如何从 xsd 生成 xpath？

Question

如何从 xsd 生成 xpath？ XSD 验证 xml。 I am working in a project where I am generating a sample XML from the xsd using java and then generating xpath from that XML. 如果有任何方法可以直接从 xsd 生成 xpath，请告诉我。

Answer 1

这可能有用：

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);
    }
}

Answer 2

此类工具存在许多问题：

很少生成的 XPath 表达式是一个很好的表达式。 除了 position 信息之外，没有这样的工具会产生有意义的谓词。

没有工具（据我所知）会生成一个 XPath 表达式，该表达式恰好选择一组选定的节点。

除此之外，没有学习XPath就使用这种工具确实有害-它们支持无知。

我建议使用书籍和其他资源，例如以下资源，认真学习 XPath。

https://stackoverflow.com/questions/339930/any-good-xslt-tutorial-book-blog-site-online/341589#341589

有关更多信息，请参阅以下答案。

是否有 xPath 选择器的在线测试仪？

Answer 3

我一直在开发一个小型库来做到这一点，尽管对于更大和更复杂的模式，您需要根据具体情况解决一些问题（例如，某些节点的过滤器）。 有关解决方案的说明，请参阅https://stackoverflow.com/a/45020739/3096687 。