Just by reading the syntax class name/function. Try following statement:
<xsl:value-of select="ExtensionTest:test()"
Cheers,
Juergen
-----Ursprüngliche Nachricht-----
Von: Spencer Bruce [mailto:spencer(_at_)ubaccess(_dot_)com]
Gesendet am: Montag, 17. Februar 2003 10:17
An: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Betreff: Re: [xsl] JD and Java Extension
If you want to call a Java method from jd.xslt then you need a namespace
declaration to direct the processor to the java class.
<xsl:stylesheet xmlns:math="java:java.lang.Math" ...>
...
<xsl:value-of select="math:sin(3.14)"/>
...
</xsl:stylesheet>
Feel free to mail me if you experience more problems.
Johannes Döbler
This works fine for a Java class from the standard library such as
java.lang.Math
However in our case we have our own Java class called for example
ExtensionTest. The class has one public static method:
public static String test(String s)
{
return s;
}
We then have a stylesheet such as:
<?xml version="1.0"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
version="1.0"
xmlns:java="ExtensionTest">
<xsl:template match="/">
<xsl:value-of select="java:test(.)" />
</xsl:template>
</xsl:stylesheet>
When we try to call this class from jd.xslt it reports the message:
jd.xml.xslt.XsltException: no extension function
'java:org.ubaccess.wizard.ExtensionTest:test' defined
We have tried to put the extension in different directories, in packages etc
but we cannot get the parser to recognise it!
Thanks for your help
Spencer Bruce
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