Question

I am using the below shown snippet of choice element in my Mule 3.3 flow. XSL Transformer feeds the choice element. XSL Transformer is supposed to return a String (name of an entity) and on the basis of string value, I use choice router to push it to different jms queues.

<flow name="ProcessOrder">
    .
    .  
    <xm:xslt-transformer xsl-file="xsl/getEntity.xslt" returnClass="java.lang.String"/>
    <choice>
        <when expression="payload.contains('ABC')">             
            <jms:outbound-endpoint queue="order.queue1" />
        </when>
        <when>
        </when>
        <otherwise>         
        </otherwise>
    </choice>
</flow>

XSL Transformer returns this payload
<?xml version="1.0" encoding="UTF-8"?>ABC

My question is how do I compare the String returned. I don’t think payload.contains() is the best way to do this, though it solves my purpose and also we won’t have matching entities returned which are ever like ABCxy but still this is not a full proof solution.

Answer 1

Add the omit-xml-declaration part in your xslt as shown below. This will give you the raw string without the prolog.

<xsl:stylesheet version="2.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform" 
xmlns:xsd="http://www.w3.org/2001/XMLSchema"    >

<xsl:output omit-xml-declaration="yes" indent="yes" />

<xsl:template match="/">
...
...

This will give

"ABC" as output instead of     "<?xml version="1.0" encoding="UTF-8"?>ABC"

Then in the expression use it like

<when expression="#[message.payload.contains('ABC')]">

This way it should work.