Parse XML Simple String using Java XPath

Question

I have XML String like this

<resp><status>good</status><msg>hi</msg></resp>

I follow this help

Simplest way to query XML in Java

MyCode:

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);


}

I want to get msg node value but i got exception

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

I am not going to use some external library for this simple task. please guide me how to get other node's values. Thanks

Don't expect to get much help by sharing just one line of code. — Ravi K Thapliyal, May 23 '13 at 11:21

score 56 · Accepted Answer · edited Jul 05 '23 at 15:27

You can't reuse the same InputSource for multiple evaluate() invocations because it's automatically closed. Hence you're getting the Stream closed IO exception. Try this

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

String msg = xpath.evaluate("/resp/msg", source1);
String status = xpath.evaluate("/resp/status", source2);

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

EDIT:
A better approach would be to use a DocumentBuilderFactory to parse your XML and build a Document first (using JAXP's DOM APIs) which can then be reused across several XPath evaluations.

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);

so if i have many nodes then i have to create source for each of them? is there any better solution? — Waqas Ali, May 23 '13 at 11:37
Use a `DocumentBuilderFactory` to build a `Document` (part of JAXP DOM APIs) that can be reused across multiple Xpath evaluations. — Ravi K Thapliyal, May 23 '13 at 11:47

score 14 · Answer 2 · edited May 23 '17 at 10:31

14

Ravi's solution can also be expressed as:

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);

edited May 23 '17 at 10:31

Community

1
1

answered May 23 '13 at 12:39

McDowell

107,573
31
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2

out of interest, if the XML string was good, ok is there anyway of just selecting the first word in the string before the comma? or just selecting the word after the comma? – Paul Alexander Mar 31 '14 at 11:26
@brookman Not with Java's XPath support. You might be able to with XPath 2 - see the Saxon API. – McDowell Apr 01 '14 at 11:42

yegor256 · Answer 3 · 2015-02-16T16:38:03.433

8

You can try jcabi-xml, which does DOM manipulations behind the scene:

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);

edited Feb 16 '15 at 16:38

answered Oct 20 '13 at 18:51

yegor256

102,010
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446
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This library is very nice, but what I don't like is that it brings aspectj with it - which I don't think is really necessary – Alexey Grigorev Mar 19 '15 at 10:06
Yeah, we're planning to get rid of that very soon – yegor256 Mar 19 '15 at 19:32
that would be very nice! any RSS I can subscribe for to keep myself up-to-date? – Alexey Grigorev Mar 19 '15 at 21:32
1

@AlexeyGrigorev follow/watch this ticket: https://github.com/jcabi/jcabi-xml/issues/61 – yegor256 Mar 20 '15 at 19:45

Parse XML Simple String using Java XPath

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