How to get the elements <and> and <or> in XPath with Jre default Processor (JAXPSAXProcessor)

I am working with some xml files which contain and/or tags. I want to transform them to html. In my XSL I am using two templates

<xsl:template match="and">
    (
    <xsl:apply-templates select="./*[1]" />
    <xsl:text> </xsl:text>
    <xsl:value-of select="name(.)" />
    <xsl:text> </xsl:text>
    <xsl:apply-templates select="./*[2]" />
    )
</xsl:template>
<xsl:template match="or">
    (
    <xsl:apply-templates select="./*[1]" />
    <xsl:text> </xsl:text>
    <xsl:value-of select="name(.)" />
    <xsl:text> </xsl:text>
    <xsl:apply-templates select="./*[2]" />
    )
</xsl:template>

It Works when i am using Xalan as a processor but when I am using开发者_JS百科 JAXPSAXProcessor I am getting errors: ERROR [main] JAXPSAXProcessorInvoker - Syntax error in 'or'. ERROR [main] JAXPSAXProcessorInvoker - Syntax error in 'and'.

I suppose that JaxPSaxProcessor translates and/or to the operators in Xpath. here you can see the list of operators I can not change the jaxb processor because it have been used in many places. Is there any other sugesstion to solve the problem?

---

If the XSLT processor is that buggy, you may try cheating it using a number of techniques, such as:

<xsl:template match="*[name()=concat('a', 'nd')">

or

<xsl:template match="*[name()=substring('land',2)">

Good luck.