Error report when all seems to work?

[wudz]

Hi,
I am a newbie to sql/php and having difficulty doing a data check, it carries out the check OK and returns the correct reply if the input email already exists in the db...if a new email is input and is not yet in the db it drops through OK and does the upload and displays the confirmation html.... ALL WELL!!!!, BUT always gives this error:- 😕

Warning: mysql_result(): Unable to jump to row 0 on MySQL result index 18 in /home/www/mysite.com/auction/intro_user.php on line 75and this is the part of the script that does the check and also gives the error:-

// check email if already invited

       $result = mysql_query("SELECT user_id from AUCTION_invitedbuyerslist WHERE list_name = '$friend_email'");
             $intro_id = mysql_result($result,0);
       if ($intro_id > "0" ){
             $friend_email="";
             $friend_name="";
             $TPL_error_text = $ERR_115c; // E-mail friend already introduced
     }[/B]

Forgive me if it is obvious, but I am a Flash graphics bloke now trying to learn SQL and PHP, but at 60+ finding it a bit slow.....a lot slow..hi

All help will be appreciated and explaination. 🙂

John

[rincewind456]

Thats because when you don't have a match in the database for the introduced email,
This line

$intro_id = mysql_result($result,0);

cannot find the first result in the sql query as there is no result.
If you replace it with this

$intro_id = mysql_num_rows($result);

it does the same check to see if there is an entry, but won't throw an error if the number of rows equals zero.