Originally Posted by stymiee

You are missing a ')'

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die(mysql_error())

die(mysql_error())

Originally Posted by boo_lolly

i don't see anything wrong with it... what's the deal? this is the results page that displays the results of a search query.
Parse error: parse error in ../../../../merchandise/testreg/results.php on line 21

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    @ $db = mysql_connect("localhost", "apache", "R3GP@SS!");
 
    if(!$db)
    {
    echo "Error: Could not connect to the database. Please try again later.";
    exit;
    }
 
    mysql_select_db("registry_DB, $db);
    $sql = mysql_query("SELECT brideLname, groomLname FROM my_search_table WHERE brideLname LIKE '%". $lname ."%' OR groomLname LIKE '%". $lname ."%'") or die(mysql_error();//<--- LINE 21 (entire SQL statement is on one line
    $result = mysql_query($sql);
    $num_result = mysql_num_rows($result);
 
 
    echo "Number of matches: ". $num_result ."<br />";
 

@ $db = mysql_connect("localhost", "apache", "R3GP@SS!"); if(!$db) { echo "Error: Could not connect to the database. Please try again later."; exit; } mysql_select_db("registry_DB, $db); $sql = mysql_query("SELECT brideLname, groomLname FROM my_search_table WHERE brideLname LIKE '%". $lname ."%' OR groomLname LIKE '%". $lname ."%'") or die(mysql_error();//<--- LINE 21 (entire SQL statement is on one line $result = mysql_query($sql); $num_result = mysql_num_rows($result); echo "Number of matches: ". $num_result ."<br />";