Cannot find sub element

riffla · March 17th, 2008, 08:22 AM

Hi

Get an error when trying to transform the xml with xsl. Error saying that <provider_id> (in XSL file) is unexpected, though I removed schema. Any suggestions of what could be wrong?

/R

XML:

Code:

<?xml version="1.0" encoding="ISO-8859-1"?>
<MESSAGE xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
    <HEADER>
        <MESSAGE_ID>584232</MESSAGE_ID>
        <REFERENCE_ID>139122</REFERENCE_ID>
        <OPP_REFERENCE_ID/>
        <MESSAGE_CODE>HK_TO_NETADMIN</MESSAGE_CODE>
        <VERSION>1</VERSION>
        <CREATION_DATE>2008/01/23 08:03:53</CREATION_DATE>
        <SENDER_NAME>Herkules</SENDER_NAME>
        <RECIPIENT_NAME>WS_NETADMIN</RECIPIENT_NAME>
    </HEADER>
    <HK_TO_NETADMIN>
        <LIST_OF_COMP_PARAMETERS>
            <provider_id>123456789</provider_id>
            <contact_info>
                <contact_info_security_number>670703-7875</contact_info_security_number>
            </contact_info>
        </LIST_OF_COMP_PARAMETERS>
        <REQUESTED_OPERATION>GetCustomers</REQUESTED_OPERATION>
    </HK_TO_NETADMIN>
</MESSAGE>

XSL:

Code:

<?xml version="1.0" encoding="ISO-8859-1"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">

    <xsl:template match="/">
        <GetCustomersIn xmlns="http://www.netadmin.se/2006/API_XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
            <xsl:apply-templates select="LIST_OF_COMP_PARAMETERS">
                <provider_id><xsl:value-of select="provider_id"/></provider_id>
                <search_arguments>
                    <contact_info>
                        <xsl:if test="string(contact_info_security_number)">
                            <security_number><xsl:value-of select="contact_info_security_number"/></security_number>            
                        </xsl:if>
                    </contact_info>
                </search_arguments>
            </xsl:apply-templates>
        </GetCustomersIn>
    </xsl:template>

</xsl:stylesheet>

Martin Honnen · March 17th, 2008, 08:33 AM

That stylesheet is not correct, the xsl:apply-templates cannot have any contents besides xsl:sort and/or xsl:with-param.
Here is a corrected stylesheet, though I am mainly guessing what you want to achieve:

Code:

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
  xmlns="http://www.netadmin.se/2006/API_XMLSchema" >
    <xsl:output method="xml" indent="yes"/>
    <xsl:template match="/">
        <GetCustomersIn xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
            <xsl:apply-templates select="MESSAGE/HK_TO_NETADMIN/LIST_OF_COMP_PARAMETERS">

            </xsl:apply-templates>
        </GetCustomersIn>
    </xsl:template>


    <xsl:template match="LIST_OF_COMP_PARAMETERS">
                <provider_id><xsl:value-of select="provider_id"/></provider_id>
                <search_arguments>
                    <contact_info>
                        <xsl:if test="contact_info/contact_info_security_number">
                            <security_number><xsl:value-of select="contact_info/contact_info_security_number"/></security_number>            
                        </xsl:if>
                    </contact_info>
                </search_arguments>
    </xsl:template>
</xsl:stylesheet>

mhkay · March 17th, 2008, 08:47 AM

The only elements allowed inside xsl:apply-templates are xsl:sort and xsl:with-param. I can't correct your code for you because I've no idea what you intended it to mean.

Michael Kay
http://www.saxonica.com/
Author, XSLT Programmer's Reference and XPath 2.0 Programmer's Reference