目标主机不能为null或在parameters.scheme = null,host = null中设置

时间:2011-10-23 17:09:55

标签: android web-services httpclient

我得到以下例外:

    Target host must not be null or set in parameters.scheme=null,
host=null,path=/webservices/tempconvert.asmx/FahrenheitToCelsius

我的源代码:

public class ParActivity extends Activity {
    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        TextView t=new TextView(this);
        String encodedUrl = null;
      //  setContentView(R.layout.main);

                HttpClient client = new DefaultHttpClient();  
               //String query = "?Fahrenheit=26";
            //  String host = "www.w3schools.com/webservices/tempconvert.asmx/";
            //   encodedUrl = host + URLEncoder.encode(query,"utf-8");
            //  int p=(Integer) null;
              // URI uri = URIUtils.createURI("vv", "www.w3schools.com", 50, "/webservices/tempconvert.asmx", "?Fahrenheit=26", null);

                try{ 
               String postURL = "/webservices/tempconvert.asmx/FahrenheitToCelsius";
                HttpPost post = new HttpPost(postURL);
              // post.addHeader("scheme","vv");
               // post.setHeader("host", "www.w3schools.com");
                String PostData="36";
                StringEntity httpPostEntity = new StringEntity(PostData, HTTP.UTF_8);
                post.setEntity(httpPostEntity);
                post.setHeader("host", "www.w3schools.com");
                post.setHeader("Content-Length", new Integer(PostData.length()).toString());
                post.setHeader("Content-Type", "application/x-www-form-urlencoded");


                List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(2);
               // nameValuePairs.add(new BasicNameValuePair("host", "www.w3schools.com"));
                nameValuePairs.add(new BasicNameValuePair("Fahrenheit", "26"));
               post.setEntity(new UrlEncodedFormEntity(nameValuePairs));


               HttpResponse responsePOST = null;
                // Execute HTTP Post Request
               // HttpResponse response = httpclient.execute(post);
                client.getConnectionManager();



                   responsePOST = client.execute(post); 
                HttpEntity resEntity = responsePOST.getEntity();  
                String response=EntityUtils.toString(resEntity);
                response=response.trim();
              //  Log.i("RESPONSE=",response);
                t.setText("response"+response);
                setContentView(t);
        } catch (Exception e) {
            // TODO Auto-generated catch block
           //e.printStackTrace();
            t.setText("ex"+e.getMessage());
        setContentView(t);
        }
    }

}

我想在以下位置调用webservice: http://www.w3schools.com/webservices/tempconvert.asmx?op=CelsiusToFahrenheit

使用HttpClient。我应该将主机和方案作为输入吗?

请帮忙......

1 个答案:

答案 0 :(得分:32)

这是主机错误 您在"www.w3schools.com"行中传递了post.setHeader("host", "www.w3schools.com");,而您必须通过"http://www.w3schools.com"

您可以在here

中引用相同的问题