I have been building the shopping cart chapter 14 p 467, and cant figure out the code to include the images for the products i want to appear when viewing the store front.I have lookes through the forums but to no avail. Help!!

echo "<a href = 'getprod.php?prodid=" . $prodnum ."'>";
         echo "THUMBNAIL<br>IMAGE</td></a>";

In reply to your concerns and assuming that you have all of your other code in place where it should be, one thing that I see is that you may be having trouble with your syntax. You may want to try different variations of the code you have and change it to what I have put below...

This is the code you have...

echo "<a href = 'getprod.php?prodid=" . $prodnum ."'>";
         echo "THUMBNAIL<br>IMAGE</td></a>";

You may want to try this...

echo "<a href='getprod.php?prodid=$prodnum'>";
         echo "THUMBNAIL<br>IMAGE</td></a>";

or this...

echo "<a href='getprod.php?prodid="$prodnum"'>";
         echo "THUMBNAIL<br>IMAGE</td></a>";

Hope this helps!

Another thing that I have done to help with setting variables is that I set a variable equal to a value, where the value is also a variable. Pay close attention to the variable $prodid, positioned just under $db in the code below.

EXAMPLE...

<?php
    include("connectme.inc");
    $connection = mysql_connect($host,$user,$password)
        or die ("Couldn't connect to server");
    $db = mysql_select_db($database,$connection)
        or die ("Couldn't select database");
    $prodid = "$prodid";
    $query = "SELECT * FROM tblProducts WHERE prodid='$prodid'";
    $result = mysql_query($query)
        or die ("Couldn't execute query")
        //****The rest of your PHP code goes here****//
?>

Correction!

I'm sorry. I'm just a beginner at this myself, so please bare with me. I don't think I put the right variable and value.

I believe it should be like this...

<?php
    include("connectme.inc");
    $connection = mysql_connect($host,$user,$password)
        or die ("Couldn't connect to server");
    $db = mysql_select_db($database,$connection)
        or die ("Couldn't select database");
    $prodnum = "$prodnum";
    $query = "SELECT * FROM tblProducts WHERE prodnum='$prodnum'";
    $result = mysql_query($query)
        or die ("Couldn't execute query")
        //****The rest of your PHP code goes here****//
?>

Good Luck!