How to display query results in a table

[Agreaves]

I created this simple shopping system but it wont show the result of the query in the table, what am I doing wrong?

<?php
// Product Id from URL
$product_id = $_GET['Id'];
// Action from URL 
$action = $_GET['action'];

//if there is an product_id and that product_id doesn't exist display an error message
if($product_id && !productExists($product_id)) {
    die("Error. Product Doesn't Exist");
}

switch($action) { //decide what to do 

    case "add":
         $_SESSION['cart'][$product_id]++; //add one to the quantity of the product with id $product_id 
    break;

    case "remove":
         $_SESSION['cart'][$product_id]--; //remove one from the quantity of the product with id $product_id 
        if($_SESSION['cart'][$product_id] == 0) unset($_SESSION['cart'][$product_id]); //if the quantity is zero, remove it completely (using the 'unset' function) - otherwise it will show zero, then -1, -2 etc when the user keeps removing items. 
    break;

    case "empty":
        unset($_SESSION['cart']); //unset the whole cart, i.e. empty the cart. 
    break;

}
?>
<?php 

if($_SESSION['cart']) { //if the cart isn't empty
//show the cart

    echo "<table border=\"1\" padding=\"3\" width=\"40%\">"; //format the cart using a HTML table

    //iterate through the cart, the $product_id is the key and $quantity is the value
     foreach($_SESSION['cart'] as $product_id => $quantity) { 

        //get the name, description and price from the database - this will depend on your database implementation.
        //use sprintf to make sure that $product_id is inserted into the query as a number - to prevent SQL injection
         $sql = sprintf("SELECT Name, Description, Price FROM products WHERE Id = %d;", $product_id);

        $result = mysqli_query($con,$sql) or die("couldnt run query");

	 $row = mysqli_num_rows($result);

	 $row2 = mysqli_fetch_assoc($result);

	 $name = $row2['Name'];
	 $description = $row2['Description'];
	 $price = $row2['Price'];

        //Only display the row if there is a product (though there should always be as we have already checked)
         if($row > 0) {

            list($name, $description, $price) = mysqli_fetch_row($result);

            $line_cost = $price * $quantity; //work out the line cost
             $total = $total + $line_cost; //add to the total cost

            echo "<tr>";
             //show this information in table cells
             echo "<td align=\"center\">$name</td>";
             //along with a 'remove' link next to the quantity - which links to this page, but with an action of remove, and the id of the current product
             echo "<td align=\"center\">$quantity <a href=\"$_SERVER[php_SELF]?action=remove&id=$product_id\">X</a></td>";
            echo "<td align=\"center\">$line_cost</td>";

            echo "</tr>";

        }

    }

    //show the total
     echo "<tr>";
    echo "<td colspan=\"2\" align=\"right\">Total</td>";
    echo "<td align=\"right\">$total</td>";
    echo "</tr>";

    //show the empty cart link - which links to this page, but with an action of empty. A simple bit of javascript in the onlick event of the link asks the user for confirmation
     echo "<tr>";
    echo "<td colspan=\"3\" align=\"right\"><a href=\"$_SERVER[php_SELF]?action=empty\" onclick=\"return confirm('Are you sure?');\">Empty Cart</a></td>";
    echo "</tr>"; 
    echo "</table>";



}else{
//otherwise tell the user they have no items in their cart
     echo "You have no items in your shopping cart.";

}
 //function to check if a product exists
 function productExists($product_id) {
		 //use sprintf to make sure that $product_id is inserted into the query as a number - to prevent SQL injection
		 $sql = sprintf("SELECT * FROM php_shop_products WHERE id = %d;",
						 $product_id); 

		 return mysql_num_rows(mysql_query($sql)) > 0;
 }
?>

[jazzman1]

Hi Agreaves, your topic sounds to me like - How to make a good song?

I don't want to be rude, but you posted here tons of code without to make even little debugging and expect from us to give you a correct answer.

My question is, have you tried dumping the code, if so what errors did you get?

[Agreaves]

I dont know how to dump the code, im not receiving any error messages, it seems to run the query properly but right where it is suppose to echo with result instead it echos a blank table.

[Jessica]

Turn on error reporting set to E_ALL. Also show us the output of the generated page.

I'm going to guess $row=0, so it's not getting into the cells part.

[Agreaves]

I just did a print_r on the rows found and the result returned 0 so I dont know whats wrong with the query.

[Jessica]

$sql = sprintf("SELECT Name, Description, Price FROM products WHERE Id = %d;", $product_id);

echo "SQL: $sql";

 

What's it output?

[Agreaves]

the results returned "SQL: SELECT Name, Description, Price FROM products WHERE Id = 0;"

[Jessica]

Well I can only assume you don't have any products with an ID of 0 right?

When you added the products to your cart in the session, you set the ID incorrectly, or not at all. Go back there.

 

And start trying to debug it yourself. Whenever you set a variable, echo it to see if it's what you expected.