamespaceAware(true);
        spf.setValidating(false);

        // Create a JAXP SAXParser
        SAXParser saxParser = spf.newSAXParser();

        // Get the encapsulated SAX XMLReader
        XMLReader xmlReader = saxParser.getXMLReader();

        // Set the ContentHandler of the XMLReader
        xmlReader.setContentHandler(new SAXCreateXPath());

        String filename = args[0];
        String path = new File(filename).getAbsolutePath();
        if (File.separatorChar != '/') {
            path = path.replace(File.separatorChar, '/');
        }
        if (!path.startsWith("/")) {
            path = "/" + path;
        }

        // Tell the XMLReader to parse the XML document
        xmlReader.parse("file:"+path);
    }
}

致谢： http ： //www.coderanch.com/how-to/java/SAXCreateXPath

Answer 3

没有开箱即用的解决方案可满足您的需求; 你必须使用访客模式自己编写。 或者，如果这是为运行时使用而不是在运行时动态生成的设计时工件，则可以使用类似于此SO帖子的解决方案。

Answer 4

阅读某些行之间的某些内容，我试图回答这个问题。 您可以查看JXPath ，它允许您使用XPath表达式遍历Java中的Object图。 我有一种预感，这就是你想要实现的目标。 我可能会弄错:)