What's wrong with this query?

I did not…thanks. Been kind of busy at work and not posting a lot. I tried a few new things and I’ll post them to update everyone.

Here is the validation form for the admin login:

[php]<?php
session_start();
include ("…login.php");
login();
$userid = $_POST[‘userid’];
$password = $_POST[‘password’];
$query = “SELECT * from admins where userid = ‘$userid’ and password = (’$password’)”;
$result = mysql_query($query);
if (mysql_num_rows($result) == 0)
{
echo “

Sorry, your account was not validated.


\n”;
echo “<a href=“admin.php”>Try again
\n”;
} else
{
$_SESSION[‘store_smalladmin’] = $userid;
//THE NEXT LINE DIRECTS SUCCESSFUL LOGINS WHERE TO GO NEXT, IN THIS CASE TO THE ADMIN PAGE
//This function must appear before any HTML code is sent to the browser.
//You can’t have any echo statements before the header() function.
header(“Location: admin.php”);
}
?>[/php]

As you can see, the cookie is set holding the userid. Here is the page that is referred to after login:

[php]<?php
session_start();
?>

<?php include("../adminlogin.php"); include("../showproducts3.php"); login(); ?>
<?php if (!isset($_REQUEST['content'])) {

if (!isset($_SESSION[‘store_admin’]))
include(“adminlogin.html”);
else
include(“adminmain.inc.php”);
}
else
{
$content = $_REQUEST[‘content’];
$nextpage = $content . “.inc.php”;
include($nextpage);
}
?>

[/php]

I’m having an error that states that the ‘nextpage’ is not defined. I’ve included a page that is referenced in the above page that may or may not have something to do with it. It’s the showproducts3 page that is referenced:

[php]<?php

function showproducts3($custid, $page, $currentpage, $newpage)
{
$query = “Select count(prodid) from products where custid = $custid”;
$result = mysql_query($query);
$row = mysql_fetch_array($result);
if ($row[0] == 0)
{
echo “


Sorry, there are no entrees in this category

\n”;
}
else
{
$thispage = $page;
     $recordsperpage = 5;
     $offset = ($thispage - 1) * $recordsperpage;
     $totpages = ceil($totrecords / $recordsperpage);

     echo "<table width=\"100%\" cellpadding=\"1\" border=\"1\">\n";


     echo "<tr><td><h2>Image</h2></td>\n";
     echo "<td><h2>Name</h2></td>\n";
     echo "<td><h2>Price</h2></td>\n";
     echo "<td><h2>Store </h2></td>\n";
 echo "<td><h2>Phone</h2></td>\n";
     echo "<td><h2>Description</h2></td>\n";
     echo "<td><h2>Special</h2></td></tr>\n";

     $query = "SELECT * from products WHERE custid=$custid LIMIT $offset,$recordsperpage";
     $result = mysql_query($query);
     while($row=mysql_fetch_array($result, MYSQL_ASSOC))
     {

//not really concerned with this area right now as these below can change
$custid = $row[‘custid’];
$prodid = $row[‘prodid’];
$description = $row[‘description’];
$price = $row[‘price’];
$name = $row[‘name’];
$onsale = $row[‘onsale’];

        echo "<tr><td>\n";
           echo "<img src=\"showimage.php?id=$prodid\" width=\"80\" height=\"60\">";
        echo "</td><td>\n";


           echo "<b><a href=\"$newpage&id=$prodid\">$name\n</b>";
        echo "</td><td>\n";
        echo "$" . $price . "\n";
        echo "</td><td>\n";
         echo "<a href=\"index.php?&content=showstoreinfo&custid=$custid\">";

        echo "<font size=\"2\"><u>$storename</u></font>\n";


        echo "</td><td>\n";
        echo $phone . "\n";
        echo "</td><td>\n";
        echo $description . "\n";
        
        echo "</td><td>\n";
        if ($onsale)
           echo "On sale!\n";
        else
           echo "&nbsp;\n";
        
     }


     echo "</table>\n";

// Code to implement paging
if ($thispage > 1)
{
$page = $thispage - 1;
$prevpage = “<a href=”$currentpage&custid=$custid&page=$page">Previous page";
} else
{
$prevpage = " ";
}

     if ($thispage < $totpages)
     {
        $page = $thispage + 1;
        $nextpage = "<a href=\"$currentpage&custid=$custid&page=$page\">Next page</a>";
     } else
     {
        $nextpage = " ";
     }

     if ($totpages > 1)
        echo $prevpage . "  " . $nextpage;

  }

}
?>[/php]

The error says that next page is not defined…?

Thanks for any help.

Well, traydavid, the following code is where the error is (as you stated!):
{
$content = $_REQUEST[‘content’];
$nextpage = $content . “.inc.php”;
include($nextpage);
}
So, just change this for now so it shows the value of $nextpage and then we can see what is wrong.
(Also, there are no () needed for includes…)
{
$content = $_REQUEST[‘content’];
$nextpage = $content . “.inc.php”;
die ("
" . $nextpage . “
”);
include $nextpage;
}

Tell us the results of where the error is…

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