在jquery中解析xml - ajax

Question

此代码有效。

    var xml = "<rss version='2.0'><channel><title>RSS Title</title></channel></rss>",
    xmlDoc = $.parseXML( xml ),
    $xml = $( xmlDoc ),
    $title = $xml.find( "title" );

    console.log($title.text()); // RSS Title

这不是：

        success: function( xml ) {

        // Extract relevant data from XML
        var xmlDoc = $.parseXML( xml ),
        $xml = $( xmlDoc ),
        $title = $xml.find( "Country" );

        console.log($title.text());

        }

我的xml表单：

object(stdClass)[2]
  public 'GetCitiesByCountryResult' => string '<NewDataSet>
  <Table>
    <Country>Spain</Country>
    <City>Fuerteventura / Aeropuerto</City>
  </Table>
  <Table>
    <Country>Spain</Country>
    <City>Hierro / Aeropuerto</City>
  </Table>
  <Table>
    <Country>Spain</Country>
    <City>La Palma / Aeropuerto</City>
  </Table>
  <Table>
    <Country>Spain</Country>
    <City>Las Palmas De Gran Canaria / Gando</City>
  </Table>
  <Table>
    <Country>Spain</Country>
    <City>Lanzarote / Aeropuerto</City>
  </Table>
  <Table>
    <Country>Spain</Country'... (length=4383)

PHP：

<?php

if ( isset( $_POST[ 'facultyname' ] ) ) {
    //Create the client object
    $soapclient = new SoapClient( 'http://www.webservicex.net/globalweather.asmx?WSDL' );

    //Use the functions of the client, the params of the function are in the associative array.
    $params = array( 'CountryName' => $_POST[ 'facultyname' ] );
    $response = $soapclient->GetCitiesByCountry( $params );

    var_dump( $response );
}
else {
    echo 'Some error on php receiving side.';
}

?>

我的XML正在返回对象，也许这就是为什么我无法获取我的国家名称。我尝试了很多但没有工作。 我正在使用网络服务Link作为演示测试。