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我想简单地将一个 Android 应用程序连接到 PHP Mysql，我阅读了许多 PHP 和 Mysql 教程以将 PHP 与 Android 应用程序连接，但每个都在 emmulor 上运行，我没有真正的设备，所以我使用 emulator 进行测试。

为什么此错误来自此代码

我的 Java 文件

import java.io.BufferedReader;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.ArrayList;

import org.apache.http.HttpEntity;
import org.apache.http.HttpResponse;
import org.apache.http.NameValuePair;
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 org.json.JSONObject;

import android.support.v7.app.ActionBarActivity;
import android.support.v7.app.ActionBar;
import android.support.v4.app.Fragment;
import android.os.Bundle;
import android.util.Log;
import android.view.LayoutInflater;
import android.view.Menu;
import android.view.MenuItem;
import android.view.View;
import android.view.ViewGroup;
import android.widget.Button;
import android.widget.EditText;
import android.widget.Toast;
import android.os.Build;

public class MainActivity extends ActionBarActivity {

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

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

    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://127.0.0.1/android/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());
}
}


@Override
public boolean onCreateOptionsMenu(Menu menu) {
    getMenuInflater().inflate(R.menu.main, menu);
    return true;

}

}

这是我的 PHP 文件

<?php
$host="127.0.0.1";
    $db="android";
$uname="root";
$pwd="";

    $con = mysqli_connect("$host","$uname","$pwd") or die("connection failed");
mysqli_select_db($con,$db) or die("db selection failed");

$id=(isset($_REQUEST['id']) )?$_REQUEST['id']: '';
$name=(isset($_REQUEST['name']) ) ? $_REQEST['name']: '';

$flag['code']=0;

if($r=mysqli_query("insert into sample values('$id','$name')   ,$con"))
{
    $flag['code']=1;
    echo"hi";
}

print(json_encode($flag));
mysqli_close($con);
?>

这是 Logcat 窗口

09-30 03:35:16.800: E/Fail 1(1064): android.os.NetworkOnMainThreadException
09-30 03:35:16.870: E/Fail 2(1064): java.lang.NullPointerException: lock ==  
 null
09-30 03:35:16.870: E/Fail 3(1064): java.lang.NullPointerException
4

1 回答 1

0

当应用程序尝试在其主线程上执行网络操作时，将引发此异常。在以下位置运行您的代码AsyncTask：

 private class NetworkTask extends AsyncTask<String, Void, Void> {

        protected void doInBackground(String... urls) {
            try {

            <!--Your network stuff goes here -->

            } catch (Exception e) {

            }
        }

        protected void onPostExecute(Void resut) {
            // TODO: do something with the feed
        }
    }
于 2015-09-30T08:12:16.677 回答