Re: Problem With mysql_query

Brian Welter wrote:

I have the following code and it doesn't behave properly but I can't find my error. The code is designed to validate user login. When I pass in an invalid email name to the checkLogin function it returns a result, not FALSE as the documentation describes. I can SELECT * from the table being used and read out all of the data so I know that the DB connection is valid and that I can compose a valid query. I have looked and looked at this code and can find no error, does anyone see an error here. When the check if (!$result) fails i.e. even thought the email/password are not in the db a none zero result is returned, the calls to mysql_fetch_assoc () return empty strings. The only clue I may have is that when I call mysql_free_result($result) I get a warning saying that $result is not a valid resource. I sure I'm doing something stupid but I just can't see it. Any help would be great! <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Untitled Document</title> </head> <body> <?php

            global $db;

                        
            $db = mysql_connect('localhost', 'xxxx', 'xxxx', 'xxxx'); 
            if (!$db) {

                        die("Unable to connect to database: ".

mysql_connect_error());

            }

            $selected = mysql_select_db("xxxx", $db) or die( "Unable to

select database");

            if (checkLogin(sam@123.com', '1')) {

                        echo "Success";

            } else {

                        echo "Failure";

            }

            if(checkLogin('duck', '2') )) {

                        echo "Success";

            } else {

                        echo "Failure";

            }

            function checkLogin($e, $p) {

                        $ans = 0;

                        $emailcheck = $e;

                        $passwdcheck = $p;

                        $result = 0;

                        $query = "SELECT * FROM maillist WHERE
email='$emailcheck' AND passwd='$passcheck'"; 
                        $result = mysql_query($query);
  

Idon't think it's your problem, but you have two different variable--$passwdcheck and $passcheck. After getting the results, you can check how many valid users are in the result, as follows: $n = mysql_num_rows($result); if($n < 1) etc. Janet

                        if (!$result) {

                                    echo "<h4>UserName is FALSE name is:

$e<h4>";

                                    echo "<h4>Password is FALSE name is:

$p<h4>";

                                    $ans = 0;

                        } else {

                                    $row = mysql_fetch_assoc($result);

                                    $em = $row['email'];

                                    $pw = $row['passwd'];

                                    echo "<p>Accepted ";

                                    echo "User $em ";

                                    echo "Password $pw</p>";

                                    $ans = 1;

                        }

                        return($ans);

            } //checkLogin

            mysql_free_result($result);

            mysql_close($db);

?> </body> </html>