Xpath - Java - 从XML中提取多个名称空间(Xpath - Java - Extracting multiple namespaces from XML)

 我正在研究用Java编写的解析器。 我可以从各个位置接收带有各种内容的XML提要。 我需要从feed中提取所有名称空间，根据feed调用this或that。 我在使用Java获取此功能时遇到了一些麻烦，我不确定问题出在哪里。 
 
 让我们考虑一下这个XML： 
 
<?xml version="1.0"?>
        <?xml-stylesheet type='text/xsl' href='new.xsl'?>
<test xmlns:mynsone="http://www.ns.com/test" xmlns:demons="http://www.demons.com/test">
    <p xmlns:domain="http://www.toto.com/test">
        this is a test.
    </p>
</test>
 
 为了测试我的xPath表达式（我相当新），我写了一个应用于该XML的.xsl脚本： 
 
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">
    <xsl:output
            method="html"
            encoding="ISO-8859-1"
            doctype-public="-//W3C//DTD XHTML//EN"
            doctype-system="http://www.w3.org/TR/2001/REC-xhtml11-20010531"
            indent="yes" />
    <xsl:template match="/">
        <xsl:for-each select="//namespace::*">
            <xsl:value-of select="." />
            <xsl:text> </xsl:text><br />
        </xsl:for-each>
    </xsl:template>
</xsl:stylesheet>
 
 这正确地为我提供了迭代节点时遇到的命名空间列表： 
 
http://www.w3.org/XML/1998/namespace 
http://www.demons.com/test 
http://www.ns.com/test 
http://www.w3.org/XML/1998/namespace 
http://www.demons.com/test 
http://www.ns.com/test 
http://www.toto.com/test 
 
 现在我回到Java：这是我使用的代码。 
 
    InputStream file = url.openStream();
    DocumentBuilderFactory builderFactory = DocumentBuilderFactory.newInstance();
    DocumentBuilder builder =  builderFactory.newDocumentBuilder();
    org.w3c.dom.Document xmlDocument = builder.parse(file);

    XPath xPath = XPathFactory.newInstance().newXPath();
    String expression = "//namespace::*";
    System.out.println(expression);

    NodeList nodelist = (NodeList) xPath.compile(expression).evaluate(xmlDocument, XPathConstants.NODESET);

    for (int k = 0; k < nodelist.getLength(); k++)
    {
        Node mynode = nodelist.item(k);
        System.out.println(mynode.toString());
    }  
 
 这是我获得的结果： 
 
xmlns:mynsone="http://www.ns.com/test"
org.apache.xml.dtm.ref.dom2dtm.DOM2DTMdefaultNamespaceDeclarationNode@7dbb8ca4
xmlns:domain="http://www.toto.com/test"
 
 因此，不会返回“demons”命名空间。 问题是，如果我在一个节点上放置几个名称空间，则在Java中只返回1，而在XSL脚本上则显示所有名称空间。 
 
 我希望我能清楚自己。 我花了几天时间在网页上浏览一些例子，我不知道我是否真的很接近，但只是遗漏了一些东西，或者我的表情根本不合适...... 
 
 提前致谢。 
 
 好的，所以我最终使用xPath 2.0来完成它，使用saxon-HE 9.4： 
 
public static boolean detectGeoRssNamespace(InputStream sourceFeed) {
    try {
        if (sourceFeed.markSupported()) {
            sourceFeed.reset();
        }

        String objectModel = NamespaceConstant.OBJECT_MODEL_SAXON;
        System.setProperty("javax.xml.xpath.XPathFactory:"+NamespaceConstant.OBJECT_MODEL_SAXON, "net.sf.saxon.xpath.XPathFactoryImpl");
        XPathFactory xpathFactory = XPathFactory.newInstance(objectModel);
        XPath xpath = xpathFactory.newXPath();

        InputSource is = new InputSource(sourceFeed);
        SAXSource ss = new SAXSource(is);
        NodeInfo doc = ((XPathEvaluator)xpath).setSource(ss);       

        String xpathExpressionStr = "distinct-values(//*[name()!=local-name()]/ concat('prefix=', substring-before(name(), ':'), '&uri=', namespace-uri()))";
        XPathExpression xpathExpression = xpath.compile(xpathExpressionStr);

        List nodelist = (List)xpathExpression.evaluate(doc, XPathConstants.NODESET);

         System.out.println("<output>");
         Iterator iter = nodelist.iterator();
         while ( iter.hasNext() ) {
             Object line = (Object)iter.next();
             System.out.println(line.toString());
         }
         System.out.println("</output>");

    } catch (XPathFactoryConfigurationException e) {
        e.printStackTrace();
    } catch (XPathException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    } catch (Exception e) {
        e.printStackTrace();                

    }  

I am working on a parser written in Java. I can receive XML feeds from various locations, with various contents. I need to extract all the namespaces from the feed, to call this or that according to the feed. I have some trouble obtaining this in Java, and i am not really sure where the issue is.
 
