我无法在php中找到链接中的数据

Question

I try to make linke and when I click to that link I want to get some data from database in new page (details.php) but when I try to do this the (details.php)page it open empty without anything!!

the link code is:

         <a href='details.php?pro_id=$pro_id' sytle='float:left;'>Details</a>

and the php code in details.php page is:

<?php


    if(isset($_Get['pro_id'])){

        $Product_id= $_Get['pro_id'];

    $get_pro= "select * from products where Product_id = '$Product_id'";

 $run_pro= mysqli_query($con, $get_pro);

 while($row_pro= mysqli_fetch_array($run_pro)){

     $pro_id=$row_pro['Product_id'];
     $pro_title=$row_pro['Product_title'];
     $pro_price=$row_pro['Product_price'];
     $pro_image=$row_pro['Product_image'];
     $pro_desc=$row_pro['Product_desc'];

     echo "

     <div id='single_product'>

     <h3>$pro_title</h3>

     <img src='admin_area/product_images/$pro_image' width='400px' height='300px' />

     <p><b> Price  $pro_price RO</b></p>

     <a href='index.php?pro_id=$pro_id' sytle='float:left;'>Go Back</a>

     <a href='index.php?pro_id=$pro_id'><button style='float:right';>Add to Cart</button></a>

     </div>
     ";

 }
    }


 ?>

Ican not find where is my mistake

Answer 1

Variables are case sensitive in PHP and the correct name for the get array is $_GET so amend you code like this

if(isset($_GET['pro_id'])){