如何在php中獲取用戶的詳細信息

Question

我使用PHP開發了一個PayPal集成頁面。付款和成功頁面也在工作，但我沒有收到用戶的電子郵件。

home.php頁面

<?php 
    ob_start(); 
     session_start(); 
     require_once 'dbconfig.php'; 
     error_reporting(0); 
    // if session is not set this will redirect to login page 
    if(!isset($_SESSION['user_id'])) { 
     header("Location: index"); 
     exit; 
      } 
    $user_id=$_SESSION['user_id']; 
    $row=mysqli_fetch_array(mysqli_query($conn,"select *from `at_reg_user` 
    where `email`='$user_id'")); 
    $name=$row['fname'].' ' .$row['lname']; 
     ?> 

在home.php頁面

PayPal按鈕

 <form action="https://www.sandbox.paypal.com/cgi-bin/webscr" method="post" target="_top"> 
           <input type="hidden" name="cmd" value="_s-xclick"> 
           <input type="hidden" name="hosted_button_id" value="YYHR897NQVQCY"> 
           <input type="image" src="https://www.sandbox.paypal.com/en_US/i/btn/btn_buynowCC_LG.gif" border="0" name="submit" alt="PayPal - The safer, easier way to pay online!"> 
           <img alt="" border="0" src="https://www.sandbox.paypal.com/en_US/i/scr/pixel.gif" width="1" height="1"> 
        </form> 

這是我的網頁成功頁面：

<?php 
include 'dbconfig.php'; 
ob_start(); 
session_start(); 
if(!isset($_SESSION['user_id'])) { 
header("Location: index"); 
exit; 
    } else { 
$user_id = $_SESSION['user_id']; 
$row = mysqli_fetch_array(mysqli_query($conn, "select * from `at_reg_user` where `email`='$user_id")); 
$email = $row['email']; 
    } 


$txn_id = $_GET['tx']; 
$payment_gross = $_GET['amt']; 
$currency_code = $_GET['cc']; 
$payment_status = $_GET['st']; 

if($_GET['st']=="Completed") { 

$txn_id = $_GET['tx']; 
$payment_gross = $_GET['amt']; 
$currency_code = $_GET['cc']; 
$payment_status = $_GET['st']; 

//select statement for disabling user to perform back button 

$result = mysqli_fetch_array(mysqli_query($conn, "SELECT txn_id FROM `payments` where `txn_id`='$txn_id'")); 
if ($result > 0) { 
    echo("<SCRIPT LANGUAGE='JavaScript'> 
         window.alert(' Sorry you have already done payment') 
         </SCRIPT>"); 
    } else { 

    $query = mysqli_query($conn, "INSERT INTO `payments`(`txn_id`, `payment_gross`, `currency_code`, `payment_status`) VALUES ('$txn_id','$payment_gross','$currency_code','$payment_status')"); 

     if ($query) { 
     echo $email; 
     echo "<br/>"; 
     echo $txn_id; 
     echo "<br/>"; 
     echo $payment_gross; 
     echo "<br/>"; 
     echo $currency_code; 
     echo "<br/>"; 
     echo $payment_status; 
     echo "<br/>"; 
    } else { 
     echo("<SCRIPT LANGUAGE='JavaScript'> 
         window.alert(' Payment Successful but not inserted') 
         </SCRIPT>"); 
    } 
    } 

} 


     echo "<h4>Payment done Successfully. Note down the above transaction details for further assistance </h4>"; 
     header("Refresh:10; url=home"); 
    ?> 

當我嘗試打印電子郵件回聲$email，它不打印。我需要獲得emailid才能將其插入到付款表中。

Answer 1

我認爲，MySQL查詢中存在語法錯誤。看看你的查詢結束。我想，有一個單引號缺失。這是您的查詢：

$row = mysqli_fetch_array(mysqli_query($conn, "select * from `at_reg_user` where `email`='$user_id")); 

嘗試此查詢：

$row = mysqli_fetch_array(mysqli_query($conn, "select * from `at_reg_user` where `email`='$user_id'")); 

提示：未經測試。 我認爲你應該重寫你的整個代碼，以確保它的安全。它充滿了錯誤。

Answer 2

我建議看看PayPal的Payment Data Transfer。它更安全，因爲您不依賴$ _GET數據。