CODEWITHSUNDEEP
javaxpath
Patryk
Patryk

Reputation: 3152

How to get every descendant of node using XPath in java?

I'm using this code, but I'm not sure what should the XPath expression look like.

XPath xpath = XPathFactory.newInstance().newXPath();
XPathExpression expression = xpath.compile("XXXX");
NodeList nodes = (NodeList) expression.evaluate(doc.getDocumentElement(), XPathConstants.NODESET);

Let's say there's a tree:

<animal>
    <big>
        <elephant>
            <white_elephant>

            </white_elephant>
        </elephant>
    </big>
    <small>
    </small>
</animal>

I would like to get every descendant of node big: (elephant, white_elephant) + big itself

Upvotes: 0

Views: 1087

Answers (1)

stuXnet
stuXnet

Reputation: 4349

You can use the descendant-or-self-axis:

XPath xpath = XPathFactory.newInstance().newXPath();
XPathExpression expression = xpath.compile("//big/descendant-or-self::*");
NodeList nodes = (NodeList) expression.evaluate(doc.getDocumentElement(), XPathConstants.NODESET);

Wikipedia is a good source to get a overview over different axis: http://en.wikipedia.org/wiki/XPath#Axis_specifiers

Edit: As Jens Erat mentions in the comments, you can shorten /descendant-or-self:: to // - it's just an abbreviation:

XPathExpression expression = xpath.compile("//big//*");

Upvotes: 3

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