luna3956

select *
from Messages
where
    user_from_id = $id_пользователя or
    user_to_id = $id_пользователя
order by id

select *
from posts p
left join (
  select post_id
  from actions a
  where
      a.user_id = 8 and
      a.action_type = 1
) a on
    a.post_id = p.id
where
    p.author_id = 8 or
    a.post_id is not null

select 
  p.id,
  (CASE WHEN t1.num = 0 THEN 'no' ELSE 'yes' END) as 'match_res' 
from product p
inner join (
  SELECT product_id, COUNT(product_detail_id) as num
  FROM products_details 
  WHERE 
    product_detail_type_id IN (1, 4, 123)
  group by product_id
) t1 on
  t1.product_id = p.id

SELECT 
    SUM(CASE WHEN day='monday' THEN 1 ELSE 0 END) as "понедельники"
from (select 'monday'::varchar as day) t1

`post_id`=73 AND (`meta_key`='_regular_price' OR `meta_key`='_sale_price')

select
      c.cname, o.amt
  from Orders o
  inner join Customers c on
      c.cnum = o.cnum
  where
      amt = (select max(amt) from Orders)

class House {

  string address;
  int numOfRooms;
  ...

  House(string address, int numOfRooms)
  {
    this.address = address;
    this.numOfRooms = numOfRooms;
  }

}

class Dog {

  string name;
  string breed;

  Dog(string name, string breed)
  {
    this.name = name;
    this.breed = breed;
  }

}

ref.OrderBy(KEY_PRIORITY).whereGreaterThan(key_priority, 1)

collectionReference.whereGreaterThan(KEY_PRIORITY, 1)

collectionReference.whereEqualTo("firstLetter", "A").whereGreaterThan(KEY_PRIORITY, 1);

...
inner join (select distinct hotelId,price from Tours)

select
    id, count(sex) as cnt
from Tab
where room=2 and sex=2
group by id
having count(sex) <= 6

tasks
 taskKey1
    taskID
    taskName
    solved
    ...
  taskKey2
    taskID
    taskName
    solved
     ....