0

我有问题可以请任何人帮助我。

我有一个 php 文件，我从我的 java 类文件的 Asynctask 中调用它。在异步任务中，我发送三个变量电子邮件、密码和密码。

正在发生的事情是，当我使用硬编码值运行我的 php 时，它会给我正确的结果。

结果是：

Inside 1st if 
Inside 2nd if 
Verified 
main if

但是当我尝试通过代码运行我的php时，它给了我错误的结果

结果是：

Inside 1st if 
Invalid
main if

我无法理解为什么会发生这种情况，请指导我。

我的 PHP 文件

<?php

require 'DbConnect.php';

    $i=1;
    $Password = $_POST["password"];
    $Email = $_POST["email"];
    $Pin = $_POST["pin"];
    //$KeyCode = $_REQUEST["key"];


        if((isset($_POST["password"])) && (isset($_POST["email"])) && (isset($_POST["pin"])))
        {
            $query4 = ("SELECT seller_id, name, email, password, verification_Pin, verification_Code, created_Date  FROM `seller` WHERE email = '$Email'");
            $query_run = mysql_query($query4);
            $row=mysql_fetch_row($query_run);

            $int=$row[0];
            $strName=$row[1];
            $strEmail=$row[2];
            $strPwd=$row[3];                
            $strPin=$row[4];

            echo $Pin;
            echo $Password;
            echo $Email;
            echo $int;
            echo $strEmail;
            echo $strPwd;
            echo $strPin;




            if(($Email==$strEmail) && ($Password==$strPwd) && ($Pin==$strPin))
            {
                global $i;
                $i=2;
                $id=updateValidation($int);
                echo $id;
                if($id==1)
                {
                    echo "Verified";
                }
                else
                {
                    echo "Not Verified";
                }
            }
            else
            {
                echo "Invaild";
            }
        }
        else
        {
            echo "Values not set";
        }



function updateValidation($sid)
{
    global $i;
    if($i==2)
    {
        echo "Inside Update vAlidation";
        $queryUpdate = ("UPDATE `seller` SET verification_Pin = 0, verification_Code = 'Verified', created_Date = CURDATE() where seller_id='$sid'");

        if(mysql_query($queryUpdate))
        {
            return 1; 
        }
        else
        {
            return 2; 
        }
    }
    else
    {
        echo "i not 2";
    }
}

?>

我的班级文件：

Button ok = (Button) myDialog
                            .findViewById(R.id.button1);
                    et_pin = (EditText) myDialog
                            .findViewById(R.id.editText1);
                    ok.setOnClickListener(new OnClickListener() {
                        public void onClick(View v) {
                            Toast.makeText(getApplicationContext(),
                                    "CLICKED OK", Toast.LENGTH_LONG).show();
                            pin = et_pin.getText().toString();
                            Toast.makeText(
                                    getApplicationContext(),
                                    "email,pass,pin= " + str1 + "," + str2
                                            + "," + pin, Toast.LENGTH_LONG)
                                    .show();
                            new App_pin_Task().execute(FILENAME_pin);
                            // Intent intent = new
                            // Intent(Dealer_details.this,
                            // Login.class);
                            // startActivity(intent);
                        }
                    });

public class App_pin_Task extends AsyncTask<String, Integer, String> {

    @Override
    protected void onPreExecute() {
        super.onPreExecute();

    }

    @SuppressLint("NewApi")
    @Override
    protected void onPostExecute(String result) {
        super.onPostExecute(result);

        Toast.makeText(getApplicationContext(),
                "Inside App_pin_Task post Execute(Result)=" + result,
                Toast.LENGTH_LONG).show();

        if (result.contains("Invalid")) {
            et_pin.setText("");
        } else {
            Intent myIntent = new Intent(Login.this, UserActivity.class);
            startActivity(myIntent);
        }

    }

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

    @Override
    protected String doInBackground(String... params) {
        // String is = null;

        HttpClient httpclient = new DefaultHttpClient();
        HttpPost httppost = new HttpPost(
                "http://animsinc.com/verifyEmail.php");
        try {
            List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(
                    4);
            nameValuePairs.add(new BasicNameValuePair("email", str1));
            nameValuePairs.add(new BasicNameValuePair("password", str2));
            nameValuePairs.add(new BasicNameValuePair("pin", pin));
            httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
            httpclient.execute(httppost);

            HttpResponse response = httpclient.execute(httppost);
            HttpEntity entity = response.getEntity();
            is = EntityUtils.toString(entity);

        } catch (ClientProtocolException e) {
            // TODO Auto-generated catch block
        } catch (UnsupportedEncodingException e) {
            e.printStackTrace();
        } catch (IOException e) {
            // TODO Auto-generated catch block
        }

        return is;
    }

}
4

3 回答 3

0

补充fayeq-ali-khan的答案，我还将执行以下操作；

首先使用 post 并添加 html 特殊字符以确保安全。

htmlspecialchars => 将特殊字符转换为 HTML 实体

$Password = htmlspecialchars($_POST['password'],ENT_QUOTES);
$Email = htmlspecialchars($_POST['email'],ENT_QUOTES);
$Pin = htmlspecialchars($_POST['pin'],ENT_QUOTES);        
$KeyCode = htmlspecialchars($_POST['key'],ENT_QUOTES);

同样在您的 android 活动中，在您发送字符串之前，修剪该值以确保您没有传输也可能与 PHP 混淆的空字符是一个好主意

nameValuePairs.add(new BasicNameValuePair("email", str1.trim()));
nameValuePairs.add(new BasicNameValuePair("password", str2.trim()));
nameValuePairs.add(new BasicNameValuePair("pin", pin.trim()));

我希望它对你有所帮助

于 2013-09-17T06:07:12.100 回答
0

尝试这个！

改变这个

$Password = $_REQUEST["password"];
$Email = $_REQUEST["email"];
$Pin = $_REQUEST["pin"];
$KeyCode = $_REQUEST["key"];

获取请求

$Password = $_GET["password"];
$Email = $_GET["email"];
$Pin = $_GET["pin"];
$KeyCode = $_GET["key"];

或用于发布请求

$Password = $_POST["password"];
$Email = $_POST["email"];
$Pin = $_POST["pin"];
$KeyCode = $_POST["key"];

或者

"SELECT seller_id, name, email, password, verification_Pin,
verification_Code, created_Date  FROM seller WHERE email = '".$Email."'"
于 2013-09-17T05:13:07.350 回答
0

什么被退回

$Password = $_REQUEST["password"];
$Email = $_REQUEST["email"];
$Pin = $_REQUEST["pin"];
$KeyCode = $_REQUEST["key"];

您的服务器配置可能禁用了此选项并且不返回任何内容。

无论如何，您都不应该使用 $_REQUEST，因为您永远无法确定数据的实际来源：$_POST、$_GET 或 $_cookie。

于 2013-09-17T05:10:39.360 回答