使用Java XPath解析XML Simple String

Waq*_*Ali 31 java xml xpath

我有像这样的XML字符串

<resp><status>good</status><msg>hi</msg></resp>
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我听从了这个帮助

在Java中查询XML的最简单方法

mycode的:

public static void main(String args[]) {

    String xml = "<resp><status>good</status><msg>hi</msg></resp>";

    XPathFactory xpathFactory = XPathFactory.newInstance();
    XPath xpath = xpathFactory.newXPath();

    InputSource source = new InputSource(new StringReader(xml));

    String status = "";
    String msg = "";
    try {
        status = (String) xpath.evaluate("/resp/status", source,XPathConstants.STRING);
        msg = (String) xpath.evaluate("/resp/msg", source,XPathConstants.STRING);
    } catch (Exception e) {
        e.printStackTrace();
    }

    System.out.println("status=" + status);
    System.out.println("Message=" + msg);


}
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我想得到msg节点值,但我得到了异常

java.io.IOException: Stream closed
at java.io.StringReader.ensureOpen(StringReader.java:39)
at java.io.StringReader.read(StringReader.java:73)
at com.sun.org.apache.xerces.internal.impl.XMLEntityScanner.load(XMLEntityScanner.java:1742)
at com.sun.org.apache.xerces.internal.impl.XMLEntityScanner.arrangeCapacity(XMLEntityScanner.java:1619)
at com.sun.org.apache.xerces.internal.impl.XMLEntityScanner.skipString(XMLEntityScanner.java:1657)
at com.sun.org.apache.xerces.internal.impl.XMLVersionDetector.determineDocVersion(XMLVersionDetector.java:193)
at com.sun.org.apache.xerces.internal.parsers.XML11Configuration.parse(XML11Configuration.java:771)
at com.sun.org.apache.xerces.internal.parsers.XML11Configuration.parse(XML11Configuration.java:737)
at com.sun.org.apache.xerces.internal.parsers.XMLParser.parse(XMLParser.java:107)
at com.sun.org.apache.xerces.internal.parsers.DOMParser.parse(DOMParser.java:225)
at com.sun.org.apache.xerces.internal.jaxp.DocumentBuilderImpl.parse(DocumentBuilderImpl.java:283)
at com.sun.org.apache.xpath.internal.jaxp.XPathImpl.evaluate(XPathImpl.java:468)
at com.sun.org.apache.xpath.internal.jaxp.XPathImpl.evaluate(XPathImpl.java:515)
at Parsing.main(Parsing.java:25)--------------- linked to ------------------
javax.xml.xpath.XPathExpressionException
at com.sun.org.apache.xpath.internal.jaxp.XPathImpl.evaluate(XPathImpl.java:475)
at com.sun.org.apache.xpath.internal.jaxp.XPathImpl.evaluate(XPathImpl.java:515)
at Parsing.main(Parsing.java:25)
Caused by: java.io.IOException: Stream closed
at java.io.StringReader.ensureOpen(StringReader.java:39)
at java.io.StringReader.read(StringReader.java:73)
at     com.sun.org.apache.xerces.internal.impl.XMLEntityScanner.load(XMLEntityScanner.java:1742)
at com.sun.org.apache.xerces.internal.impl.XMLEntityScanner.arrangeCapacity(XMLEntityScanner.java:1619)
at com.sun.org.apache.xerces.internal.impl.XMLEntityScanner.skipString(XMLEntityScanner.java:1657)
at com.sun.org.apache.xerces.internal.impl.XMLVersionDetector.determineDocVersion(XMLVersionDetector.java:193)
at com.sun.org.apache.xerces.internal.parsers.XML11Configuration.parse(XML11Configuration.java:771)
at com.sun.org.apache.xerces.internal.parsers.XML11Configuration.parse(XML11Configuration.java:737)
at com.sun.org.apache.xerces.internal.parsers.XMLParser.parse(XMLParser.java:107)
at com.sun.org.apache.xerces.internal.parsers.DOMParser.parse(DOMParser.java:225)
at com.sun.org.apache.xerces.internal.jaxp.DocumentBuilderImpl.parse(DocumentBuilderImpl.java:283)
at com.sun.org.apache.xpath.internal.jaxp.XPathImpl.evaluate(XPathImpl.java:468)
... 2 more
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我不打算使用一些外部库来完成这个简单的任务.请指导我如何获取其他节点的值.谢谢

Rav*_*yal 47

您不能在InputSource多次evaluate()调用中重复使用它,因为它会自动关闭.因此,您将获得Stream closedIO异常.试试这个

InputSource source1 = new InputSource(new StringReader(xml));
InputSource source2 = new InputSource(new StringReader(xml));

String msg = xpath.evaluate("/resp/msg", source);
String status = xpath.evaluate("/resp/status", source2);

System.out.println("msg=" + msg + ";" + "status=" + status);
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编辑:
更好的方法是使用一个DocumentBuilderFactory解析您的XML,并建立一个Document第一(使用JAXP的DOM API)的,然后可以在几个XPath计算被重用.

String xml = "<resp><status>good</status><msg>hi</msg></resp>";

InputSource source = new InputSource(new StringReader(xml));

DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
DocumentBuilder db = dbf.newDocumentBuilder();
Document document = db.parse(source);

XPathFactory xpathFactory = XPathFactory.newInstance();
XPath xpath = xpathFactory.newXPath();

String msg = xpath.evaluate("/resp/msg", document);
String status = xpath.evaluate("/resp/status", document);

System.out.println("msg=" + msg + ";" + "status=" + status);
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McD*_*ell 12

Ravi的解决方案也可以表达为:

String xml = "<resp><status>good</status><msg>hi</msg></resp>";

XPathFactory xpathFactory = XPathFactory.newInstance();
XPath xpath = xpathFactory.newXPath();

InputSource source = new InputSource(new StringReader(xml));
Document doc = (Document) xpath.evaluate("/", source, XPathConstants.NODE);
String status = xpath.evaluate("/resp/status", doc);
String msg = xpath.evaluate("/resp/msg", doc);

System.out.println("status=" + status);
System.out.println("Message=" + msg);
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  • 出于兴趣,如果XML字符串<status> good,ok </ status>无论如何只是在逗号之前选择字符串中的第一个单词?或者只是选择逗号后面的单词? (2认同)

yeg*_*256 6

您可以尝试jcabi-xml,它在场景后面进行DOM操作:

import com.jcabi.xml.XML;
import com.jcabi.xml.XMLDocument;
XML xml = new XMLDocument("<resp>...</resp>");
String status = xml.xpath("/resp/status/text()").get(0);
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