[groovy-user] Sort XML NodeList and replace old List

[RelleschB]

Hello,

I want to sort some Elements in a list of XML Elements.
The sorting works.

The statement "*records.value = sorted*" works fine.
But I want "*records.records.value = sorted*" I get the following exception:
Caught: java.lang.IllegalArgumentException: argument type mismatch
java.lang.IllegalArgumentException: argument type mismatch
at
com.bsp.workflow.NormalizeWorkflow.nomalize(NormalizeWorkflow.groovy:44)
at com.bsp.workflow.NormalizeWorkflow$nomalize.call(Unknown Source)
at
com.bsp.workflow.NormalizeWorkflow.main(NormalizeWorkflow.groovy:5)

Here is the code:
def records = new XmlParser().parseText(XmlExamples.CAR_RECORDS)
def sorted = records.records.car.sort{ it.'@year'.toInteger() }
//records.value = sorted
records.records.value = sorted
println(records.toString())

static def CAR_RECORDS = '''
<records>
<records>
<car name='HSV Maloo' make='Holden' year='2006'>
<country>Australia</country>
<record type='speed'>Production Pickup Truck</record>
</car>
<car name='P50' make='Peel' year='1962'>
<country>Isle of Man</country>
<record type='size'>Smallest Street-Legal Car</record>
</car>
<car name='Royale' make='Bugatti' year='1931'>
<country>France</country>
<record type='price'>Most Valuable Car</record>
</car>
</records>
</records>
'''

Greets Björn



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[Tim Yates]

You could try:

def xml = new XmlParser().parseText( CAR_RECORDS )

cars = xml.records.car.sort { it.@year.toInteger() }

xml.records.head().children().clear()

cars.each { xml.records.head().append( it ) }

println XmlUtil.serialize( xml ) 

Tim

[Tim Yates]

Or, you could use XmlSlurper and create a new document with StreamingMarkupBuilder:

def xml = new XmlSlurper().parseText( CAR_RECORDS )

new StreamingMarkupBuilder().bind {

  records {

    records {

      xml.records.car.collect().sort { it.@year.toInteger() }.each {

        mkp.yield it

      }

    }

  }

}

Tim

[RelleschB]

Hello,

thanks. Your first suggestion is what I need.

Greets Björn



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