Question

可能重复：
passing parameters to url in android

我已经将参数传递给android中的url。我怎么知道我传递的值是否被发送到url.Below是我的java代码

package com.xib;


import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.Reader;
import java.io.StringWriter;
import java.io.Writer;
import java.util.ArrayList;
import java.util.List;
import org.apache.http.HttpResponse;
import org.apache.http.NameValuePair;
import org.apache.http.client.ClientProtocolException;
import org.apache.http.client.HttpClient;
import org.apache.http.client.entity.UrlEncodedFormEntity;
import org.apache.http.client.methods.HttpPost;
import org.apache.http.impl.client.DefaultHttpClient;
import org.apache.http.message.BasicNameValuePair;
import android.app.Activity;
import android.os.Bundle;
import android.util.Log;
public class TestHttpPost extends Activity {


@Override
public void onCreate(Bundle savedInstanceState) 
{
    super.onCreate(savedInstanceState);
    postData();

}
public void postData() { 
    Log.d("hi", "value of array is ");

    HttpClient httpclient = new DefaultHttpClient();
    HttpPost httppost = new HttpPost("http://services.mascus.com/api/mascusapi.asmx?op=OpenSession");


    try 
    {
        List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(2);
        nameValuePairs.add(new BasicNameValuePair("username", "xib"));
        nameValuePairs.add(new BasicNameValuePair("password", "efi99LKW"));

        httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
        Log.d("hi", "value of array is ");

        HttpResponse response = httpclient.execute(httppost);
        Log.e("hi", "value of array is ->"+convertStreamToString(response.getEntity().getContent()));

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

    }
} 
public String convertStreamToString(InputStream is)
throws IOException {
    /*
     * To convert the InputStream to String we use the
     * Reader.read(char[] buffer) method. We iterate until the
     * Reader return -1 which means there's no more data to
     * read. We use the StringWriter class to produce the string.
     */
    if (is != null)
    {
        Writer writer = new StringWriter();
      char[] buffer = new char[1024];
        try {
            Reader reader = new BufferedReader(new InputStreamReader(is, "UTF-8"));
            int n;
            while ((n = reader.read(buffer)) != -1) {
                writer.write(buffer, 0, n);
            }
        } finally {
            is.close();
        }
        return writer.toString();
    } else {        
        return "";
    }
  }

}

任何人都可以帮助我提前致谢图沙尔

Answer 1

只需添加此行

即可

   String data = EntityUtils.toString(response.getEntity()); 
     System.out.println("Data..."+data);

在这一行之后..

 httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));

并检查日志.... 你会得到服务器的响应..

Answer 2

   String link_push ="Your link"
DefaultHttpClient hc_push=new DefaultHttpClient();  
ResponseHandler <String> res_push=new BasicResponseHandler();  
HttpPost postMethod_push=new HttpPost(link_push);  
List<NameValuePair> nameValuePairs_push= new ArrayList<NameValuePair>(3);  
nameValuePairs_push.add(newBasicNameValuePair("city_id","0000")));    
   nameValuePairs_push.add(new BasicNameValuePair("user_id","value"));  
    postMethod_push.setEntity(new UrlEncodedFormEntity(nameValuePairs_push));
String response_push=hc_push.execute(postMethod_push,res_push);
Log.v("Response for Push Notification",""+response_push);

在这些respones_push中，您将从服务器获得响应..试试这个..

在android中传递参数到url的问题

2 个答案: