Question

我正在学习如何构建Android应用，并尝试使用Java在我的数据库中插入数据。

但我没有成功，我收到了这个错误：

 android.os.NetworkOnMainThreadException

我对此进行了更多搜索，并且每个人都说AsynkTask删除了此错误。如何使我的代码适应AsynkTask？你能解释一下它是如何运作的吗？

由于

代码

package com.example.turanja;
....
public class register extends Activity {

    String name;
    String id;
    InputStream is=null;
    String result=null;
    String line=null;
    int code;

    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.register);

        final EditText e_id=(EditText) findViewById(R.id.editText1);
        final EditText e_name=(EditText) findViewById(R.id.editText2);
        Button insert=(Button) findViewById(R.id.button1);

        insert.setOnClickListener(new View.OnClickListener() {

        @Override
        public void onClick(View v) {
            // TODO Auto-generated method stub

            id = e_id.getText().toString();
            name = e_name.getText().toString();
            insert();

        }
    });
    }


    public void insert()
    {
        ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();

    nameValuePairs.add(new BasicNameValuePair("id",id));
    nameValuePairs.add(new BasicNameValuePair("name",name));

        try
        {
        HttpClient httpclient = new DefaultHttpClient();
            HttpPost httppost = new HttpPost("http://10.0.2.2/example/insert.php");
            httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
            HttpResponse response = httpclient.execute(httppost); 
            HttpEntity entity = response.getEntity();
            is = entity.getContent();
            Log.e("pass 1", "connection success ");
    }
        catch(Exception e)
    {
            Log.e("Fail 1", e.toString());
            Toast.makeText(getApplicationContext(), "Invalid IP Address",
            Toast.LENGTH_LONG).show();
    }     

        try
        {
            BufferedReader reader = new BufferedReader
            (new InputStreamReader(is,"iso-8859-1"),8);
            StringBuilder sb = new StringBuilder();
            while ((line = reader.readLine()) != null)
        {
                sb.append(line + "\n");
            }
            is.close();
            result = sb.toString();
        Log.e("pass 2", "connection success ");
    }
        catch(Exception e)
    {
            Log.e("Fail 2", e.toString());
    }     

    try
    {
            JSONObject json_data = new JSONObject(result);
            code=(json_data.getInt("code"));

            if(code==1)
            {
        Toast.makeText(getBaseContext(), "Inserted Successfully",
            Toast.LENGTH_SHORT).show();
            }
            else
            {
         Toast.makeText(getBaseContext(), "Sorry, Try Again",
            Toast.LENGTH_LONG).show();
            }
    }
    catch(Exception e)
    {
            Log.e("Fail 3", e.toString());
    }
    }


}

Answer 1

请在此处查看answer。它会显示一个类似

的流程图

enter image description here

记住这一点..i将向您展示如何在您的代码中执行此操作。

创建一个扩展AsyncTask的新类，如..

private class MyTask extends AsyncTask<String, Integer, String> {

    // Runs in UI before background thread is called
    @Override
    protected void onPreExecute() {
        super.onPreExecute();

        // Do something like display a progress bar
    }
    // This is run in a background thread
    @Override
    protected String doInBackground(String... params) {
        String msg="";
        ArrayList<NameValuePair> nameValuePairs = new 
        ArrayList<NameValuePair>();
        nameValuePairs.add(new BasicNameValuePair("id",id));
        nameValuePairs.add(new BasicNameValuePair("name",name));
        try
        {
             HttpClient httpclient = new DefaultHttpClient();
             HttpPost httppost = new HttpPost("http://10.0.2.2/example/insert.php");
             httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
             HttpResponse response = httpclient.execute(httppost); 
             HttpEntity entity = response.getEntity();
             is = entity.getContent();
             Log.e("pass 1", "connection success ");
         }
         catch(Exception e)
         {
             Log.e("Fail 1", e.toString());
             msg="Invalid IP Address";
         }     

         try
        {
             BufferedReader reader = new BufferedReader(new InputStreamReader(is,"iso-8859-1"),8);
             StringBuilder sb = new StringBuilder();
             while ((line = reader.readLine()) != null)
             {
                 sb.append(line + "\n");
             }
             is.close();
             result = sb.toString();
             Log.e("pass 2", "connection success ");
         }
         catch(Exception e)
         {
              Log.e("Fail 2", e.toString());
         }     
         try
         {
              JSONObject json_data = new JSONObject(result);
              code=(json_data.getInt("code"));
              if(code==1)
              {
                    msg="Inserted Successfully";
              }
              else
              {
                    msg="Sorry, Try Again";
              }
         }
         catch(Exception e)
         {
             Log.e("Fail 3", e.toString());
         }
    }
        return msg;
    }

    // This runs in UI when background thread finishes
    @Override
    protected void onPostExecute(String result) {
        super.onPostExecute(result);

        // Do things like hide the progress bar or change a TextView
             Toast.makeText(getBaseContext(),result,Toast.LENGTH_LONG).show();
    }
}

并添加new MyTask().execute();，而不是调用insert();。我已将您在insert()中完成的所有操作移至doInBackground()。

Answer 2

当您尝试执行主UI线程的网络操作时，会发生此异常。

有关例外的详情 -

当应用程序尝试在其主线程上执行网络操作时引发的异常。

仅针对Honeycomb SDK或更高版本的应用程序进行此操作。针对早期SDK版本的应用程序可以在其主要事件循环线程上进行联网，但是非常不鼓励。

Documentation

尝试在AsyncTask中执行相同操作。示例代码段看起来像 -

class InsertTask extends AsyncTask<String, Integer, HttpEntity> {

    protected HttpEntity doInBackground(String... data) {
        try {

            ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
            nameValuePairs.add(new BasicNameValuePair("id",data[0]));
            nameValuePairs.add(new BasicNameValuePair("name",data[1]));

            HttpClient httpclient = new DefaultHttpClient();
            HttpPost httppost = new HttpPost("http://10.0.2.2/example/insert.php");
            httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
            HttpResponse response = httpclient.execute(httppost); 
            HttpEntity entity = response.getEntity();
            return entity;
        } catch (Exception e) {
            //log exception
            return null;
        }
    }

    protected void onPostExecute(HttpEntity id) {
        //further processing
        //perhaps callback to main activity for success/failure
    }
}

您可以将此AsyncTask执行为

new InsertTask().execute(dataToInsert);

Answer 3

我更喜欢将服务用于此类工作，因为asyntask几乎没有限制。请参考此链接： “重新启动Activity时，AsyncTask对Activity的引用无效，因此onPostExecute（）对新Activity没有任何影响。” http://blog.danlew.net/2014/06/21/the-hidden-pitfalls-of-asynctask/

我如何使我的代码适应Asynctask

3 个答案: