从xml获取具有特定属性值java的所有节点

Question

<main-project name="" version="1.0.2">
<data name="data">
<tag>
<link-to  object="processor"/>
</tag>
</data>
<output name="output">
<tag>
<link-to  object="processor"/>
</tag>
</output>
<processor name ="processor">
<tag>
<link-to  object="data"/>
</tag>
</processor>
</main-project>

我想让所有具有属性对象=处理器的节点，我尝试使用

 XPath xPath =  XPathFactory.newInstance().newXPath();
NodeList linkageNodesEpf = (NodeList) xPath.compile("//link-to[@object = 'processor']").evaluate(Doc, XPathConstants.NODESET);

该查询给出了一个空列表，但是当我替换上面的查询的link-to link可以正常工作并给出正确的结果时，所以我认为可能是dash（-）造成了问题。

请帮我解决这个问题

Answer 1

好吧，该代码：

import org.w3c.dom.Document;
import org.w3c.dom.NodeList;
import org.xml.sax.SAXException;

import javax.xml.parsers.DocumentBuilder;
import javax.xml.parsers.DocumentBuilderFactory;
import javax.xml.parsers.ParserConfigurationException;
import javax.xml.xpath.XPath;
import javax.xml.xpath.XPathConstants;
import javax.xml.xpath.XPathExpressionException;
import javax.xml.xpath.XPathFactory;
import java.io.ByteArrayInputStream;
import java.io.IOException;

public class SimpleClass2 {

    public static void main(String[] args) throws InterruptedException, XPathExpressionException, ParserConfigurationException, IOException, SAXException {
        String str = "<main-project name=\"\" version=\"1.0.2\">\n" +
                "<data name=\"data\">\n" +
                "<tag>\n" +
                "<link-to  object=\"processor\"/>\n" +
                "</tag>\n" +
                "</data>\n" +
                "<output name=\"output\">\n" +
                "<tag>\n" +
                "<link-to  object=\"processor\"/>\n" +
                "</tag>\n" +
                "</output>\n" +
                "<processor name =\"processor\">\n" +
                "<tag>\n" +
                "<link-to  object=\"data\"/>\n" +
                "</tag>\n" +
                "</processor>\n" +
                "</main-project>";
        DocumentBuilder builder = DocumentBuilderFactory.newInstance().newDocumentBuilder();
        Document xmlDoc = builder.parse(new ByteArrayInputStream(str.getBytes()));
        XPath xPath =  XPathFactory.newInstance().newXPath();

        NodeList linkageNodesEpf = (NodeList) xPath.compile("//link-to[@object = 'processor']").evaluate(xmlDoc, XPathConstants.NODESET);
        for (int i = 0; i < linkageNodesEpf.getLength(); i++) {
            System.out.println(linkageNodesEpf.item(i));
        }
    }

}

在我的机器上产生sush结果（oracle jdk8_45）：

[link-to: null]
[link-to: null]

您可以尝试复制它吗，它会产生什么结果？