我该如何从XSD生成XPath？XSD用于验证XML。我正在一个项目中工作，在该项目中，我使用Java从XSD生成一个样本XML，然后从该XML生成XPath。如果有任何直接从XSD生成XPath的方法，请告诉我。
3个回答
2
这可能会有用：
这可能对您有所帮助：
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);
}
}
- chaity
1
我一直在开发一个小型库，来实现这个功能，不过对于更大更复杂的模式，你需要根据具体情况解决一些问题（例如某些节点的筛选器）。请参见https://stackoverflow.com/a/45020739/3096687以了解解决方案的描述。
- bbarker
0
这类工具存在一些问题：
生成的XPath表达式很少是好的。除了位置信息，没有任何工具能产生有意义的谓词。
据我所知，没有任何工具能生成一个精确选择一组节点的XPath表达式。
此外，如果不学习XPath而使用这类工具，实际上是有害的——它们支持无知。
我建议通过阅读书籍和其他资源来认真学习XPath，如下所示。
请参考以下答案获取更多信息：- reddyvaribabu
网页内容由stack overflow 提供, 点击上面的可以查看英文原文，
原文链接
原文链接