显示sql数据库问题的信息

我正在制作一个简单的购物车应用程序。主页面显示产品列表。当你点击产品名称时，它会把你带到一个只显示该产品信息的页面。现在，它没有显示任何内容。这是我的代码：

<?php
    $sql = "SELECT product_id, image_path, name, description, price FROM products;";
    $result = mysql_query($sql);
    while(list($product_id, $image_path, $name, $description, $price) = mysql_fetch_row($result)) {
        echo "<tr>";
            echo "<td><img src=$image_path height=200 width=160 border=1px /></td>";
            echo "<td><b><a href='"getprod.php?product=" .
                  $product_id['product_id'] . "'">$name</a></b><br/ >";
            echo "$description<br/ >";
            echo "$$price<br/ >";
            echo "<input type='button' value='Add to Cart' onclick='addtocart'/>"; 
        echo "</tr>";
    }
?>

我的productInfo方法应该显示我的信息：

function productInfo($product_id) {
    global $conn;
    if($product_id) {
        $sql= "SELECT product_id,image_path, name, description, price " . 
              "FROM products WHERE product_id = " . $product_id; " . 
        $result= mysql_query($sql, $conn) or
            die('Could not find product info: ' . mysql_error());
        while(list($product_id, $image_path, $name, $description, $price) = mysql_fetch_row($result)) {
            echo "<td><img src=$image_path height=200 width=160 border=1px /></td>";
            echo "<td><b><a href='"getprod.php?product=" .
                  $product_id['product_id'] . "'">$name</a></b><br/ >";
            echo "$long_description<br/ >";
            echo "$$price<br/ >";
            echo "<input type='button' value='Add to Cart' onclick='addtocart'/>"; 
        }
        }
    }

如有任何帮助或推动，我们将不胜感激！