nozzy

Используйте
select * from table for update;
update table ....;

select 
t.id, t.date
from table1 t
where t.date >= ( select max(date) from table1 where id = t.id )

Можно сделать на 2 таблицах.
Таблица user_transport:
user_id | transport_id

Таблица transport:
id | param | value

1 | transport_type | bus
1 | name_param1 | value1
1 | name_param2 | value2
и тд
2 | transport_type | metro
2 | name_param1 | value1
2 | name_param2 | value2
и тд

SELECT * FROM T1
LEFT JOIN
(
   SELECT 
   DISTINCT item_id 
   FROM T2 
   WHERE itype='A'
) t ON t.item_id = T1.item1
   AND t.item_id = T1.item2
   AND t.item_id = T1.item3

Это обычная проверка.
Поэтому к этой странице нельзя будет обратиться напрямую - по url, можно только из php-кода.

id, parent_id, name
в parent_id пишите id дисциплины, которую нужно изучить.

Без оконных функций:

select 
id,
Column1,
Column2
from your_table
where id in (
  select min(id) from your_table 
  group by Column2
  )
order by id;

Не проверял:

select 
u.id,
u.user_login,
p.count_posts  
from wp_users u  
inner join 
(
    select  
    post_author, 
    count(*) as count_posts  
    from wp_posts
    where post_type = 'post'
    group by post_author  
) p on p.post_author = u.id
order by u.id,
p.count_posts

На PHP и MySQL, не проверял:

//где находится клиент
$lat = клиент latitude;
$lng = клиент longitude;
//sql-запрос
SELECT
    ROUND(
		6371  * ACOS(SIN( $lat *PI()/180 ) * SIN( $lat *PI()/180 )
        + COS( $lat *PI()/180 ) * COS( $lat *PI()/180 )  *  COS( ($lng*PI()/180) - ($lng*PI()/180) )   ) 
        , 1)  AS distance,
    latitude,
    longitude
FROM
your_table
ORDER BY distance ASC   
LIMIT 8
;

Попробуйте с PIVOT, не проверял:

select *
from
(
	select 
		p.email,
		cr.character_name,
		cs.class_name
	from players p
	inner join characters cr on cr.player_id = p.id
	inner join classes cs on cs.id = cr.class_id
)
pivot
(      
 MAX(character_name)
	for class_name in ('Priest' as 	"Priest", 
								'Mage' as 	"Mage", 
								'Warrior' as "Warrior", 
								'Rogue' as "Rogue")
)

И с DECODE:

select 
    p.email,
 decode(class_name, 'Priest', character_name, null)  as "Priest",
 decode(class_name, 'Mage', character_name, null)  as "Mage",
 decode(class_name, 'Warrior', character_name, null)  as "Warrior",
 decode(class_name, 'Rogue', character_name, null)  as "Rogue"
  from players p
  inner join characters cr on cr.player_id = p.id
  inner join classes cs on cs.id = cr.class_id

Попробуйте заменить break на continue.

SELECT (@row:=@row+1) AS ROW, id
FROM Users
,(SELECT  @row := 0) r   
order by ID

SELECT
GROUP_CONCAT (id),
DATE_FORMAT(date_from, '%m-%d-%Y') AS date_from_date,
DATE_FORMAT(date_to, '%m-%d-%Y') AS date_to_date
FROM `task`
GROUP BY DATE_FORMAT(date_from, '%m-%d-%Y') ,
DATE_FORMAT(date_to, '%m-%d-%Y')