This is the result of the fact that the Java xpath classes support version 1 of
XPath, while the "if()" construct is new with XPath version 2.
--
Charles Knell
cknell(_at_)onebox(_dot_)com - email
-----Original Message-----
From: James Neff <jneff(_at_)tethyshealth(_dot_)com>
Sent: Fri, 05 Jan 2007 11:00:49 -0500
To: XSL list <xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com>
Subject: [xsl] xpath if expression in Java
Greetings,
My apologies if this is not the correct list to post this question on.
I did look for an Xpath list specific to Java but could not find one.
I have the following method in my Java application:
public String evaluateDocument(String inputDoc, String xPathExpression){
String results = "";
try{
XPathFactory factory = XPathFactory.newInstance();
XPath xPath = factory.newXPath();
InputSource inputSource = new InputSource(new
StringReader(inputDoc));
results = (String) xPath.evaluate(xPathExpression,
inputSource, XPathConstants.STRING); }
catch(Exception e){
System.out.println("XpathHelper.evaluateDocument,
xPathExpression: " + xPathExpression + " exception: " + e); }
return results;
}
The inputDoc is:
<enrollment_record>
<member_identifier>1234567890</member_identifier>
<subscriber_number>1234567890</subscriber_number>
</enrollment_record>
The XPath expression is:
if (enrollment_record/subscriber_number =
enrollment_record/member_identifier) then '1' else '2'
If I run this XPath Expression from an XSLT stylesheet from the command
line within a <xsl:value-of> element, I get the correct result.
However, from within my java applicaiton I get:
exception: javax.xml.xpath.XPathExpressionException
It does not like the "if" statement within the XPath. Other simple
XPath expressions do work. It's as if the Java application is using an
older parser that doesn't recognize the "if" statement.
How can I get my Java application to correctly process this XPath
expression?
Can someone point me in the right direction, or at least tell me which
mailing list to post my question to?
Thank you in advance,
James
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