Question

尝试执行基本的Web服务，我不断收到错误：

org.xmlpull.v1.XmlPullParserException: expected: START_TAG {http://schemas.xmlsoap.org/soap/envelope/}Envelope (position:START_TAG <html>@2:44 in java.io.InputStreamReader@21f12310)

我已经被困在这3天了。我几乎看过StackOverflow上的每一个例子，但仍然无法获得其他什么，我觉得我已经多次看过它了，我现在只是在寻找一些东西......这是我的代码;

private final String NAMESPACE   =  "http://tempuri.org/";
private final String URL         =  "http://foo/bar/Service1.asmx";
private final String SOAP_ACTION =  "http://tempuri.org/HelloWorld";
private final String METHOD_NAME =  "HelloWorld";
private  String TAG              =  "SOAP";

private static String message;

@Override
protected void onCreate(Bundle savedInstanceState) {
    ....
}

public void getMessage() {

        String dump;
        //Create request
        SoapObject request = new SoapObject(NAMESPACE, METHOD_NAME);

        //Create envelope
        SoapSerializationEnvelope envelope = new SoapSerializationEnvelope(
                SoapEnvelope.VER11);
        //envelope.encodingStyle = SoapSerializationEnvelope.ENC2001;
        envelope.dotNet = true;



        //Set output SOAP object
        envelope.setOutputSoapObject(request);
        //Create HTTP call object
        HttpTransportSE androidHttpTransport = new HttpTransportSE(URL);

        try {

            androidHttpTransport.debug=true;
            androidHttpTransport.call(SOAP_ACTION, envelope );
            dump = androidHttpTransport.responseDump;
            //Get the response
            //SoapObject result = (SoapObject) envelope.bodyIn;
            SoapPrimitive response = (SoapPrimitive) envelope.getResponse();
            //Assign it to fahren static variable
            message = response.toString();

    } catch (Exception e) {
        //tv.setText(e.getMessage());
        e.printStackTrace();
    }

}

然后，当我单步执行它并查看responseDump时，它会说：

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-   strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1"/>
<title>401 - Unauthorized: Access is denied due to invalid credentials.</title>
<style type="text/css">
<!--
body{margin:0;font-size:.7em;font-family:Verdana, Arial, Helvetica, sans-   serif;background:#EEEEEE;}
fieldset{padding:0 15px 10px 15px;} 
h1{font-size:2.4em;margin:0;color:#FFF;}
h2{font-size:1.7em;margin:0;color:#CC0000;} 
h3{font-size:1.2em;margin:10px 0 0 0;color:#000000;} 
#header{width:96%;margin:0 0 0 0;padding:6px 2% 6px 2%;font-family:"trebuchet MS", Verdana, sans-    serif;color:#FFF;
background-color:#555555;}
#content{margin:0 0 0 2%;position:relative;}
.content-container{background:#FFF;width:96%;margin-top:8px;padding:10px;position:relative;}
-->
</style>
</head>
<body>
<div id="header"><h1>Server Error</h1></div>
<div id="content">
 <div class="content-container"><fieldset>
  <h2>401 - Unauthorized: Access is denied due to invalid credentials.</h2>
  <h3>You do not have permission to view this directory or page using the credentials that you    supplied.</h3>
 </fieldset></div>
</div>
</body>
</html>

我的代码错误在这一行：

androidHttpTransport.call(SOAP_ACTION, envelope );

当Exception捕获它时，这就是错误：

org.xmlpull.v1.XmlPullParserException: expected: START_TAG {http://schemas.xmlsoap.org/soap/envelope/}Envelope (position:START_TAG <html>@2:44 in java.io.InputStreamReader@21f12310)

如果有人可以帮助我，我会非常感激..这让我很难过，我无法理解......

Answer 1

您无法使用肥皂获取html网页！如果你打电话给网络服务，你必须用wsdl

替换你的链接

富/酒吧/ Service1.asmx的？WSDL

Android org.xmlpull.v1.XmlPullParserException错误

1 个答案: