Androidphp mysql

Question

我正在为登录过程创建一个简单的测试应用程序，但当我在emulator中使用浏览器并尝试我使用的url时，我得到的错误是“给我网页不可用”我使用了localhost 127.0.0.2，但它给了我相同的消息，我使用了我电脑的ip相同的消息任何人都可以帮助我？？

package com.sencide;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
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 org.apache.http.protocol.HTTP;
import org.apache.http.util.EntityUtils;

import android.os.Bundle;
import android.os.StrictMode;
import android.app.Activity;
import android.util.Log;
import android.view.Menu;
import android.view.View;
import android.view.View.OnClickListener;
import android.widget.Button;
import android.widget.EditText;
import android.widget.TextView;

public class AndroidLogin extends Activity implements OnClickListener {


     Button ok,back,exit;
     TextView result;

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


        StrictMode.ThreadPolicy policy = new StrictMode.ThreadPolicy.Builder().permitAll().build();
        StrictMode.setThreadPolicy(policy);

     // Login button clicked
        ok = (Button)findViewById(R.id.btn_login);
        ok.setOnClickListener(this);

        result = (TextView)findViewById(R.id.lbl_result);



    }



    public void postLoginData() {
        // Create a new HttpClient and Post Header
        HttpClient httpclient = new DefaultHttpClient();
         Log.e("Responce-->","after httpclient");
        /* login.php returns true if username and password is equal to saranga */
        HttpPost httppost = new HttpPost("http://192.168.1.70/login.php");
        Log.e("Responce-->","after httppost");
        try {
            // Add user name and password
         EditText uname = (EditText)findViewById(R.id.txt_username);
         String username = uname.getText().toString();

         EditText pword = (EditText)findViewById(R.id.txt_password);
         String password = pword.getText().toString();

            List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(2);
            nameValuePairs.add(new BasicNameValuePair("username", username));
            nameValuePairs.add(new BasicNameValuePair("password", password));
            httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
            Log.e("Responce-->","after using the list name pair");

            // Execute HTTP Post Request
            Log.w("SENCIDE", "Execute HTTP Post Request");
            HttpResponse response = httpclient.execute(httppost);
            Log.e("Responce-->","after execute the http response");
          //  String str = inputStreamToString(response.getEntity().getContent()).toString();
            String str = EntityUtils.toString(response.getEntity(), HTTP.UTF_8);
            //Log.w("SENCIDE", str);

            Log.e("Responce-->",""+str);

            if(str.toString().equalsIgnoreCase("true"))
            {
             Log.w("SENCIDE", "TRUE");
             result.setText("Login successful");  
            }else
            {
             Log.w("SENCIDE", "FALSE");
             result.setText(str);            
            }

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



    private StringBuilder inputStreamToString(InputStream is) {
        String line = "";
        StringBuilder total = new StringBuilder();
        // Wrap a BufferedReader around the InputStream
        BufferedReader rd = new BufferedReader(new InputStreamReader(is));
        // Read response until the end
        try {
         while ((line = rd.readLine()) != null) {
           total.append(line);
         }
        } catch (IOException e) {
         e.printStackTrace();
        }
        // Return full string
        return total;
       }

        /* login.php returns true if username and password is equal to saranga */


    @Override
    public void onClick(View view) {
        // TODO Auto-generated method stub
        if(view == ok){
             Thread t = new Thread(){
                    public void run(){
                        postLoginData();
                    }
                };
                t.start();

          }
    }





}

<?php
$host="localhost"; // Host name
$username="root"; // Mysql username
$password="root"; // Mysql password
$db_name="testlogin"; // Database name
$tbl_name="members"; // Table name

// Connect to server and select databse.
mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");

// username and password sent from form
$myusername=$_POST['username'];
$mypassword=$_POST['password'];

// To protect MySQL injection
$myusername = stripslashes($myusername);
$mypassword = stripslashes($mypassword);
$myusername = mysql_real_escape_string($myusername);
$mypassword = mysql_real_escape_string($mypassword);

$sql="SELECT * FROM $tbl_name WHERE username='$myusername' and password='$mypassword'";
$result=mysql_query($sql);

// Mysql_num_row is counting table row
$count=mysql_num_rows($result);

// If result matched $myusername and $mypassword, table row must be 1 row
if($count==1){
echo "true";
}
else {
echo "Login Failed";
}
?>

Answer 1

尝试使用curl（如果您使用unix os）使用来自php服务器的请求访问php或制作一个简单的网页来使用。确保它为您提供了有效的（200）响应。

<html>
  <head>
  </head>
  <body>
    <form action="http://localhost/{{ your-php-script }}" method="POST">
      <input type="text" name="username">
      <input type="text" name="password">
      <input type="submit" value="submit">
    </form>
  </body>
</html>

此外，主机电脑应该10.0.2.2，而不是127.0.0.2（或127.0.0.1）