hi, can anyone help me spot the parse error?
It's throwing up on the first "if" line, and for the life of me I cannot
find it! :)

tia,
andrew

<?php

include("db_connect_params.inc");
$sql="select path from PHOTO where pid =1";
$link_id = mysql_connect($host, $usr, $pass);
$result = mysql_db_query($database, $sql, $link_id) or die("no result");

if ((mysql_num_rows($result)) = 1)

{
 $row=mysql_fetch_array($result);
 extract($row);
 echo "<img src =\"$path\"></img><br>";

} else {
         if (!isset($ii)) $ii = 1;
         $i = 1;
         while($row=mysql_fetch_array($result))
             {
                extract($row);
                $displayed[$i]=$path;
                echo "<a href=\"$PHP_SELF?ii=$i\">$i</a>&nbsp&nbsp";
                $i++;
             }
         echo "<img src =\"$displayed[$ii]\"></img><br>";
          }
?>


-- 
PHP General Mailing List (http://www.php.net/)
To unsubscribe, e-mail: [EMAIL PROTECTED]
For additional commands, e-mail: [EMAIL PROTECTED]
To contact the list administrators, e-mail: [EMAIL PROTECTED]

Reply via email to