HttpClient.execute（HttpPost）在Android 4.2的错误 [英] HttpClient.execute(HttpPost) on Android 4.2 error

问题描述

我下面这个网站 http://givemepass.blogspot.hk/2011 /12/http-server.html 尝试使用Android应用程序连接的PHP服务器以获取信息。

I am following this website http://givemepass.blogspot.hk/2011/12/http-server.html to try to use the android application connect the PHP server to get message.

GetServerMessage.java

GetServerMessage.java

import org.apache.http.HttpEntity;
import org.apache.http.HttpResponse;
import org.apache.http.client.HttpClient;
import org.apache.http.client.methods.HttpPost;
import org.apache.http.impl.client.DefaultHttpClient;
import org.apache.http.util.EntityUtils;
public class GetServerMessage {

    public String stringQuery(String url){
        try
        {
            HttpClient httpclient = new DefaultHttpClient();
            HttpPost method = new HttpPost(url);
            HttpResponse response = httpclient.execute(method);
            HttpEntity entity = response.getEntity();
            if(entity != null){
                return EntityUtils.toString(entity);
            }
            else{
                return "No string.";
            }
         }
         catch(Exception e){
             return "Network problem";
         }
    }
}

GetPhpServerMessageDemoActivity.java

GetPhpServerMessageDemoActivity.java

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

public class GetPhpServerMessageDemoActivity extends Activity {
    /** Called when the activity is first created. */
    private TextView textView;
    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.main);
        textView = (TextView)findViewById(R.id.text_view);
        GetServerMessage message = new GetServerMessage();
        String msg = message.stringQuery("http://192.168.1.88/androidtesting.php");
        textView.setText("Server message is "+msg);
    }

}

我试图从该网站 http://uploadingit.com/file下载Android应用程序项目/d4632ekfpwjiupyn/GetPhpServerMessageDemo2.zip 并在我的手机上运行，​​它的工作。 但是，当我开始一个新项目（Minimuim科目编号SDK：API8，目标SDK：API17，与编译：API17）和复制这两个Java codeS。我已经添加了权限 android.permission.INTERNET对，所以我不知道问题出在哪里，我只知道运行时的Htt presponse响应= httpclient.execute（法）; 有一个错误，返回字符串网络问题

I tried to download the Android Application Project from that site http://uploadingit.com/file/d4632ekfpwjiupyn/GetPhpServerMessageDemo2.zip and run on my phone, it's work. But when I start a new project (Minimuim Requied SDK: API8, Target SDK: API17, Compile with: API17) and copy these two java codes. I've added the permission android.permission.INTERNET, so I don't know where is the problem, I only know that when run HttpResponse response = httpclient.execute(method); there is a error and returned String "Network problem".

推荐答案

您是在UI线程上运行的网络相关的操作。您将获得 NetworkOnMainThreadException 后蜂窝。

You are running network related operation on the ui thread. You will get NetworkOnMainThreadException post honeycomb.

使用一个发或的AsyncTask

调用的AsyncTask

Invoke asynctask

   new TheTask().execute("http://192.168.1.88/androidtesting.php");

的AsyncTask

AsyncTask

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

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

@Override
protected String onPostExecute(Void result) {
    // TODO Auto-generated method stub
    super.onPostExecute(result);
    // update textview here
    textView.setText("Server message is "+result);
}

@Override
protected void onPreExecute() {
    // TODO Auto-generated method stub
    super.onPreExecute();
}

@Override
protected String doInBackground(String... params) {
     try
        {
            HttpClient httpclient = new DefaultHttpClient();
            HttpPost method = new HttpPost(params[0]);
            HttpResponse response = httpclient.execute(method);
            HttpEntity entity = response.getEntity();
            if(entity != null){
                return EntityUtils.toString(entity);
            }
            else{
                return "No string.";
            }
         }
         catch(Exception e){
             return "Network problem";
         }

}
}

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