SELECT id FROM table WHERE name='Vasya'
У многих с JOIN'ами и не одним. И если в коде встречается несколько таких конструкций, то уже неудобно.
Вот пример Invision Board
SELECT p.*,m.id,m.name,m.mgroup,m.email,m.joined,m.avatar,m.avatar_size,m.posts,m.aim_name,m.icq_number,
m.signature, m.website,m.yahoo,m.integ_msg,m.title,m.hide_email,m.msnname, m.warn_level, m.warn_lastwarn,
g.g_id, g.g_title, g.g_icon, g.g_dohtml, pc.*
FROM ibf_posts p
LEFT JOIN ibf_members m ON (p.author_id=m.id)
LEFT JOIN ibf_groups g ON (g.g_id=m.mgroup)
LEFT JOIN ibf_pfields_content pc ON (pc.member_id=p.author_id)
WHERE p.topic_id=1 AND p.queued != 1
ORDER BY p.pid LIMIT 0, 20
или вот:
SELECT on_resume.*, on_users.subscribed, on_users.user_avatar, on_resume_page.*
FROM on_resume
LEFT JOIN on_users ON (on_resume.resume_userid = on_users.user_id )
LEFT JOIN on_resume_page ON ( on_resume.resume_userid = on_resume_page.resume_userid)
WHERE on_resume.active= '1'
GROUP BY on_resume.rid
ORDER BY on_resume.rid DESC
LIMIT 0,18