Simplest way to query XML in Java

Question

I have small Strings with XML, like:

String myxml = \"goodhi\";

<

Answer 1

You could try JXPath

Answer 2

XPath using Java 1.5 and above, without external dependencies:

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

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

Answer 3

Using dom4j, similar to McDowell's solution:

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

Document document = new SAXReader().read(new StringReader(myxml));
String status = document.valueOf("/resp/msg");

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

XML handling is a bit simpler using dom4j. And several other comparable XML libraries exist. Alternatives to dom4j are discussed here.

Answer 4

@The comments of this answer:

You can create a method to make it look simpler

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

System.out.printf("satus= %s\n", getValue("/resp/status", xml ) );

The implementation:

public String getValue( String path, String xml ) { 
    return XPathFactory
               .newInstance()
               .newXPath()
               .evaluate( path , new InputSource(
                                 new StringReader(xml)));

}

Answer 5

After your done with simple ways to query XML in java. Look at XOM.

Answer 6

Here is example of how to do that with XOM:

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

Document document = new Builder().build(myxml, "test.xml");
Nodes nodes = document.query("/resp/status");

System.out.println(nodes.get(0).getValue());

I like XOM more than dom4j for its simplicity and correctness. XOM won't let you create invalid XML even if you want to ;-) (e.g. with illegal characters in character data)