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我正在尝试从 android 客户端访问 magento Web 服务。它应该返回一个会话 id。我已经使用 apache 完成了这个使用 java 客户端，并且成功调用了该方法。在尝试使用 android 客户端时，我如何得到 xmlpullparser 异常：10-24 15:25:44.409: WARN/System.err(277): org.xmlpull.v1.XmlPullParserException: 预期：START_TAG { http://www. w3.org/2001/12/soap-envelope }信封（位置：START_TAG @2:327 in java.io.InputStreamReader@44ee2268）

好吧，这是我试图从中访问登录方法的 wsdl 文件：wsdl

我的Java代码：

public class DeviceClientActivity extends Activity {
/** Called when the activity is first created. */
private static final String NAMESPACE = "urn:Magento";
private static final String URL = "http://xxx.xxx.xx.xxx/magento/index.php/api/v2_soap?wsdl";
private static final String METHOD_NAME = "login";
private static final String SOAP_ACTION ="urn:Mage_Api_Model_Server_V2_HandlerAction";
@Override
public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.main);
    Log
    .d("WS",
          "--------------------- Webservice Part Begins ---------------------");  
Log.d("WS", "1. SoapObject Construction");  
SoapObject objsoap=new SoapObject(NAMESPACE, METHOD_NAME);

objsoap.addProperty("用户名", "alokxxxx");
objsoap.addProperty("apiKey", "xxxxxx");
Log.d("WS", "构造SOAP对象结束！！！"); 

SoapSerializationEnvelope envelope = new SoapSerializationEnvelope(
      SoapEnvelope.VER11); 
Log.d("WS", "2. Envelop Created");    
envelope.setOutputSoapObject(objsoap);
Log.d("WS", "3. Request Into Envelop");
AndroidHttpTransport httpTransport = new AndroidHttpTransport(URL);
httpTransport.debug = true; 
Log.d("WS", "5. Transport Level to True");
try {
    httpTransport.call(SOAP_ACTION, envelope);

//此处出错 WARN/System.err(277): org.xmlpull.v1.XmlPullParserException: 预期：START_TAG { http://www.w3.org/2001/12/soap-envelope }信封（位置：START_TAG @2 :327 在 java.io.InputStreamReader@44ee2268) 

Log.d("WS", "6. httpTransport.call");
    if (envelope != null) 
    {
       SoapObject loresponse = (SoapObject) envelope.getResponse();
       SoapObject logObject = (SoapObject)loresponse.getProperty("sessionId");           
       Log.d("WS", "logObject: "+logObject);                      
    } 
    else 
    {
       Log.d("WS", "Response Envelop Error");
    }

} catch (IOException e) {
    e.printStackTrace();
} catch (XmlPullParserException e) {
    e.printStackTrace();
}
}

}

登录以寻求帮助：DEBUG/WS(333): --------- Webservice Part Begins -------------- -------
10-29 15:38:33.643: DEBUG/WS(333): 1. SoapObject 构造
10-29 15:38:33.673: DEBUG/WS(333): SOAP 对象的构造结束！！ ！
10-29 15:38:33.673: DEBUG/WS(333): 2. Envelop Created 10-29 15:38:33.673: DEBUG/WS(333): 3. Request into
Envelop 10-29 15:38:33.683: DEBUG/WS(333): 5. 到 Trueorg.ksoap2.transport.AndroidHttpTransport@44eeb200 的传输级别 10-29 15:38:33.683: DEBUG/Try(333): try block
10-29 15:38:34.903: WARN /System.err（333）：org.xmlpull.v1.XmlPullParserException：预期：START_TAG { http://schemas.xmlsoap.org/soap/envelope/ }信封（位置：START_TAG @2:327 in java.io.InputStreamReader @44efbe90)
有什么建议吗？谢谢。

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1 回答 1

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这就是我让登录工作的方式。诀窍是不要将 ?wsdl 放在 url 上

public boolean login()  
    {
        try
        {
            SoapObject request = new SoapObject("http://schemas.xmlsoap.org/wsdl/","login");
            request.addProperty("username",apiuid.getText().toString());
            request.addProperty("apiKey",apipwd.getText().toString());
            SoapSerializationEnvelope envelope = new SoapSerializationEnvelope(SoapEnvelope.VER11); 
            envelope.dotNet=false;
            envelope.setOutputSoapObject(request); 
            HttpTransportSE androidHttpTransport = new HttpTransportSE(apiurl.getText().toString());
            androidHttpTransport.call("http://schemas.xmlsoap.org/wsdl/login", envelope);
            String result =(String)envelope.getResponse();
            String URL = new String();
            URL=apiurl.getText().toString();
            Editor e = userDetails.edit();
                e.putString("url", URL);
                e.putString("uid", apiuid.getText().toString());
                e.putString("pwd", apipwd.getText().toString());
                e.putString("lastsession", result);
                e.commit();
               return true;
        } catch(IOException e)
             {alertbox("IO error",e.getMessage());return false;}
          catch(XmlPullParserException e)
             {alertbox("xml error",e.getMessage());return false;}
          catch(Exception e)
             {alertbox("error",e.getMessage());return false;}     
    }
于 2011-10-26T10:54:20.137 回答