Question

I’m learning PHP and MySQL to get a hang of server-side scripting and for personal interest/development. The first ‘program,’ beyond ‘hello world’ seems to be writing ones’ own blog.

This is a little beyond me yet and so thought I’d start with a contacts-database type. So far so good, but (and there’s always the but!) I’m trying to format a presentation in the form that the contact’s name is formatted as a mailto: link.

$query = mysql_query('     SELECT CONCAT(fname,', ',sname) AS name, email.address AS email     FROM contacts, emails      WHERE contacts.contactID=email.contactID');  while ($row = mysql_fetch_array($query)) {   echo '<li><a href=\'mailto' . $row['email'] . '\'> . $row['name'] . '</a></li>';  } 

This is fine until I dare to enter a contact that has no email address in the Db, at which point the obvious lack of an email.contactID to equal the contacts.contactID rears its head.

I realise that I should be able to deal with this but…I can’t think of how (yay to insomnia! @.@ ). Any suggestions would be appreciated.

Answer 1

You want a left (aka left outer) join.

select concat(c.fname, ' ', c.sname) as name, e.address as email from contacts c left join emails e on (c.contactID = e.contactID)