Question

我正在学习android。我开始研究与android的网络。我面临的问题是在try语句下httpclient.execute（request）行无法正常工作。而是每次都执行catch子句。我搜索了很长时间但我没有得到关于这个问题的答案。我很乐意从你这边得到任何帮助。

这是主要活动

package com.example.androidaspect;

import android.app.Activity;
import android.content.Context;
import android.os.Bundle;
import android.widget.TextView;

public class MainActivity extends Activity {

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);

        Context context = getApplicationContext();
        String url = "http://www.google.com";
        TextView tv = (TextView) findViewById(R.id.tvres);

        AsyncTaskRunner runner = new AsyncTaskRunner();
        runner.execute(url, context, tv);

    }

}

这是Asynctaskrunner

package com.example.androidaspect;

import java.io.BufferedReader;
import java.io.InputStreamReader;

import org.apache.http.HttpResponse;
import org.apache.http.client.HttpClient;
import org.apache.http.client.methods.HttpGet;
import org.apache.http.impl.client.DefaultHttpClient;

import android.content.Context;
import android.os.AsyncTask;
import android.widget.TextView;
import android.widget.Toast;

public class AsyncTaskRunner extends AsyncTask<String, String, String>{

    Context context;
    TextView tv;
    String url;
    String result="";

    public void execute(String url, Context context, TextView tv) {
        // TODO Auto-generated method stub
        this.tv= tv;
        this.context = context;
        this.url = url;
    }

    @Override
    protected String doInBackground(String... params) {
        // TODO Auto-generated method stub
        try{
            publishProgress("Loading");
            HttpClient hc = new DefaultHttpClient();
            HttpGet request = new HttpGet(url);
            HttpResponse res = hc.execute(request);

            BufferedReader rd = new BufferedReader(new InputStreamReader(res.getEntity().getContent()));
            String line="";
            while((line=rd.readLine())!=null){
                result.concat(line);
            }

        }catch(Exception e){
            e.printStackTrace();
        }
        return result;
    }

    @Override
    protected void onPostExecute(String result) {
        // TODO Auto-generated method stub
        tv.setText(result);
    }

    @Override
    protected void onProgressUpdate(String... values) {
        // TODO Auto-generated method stub
        Toast.makeText(context, values[0], Toast.LENGTH_SHORT).show();
    }

}

我已添加了互联网权限。我的xml中有一个textview，它从网站获得响应。但是当我在模拟器上运行应用程序时，它会运行，但textview不会加载数据。

Answer 1

正如我在评论中提到的，使用AsyncTask。

引用文档

应用程序尝试执行时抛出的异常   在其主线程上进行网络操作。

这仅适用于针对Honeycomb SDK或。的应用程序   更高。允许使用早期SDK版本的应用程序   他们的主要事件循环线程上的网络，但它很重要   泄气

http://developer.android.com/reference/android/os/AsyncTask.html

public class MainActivity extends Activity {

    TextView tv;
    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.tvres);
        tv = (TextView) findViewById(R.id.textView1);
        new TheTask().execute();

    }
    class TheTask extends AsyncTask<Void,Void,String>
    {

        @Override
        protected String doInBackground(Void... arg0) {
            // TODO Auto-generated method stub
             String response=null; 
             try {

                    HttpClient hc = new DefaultHttpClient();
                    HttpGet hg = new HttpGet("http://www.google.com");
                    HttpResponse res = hc.execute(hg);
                    HttpEntity entity = res.getEntity();
                    response = EntityUtils.toString(entity);

                } catch (Exception e) {
                    // TODO Auto-generated catch block
                    e.printStackTrace();
                }
            return response;
        }

        @Override
        protected void onPostExecute(String result) {
            // TODO Auto-generated method stub
            super.onPostExecute(result);
            tv.setText(result);
        }

    }

你有AsyncTask作为内部类。

 class TheTask extends AsyncTask<Void,Void,String>

你没有将任何内容传递给doInbackground。所以第一个arg类型是Void。第二个arg也是Void因为你没有进度指示器。第三个是背景计算结果的类型。所以它的字符串是因为我们return response;是字符串。

现在在后台线程上调用doInbackground。所以在那里进行背景计算。

在ui线程上调用

onPostExecute。所以在那里更新ui。

编辑：

你的方式

public class MainActivity extends Activity {

    TextView tv;
    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);
        String url = "http://www.google.com";
         tv = (TextView) findViewById(R.id.tvres);
        AsyncTaskRunner runner = new AsyncTaskRunner(url,this,tv);
        runner.execute();


    }
    }

然后

public class AsyncTaskRunner extends AsyncTask<Void, String, String>{

    Context context;
    TextView tv;
    String url;
    String result="";

    public  AsyncTaskRunner(String url, Context context, TextView tv) {
        // TODO Auto-generated method stub
        this.tv= tv;
        this.context = context;
        this.url = url;
    }

    @Override
    protected String doInBackground(Void... params) {
        // TODO Auto-generated method stub

            publishProgress("Loading");
            String response=null; 
            try {

                   HttpClient hc = new DefaultHttpClient();
                   HttpGet hg = new HttpGet(url);
                   HttpResponse res = hc.execute(hg);
                   HttpEntity entity = res.getEntity();
                   response = EntityUtils.toString(entity);

        }catch(Exception e){
            e.printStackTrace();
        }
          return response;
    }

    @Override
    protected void onPostExecute(String result) {
        // TODO Auto-generated method stub
        tv.setText(result);
    }

    @Override
    protected void onProgressUpdate(String... values) {
        // TODO Auto-generated method stub
        Toast.makeText(context, values[0], Toast.LENGTH_SHORT).show();
    }

}

Answer 2

您需要使用AsyncTask＆lt;＆gt;或IntentService来处理HttpQueries

httpclient无法执行httpget对象

2 个答案: