Question

从这个给定的XML中，我想提取标签的值并将其分配给变量。我正在使用Scala。

<soap:Envelope xmlns:wsdlns="http://schemas.xmlsoap.org/wsdl/" xmlns:soap="http://www.w3.org/2003/05/soap-envelope" xmlns:iav="http://www.example.com/xxx/yy/ABC_DE_Fghijklmn">
<soap:Body>
<iav:OperationResponse>
<iav:XYZ>AAA</iav:XYZ>
<iav:Code/>
</iav:OperationResponse>
</soap:Body>
</soap:Envelope>

将给定的XML分配给val resp。我尝试过以下选项：

(resp.get \\"iav:XYZ")

(resp.get \"@iav:XYZ")

(resp.get \\"@iav:XYZ")

(resp.get \"@iav:XYZ")

但是他们都不返回任何东西。

我将如何实现？我将收到许多这样的响应集，并且必须对所有集合进行相同的提取和分配。

Answer 1

以下是提取方法：

import scala.xml.Elem

object StackOverflow {
  def main(args: Array[String]): Unit = {
    val xml = """<soap:Envelope xmlns:xml="http://www.w3.org/XML/1998/namespace" xmlns:wsdlns="http://schemas.xmlsoap.org/wsdl/" xmlns:soap="http://www.w3.org/2003/05/soap-envelope" xmlns:iav="http://www.informatica.com/xxx/yy/ABC_DE_Fghijklmn">
                <soap:Body>
                <iav:OperationResponse>
                <iav:XYZ>AAA</iav:XYZ>
                <iav:Code/>
                </iav:OperationResponse>
                </soap:Body>
                </soap:Envelope>"""
    val xmlElem: Elem = scala.xml.XML.loadString(xml)
    val operationResponse =  xmlElem \\ "OperationResponse"
    val xyz = operationResponse \ "XYZ"
    print( "Operation response is " +  xyz.text)
  }
}

从XML提取元素并分配给变量

1 个答案: