Avanceret søgning i mySQL

Kode 

// LEFT JOIN, hvis et forum_posts.users ikke matcher users.id, and får users.name værdien NULL
SELECT forum_posts.*, users.name
FROM forum_posts
LEFT JOIN(users) ON (users.id = forum_posts.users)
WHERE forum_posts.postby = '" . userid() . "'
AND forum_posts.posttime > 1200654188


// INNER JOIN : begge eksempler giver samme resultat
SELECT forum_posts.*, users.name
FROM (forum_posts)
INNER JOIN(users) ON (users.id = forum_posts.users)
WHERE forum_posts.postby = '" . userid() . "'
AND forum_posts.posttime > 1200654188

SELECT forum_posts.*, users.name
FROM (forum_posts,users)
WHERE forum_posts.postby = '" . userid() . "'
AND forum_posts.posttime > 1200654188
AND users.id = forum_posts.users

Kode 

 
mysql_query("SELECT * FROM `usertable` JOIN `forumtable` ON `usertable`.`id` = `forumtable`.`postBy` WHERE `forumtable`.`postTime` > '2008-01-13 12:28:24' AND `usertable`.`userName` LIKE '%" . $searchString . "%'");
 

Kode 

SELECT *, count(forum_posts.id) AS count
FROM users
LEFT JOIN (forum_posts) ON (users.id = forum_posts.users)
WHERE users.name LIKE '%jens%'
HAVING count >= 5

Kode 

$query = mysql_query('SELECT * FROM table)
while($row = mysql_fetch_assoc($query) {
    $result[] = $row;
}

for($i=0,$len=sizeof($result);$i<$len;$i++) {
    print $result[$i]['topic'];
}