使用jQuery从URL解析XML

Question

我想从此XML（http://stagservices.upol.cz/ws/services/rest/student/getStudentInfo?osCislo=R140742）中取回“jmeno”，并且每次都返回错误。

$(document).ready(function() { 
    $.ajax({ 
     type: 'GET', 
     url: 'http://stagservices.upol.cz/ws/services/rest/student/getStudentInfo?osCislo=R140742', 
     crossDomain: true, 
     dataType: "jsonp", 
     success: parseXml, 
     error: function() { 
      alert("Error: Something went wrong"); 
     } 
    }); 
}); 

function parseXml(xml) { 
    $(xml).find('student').each(function() { 
     $("#output").append($(this).find('jmeno').text() + "<br />"); 
    }); 
} 

Answer 1

DEMO

使用datatype: 'xml'。

$(document).ready(function() { 
     $.ajax({ 
      type: 'GET', 
      url: 'http://stagservices.upol.cz/ws/services/rest/student/getStudentInfo?osCislo=R140742', 
      crossDomain: true, 
      dataType: "xml", 
      success: function(data){ 
      console.log(data); 
      }, 
      error: function() { 
       alert("Error: Something went wrong"); 
      } 
     }); 
    });