Let's consider this XML:
 
<?xml version="1.0"?>
        <?xml-stylesheet type='text/xsl' href='new.xsl'?>
<test xmlns:mynsone="http://www.ns.com/test" xmlns:demons="http://www.demons.com/test">
    <p xmlns:domain="http://www.toto.com/test">
        this is a test.
    </p>
</test>
 
In order to test my xPath expression (i am rather new to it), i wrote a little .xsl script applied to that XML:
 
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">
    <xsl:output
            method="html"
            encoding="ISO-8859-1"
            doctype-public="-//W3C//DTD XHTML//EN"
            doctype-system="http://www.w3.org/TR/2001/REC-xhtml11-20010531"
            indent="yes" />
    <xsl:template match="/">
        <xsl:for-each select="//namespace::*">
            <xsl:value-of select="." />
            <xsl:text> </xsl:text><br />
        </xsl:for-each>
    </xsl:template>
</xsl:stylesheet>
 
And this correctly provides me the list of namespaces encountered iterating the nodes:
 
http://www.w3.org/XML/1998/namespace 
http://www.demons.com/test 
http://www.ns.com/test 
http://www.w3.org/XML/1998/namespace 
http://www.demons.com/test 
http://www.ns.com/test 
http://www.toto.com/test 
 
Now i get back to Java: here is the code i use.
 
    InputStream file = url.openStream();
    DocumentBuilderFactory builderFactory = DocumentBuilderFactory.newInstance();
    DocumentBuilder builder =  builderFactory.newDocumentBuilder();
    org.w3c.dom.Document xmlDocument = builder.parse(file);

    XPath xPath = XPathFactory.newInstance().newXPath();
    String expression = "//namespace::*";
    System.out.println(expression);

    NodeList nodelist = (NodeList) xPath.compile(expression).evaluate(xmlDocument, XPathConstants.NODESET);

    for (int k = 0; k < nodelist.getLength(); k++)
    {
        Node mynode = nodelist.item(k);
        System.out.println(mynode.toString());
    }  
 
And here is the result i obtain:
 
xmlns:mynsone="http://www.ns.com/test"
org.apache.xml.dtm.ref.dom2dtm.DOM2DTMdefaultNamespaceDeclarationNode@7dbb8ca4
xmlns:domain="http://www.toto.com/test"
 
Therefore, the "demons" namespace is not returned. The problem is that if i put several namespaces on 1 node, only 1 is return in Java, whereas on the XSL script all are displayed. 
 
I hope i maed myself clear; i spent the past days on the web browsing for examples, and i dont know if im really close but just missing a little something or if my expression is simply not proper..
 
Thanks in advance.
 
OK so i eventually used xPath 2.0 to do it, using saxon-HE 9.4:
 
public static boolean detectGeoRssNamespace(InputStream sourceFeed) {
    try {
        if (sourceFeed.markSupported()) {
            sourceFeed.reset();
        }

        String objectModel = NamespaceConstant.OBJECT_MODEL_SAXON;
        System.setProperty("javax.xml.xpath.XPathFactory:"+NamespaceConstant.OBJECT_MODEL_SAXON, "net.sf.saxon.xpath.XPathFactoryImpl");
        XPathFactory xpathFactory = XPathFactory.newInstance(objectModel);
        XPath xpath = xpathFactory.newXPath();

        InputSource is = new InputSource(sourceFeed);
        SAXSource ss = new SAXSource(is);
        NodeInfo doc = ((XPathEvaluator)xpath).setSource(ss);       

        String xpathExpressionStr = "distinct-values(//*[name()!=local-name()]/ concat('prefix=', substring-before(name(), ':'), '&uri=', namespace-uri()))";
        XPathExpression xpathExpression = xpath.compile(xpathExpressionStr);

        List nodelist = (List)xpathExpression.evaluate(doc, XPathConstants.NODESET);

         System.out.println("<output>");
         Iterator iter = nodelist.iterator();
         while ( iter.hasNext() ) {
             Object line = (Object)iter.next();
             System.out.println(line.toString());
         }
         System.out.println("</output>");

    } catch (XPathFactoryConfigurationException e) {
        e.printStackTrace();
    } catch (XPathException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    } catch (Exception e) {
        e.printStackTrace();                

    }  

原文：https://stackoverflow.com/questions/21157255