在Groovy中读取SOAP XML

我试图阅读下面的soapxml来接收ServiceID值。我不知道如何在Groovy中处理SOAP XML。我的代码如下，但我没有收到这样的财产例外。如果您能提供任何帮助，我们将不胜感激

xml付费软件

http://www.ultipro.com/dataservices/bidata/2/IBIDataService/LogOnResponse 1233 ABCD 12347654 好啊 12346484

我的代码

import com.sap.gateway.ip.core.customdev.util.Message;
import java.util.HashMap;
import groovy.util.XmlParser;
import groovy.util.slurpersupport.GPathResult;
import groovy.xml.MarkupBuilder;
def Message processData(Message message) {
      Reader reader = message.getBody(Reader)
      def request = new XmlSlurper().parse(reader)
      Writer writer = new StringWriter()
      def indentPrinter = new IndentPrinter(writer, ' ')
      def builder = new MarkupBuilder(indentPrinter)
      builder.result {
        'LogOnResult'
        Envelope.Header.Body.LogOnResponse.LogOnResult.ServiceId
      }
       message.setBody(writer.toString())

      return message
    }

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似乎缺少

信封

变量声明