read and update xml using groovy ,

read xml file in groovy to read xml file , using  XmlSlurper().parseText and display its content ;xml file: <?xml version='1.0' encoding='UTF-8'?>

<Records>

  <Line>

  <Field1>ABC</Field1>

  <Field2>123</Field2>

  <Field3>XXX</Field3>

  <Field4>567890</Field4>

  </Line>

</Records>; 

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You can use the XmlSlurper class in Groovy to parse an XML file and display its content. Here's a Groovy script that reads the XML content you provided and displays it:

groovy

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def xmlText = """<?xml version='1.0' encoding='UTF-8'?>

<Records>

  <Line>

    <Field1>ABC</Field1>

    <Field2>123</Field2>

    <Field3>XXX</Field3>

    <Field4>567890</Field4>

  </Line>

</Records>"""

def records = new XmlSlurper().parseText(xmlText)

// Display XML content

println "Root Element Name: ${records.name()}"

records.Line.each { line ->

    println "Line Element:"

    println "  Field1: ${line.Field1.text()}"

    println "  Field2: ${line.Field2.text()}"

    println "  Field3: ${line.Field3.text()}"

    println "  Field4: ${line.Field4.text()}"

}

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def xmlText = """<?xml version='1.0' encoding='UTF-8'?>

<Records>

  <Line>

    <Field1>ABC</Field1>

    <Field2>123</Field2>

    <Field3>1111</Field3>

    <Field4>567890</Field4>

  </Line>

</Records>"""

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

def xpath = "/Records/Line/Field4"

// Update the element with XPath

xml.'**'.findAll { it.name() == 'Field4' }.each { element ->

    element.replaceBody("Rama")

}

// Serialize the modified XML back to text

def updatedXmlText = groovy.xml.XmlUtil.serialize(xml)

println updatedXmlText

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https://blogs.sap.com/2016/01/26/hci-parse-xml-easily-with-groovy-scripts/

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