Extracting multiple XML values using XSLT

Question

I'm working on some XSLT to extract the values from complex XML.

The xml:

   <bean id="timingAdvice" 

class="org.springframework.aop.interceptor.PerformanceMonitorInterceptor" />

<bean id="XMLhandler" class="com.order.OrderStatusSAXHandler">
</bean>

The output I wish to achieve:

<bean>
<id>timingAdvice</id>
<class>org.springframework.aop.interceptor.PerformanceMonitorInterceptor</class>
</bean>

<bean>
<id>XMLhandler</id>
<class>com.citi.get.rio.order.OrderStatusSAXHandler</class>
</bean>

I am using this XSLT:

    <?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0"
    xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:output method="xml" version="1.0" encoding="UTF-8"
        indent="yes" />
    <xsl:template match="@* | node()">
        <xsl:copy>
            <xsl:apply-templates select="@* | node()" />
        </xsl:copy>
    </xsl:template>
        <xsl:template match="beans/bean">
        <xsl:element name="{@class}">
            <xsl:apply-templates />
        </xsl:element>
    </xsl:template>
    <xsl:template match="@* | node()">
        <xsl:copy>
            <xsl:apply-templates select="@* | node()" />
        </xsl:copy>
        </xsl:template>
    <xsl:template match="beans/bean">
        <xsl:element name="{@id}">
            <xsl:apply-templates />
        </xsl:element>
    </xsl:template>
</xsl:stylesheet>

However this outputs:

    <?xml version="1.0" encoding="UTF-8"?>
<beans>

<timingAdvice/>
<XMLhandler>
</XMLhandler>
</beans>

Which is not what I am looking for.

I am want to inspect each attribute of the xml print them like the following:

<attributeName>value<attributeName>

EDIT

I've encountered the problem with the beans tag it holds a number of spring references to the Spring Framework:

<beans xmlns="http://www.springframework.org/schema/beans" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:util="http://www.springframework.org/schema/util"
xsi:schemaLocation="
    http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-2.5.xsd 
    http://www.springframework.org/schema/util http://www.springframework.org/schema/util/spring-util-2.5.xsd"
default-lazy-init="false">

The solution provided doesn't provide the required output when this is the opening tag. Is there a way to ignore these references within the beans tag

to be honest it's just an identifier for it, so it's not of great significance. However, I have corrected it to id. — Will
– Will, Commented May 16, 2013 at 13:09

JLRishe · Accepted Answer · 2013-05-16 17:11:08Z

1

So, something like this?

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
                              xmlns:bn="http://www.springframework.org/schema/beans">
  <xsl:output method="xml" indent="yes" omit-xml-declaration="yes"/>

  <xsl:template match="@* | node()">
    <xsl:copy>
      <xsl:apply-templates select="@* | node()"/>
    </xsl:copy>
  </xsl:template>

  <xsl:template match="bn:bean/@*">
    <xsl:element name="{name()}" namespace="{namespace-uri(..)}">
      <xsl:value-of select="." />
    </xsl:element>
  </xsl:template>
</xsl:stylesheet>

When this is run on your sample input (when it's wrapped in a <beans> element), the result is:

<beans xsi:schemaLocation="     http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-2.5.xsd      http://www.springframework.org/schema/util http://www.springframework.org/schema/util/spring-util-2.5.xsd" default-lazy-init="false" xmlns="http://www.springframework.org/schema/beans" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:util="http://www.springframework.org/schema/util">
  <bean>
    <id>timingAdvice</id>
    <class>org.springframework.aop.interceptor.PerformanceMonitorInterceptor</class>
  </bean>

  <bean>
    <id>XMLhandler</id>
    <class>com.order.OrderStatusSAXHandler</class>
  </bean>
</beans>

Does the order of the elements converted from attributes matter, or can they occur in the same order as the attributes?

edited May 16, 2013 at 17:11

answered May 16, 2013 at 13:09

JLRishe

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8 Comments

Will Over a year ago

Yeah that seems to do the trick.The order doesn't matter as long as they are placed under bean. I have some more complex xml than the sample provided which I must write the xslt around. I may have to work out a way to navigate through that.

LarsH Over a year ago

+1, good answer. BTW the order of the attributes in the input is information that's not available to XSLT, so the final question here is pretty well moot.

JLRishe Over a year ago

@LarsH While that is technically true, I would presume that a lot of XSLT processors treat attributes as being in the order that they occur in the source document. I know this is definitely the case for MSXSL at least.

LarsH Over a year ago

I wouldn't presume that, since the spec says not to. I'm curious how you know this is definitely the case for MSXSL; is it documented, or have you just observed it in several instances, in which case it could change in different versions or different execution environments? In any case, I don't object to the question being asked; just wanted to point out that relying on the order of XML attributes is contrary to the XML Info model. The OP is well-served to be aware of that before building a process that relies on the order.

Will Over a year ago

@JLRishe Cheers for the answer. I am doing this from spring files and the opening bean tag has some references to the spring framework, how can I ignore these as it's causing the xslt not to work.

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