我在mysql中有一个函数，叫做：

DB::raw("count_adults(rooms.id) as adults"),

伯爵大人是这样的：

BEGIN

 DECLARE adults INT;

    SELECT  
     count(*) INTO adults 
    FROM 
     clients 
    WHERE 
     clients.room_id = r_id 
        and (age >= 18 
        or age = 0);
            RETURN adults;
    END

但是，如果此查询不返回任何结果（0），请执行另一个select或函数，如下所示：

SELECT count(*) INTO adults  FROM client_room, clients    
WHERE client_id = clients.id and client_room.room_id = r_id and (age >18 or age = 0)

我可以创建一个if条件吗？在php代码中？或者在查询中？像这样：

IF EXISTS (SELECT 1 FROM clients WHERE clients.room_id = r_id) 

BEGIN

DECLARE adults INT;

    SELECT  
        count(*) INTO adults 
    FROM 
        clients 
    WHERE 
        clients.room_id = r_id 
        and (age >= 18 
        or age = 0);

   RETURN adults;
END
ELSE 

BEGIN
DECLARE adults INT;
SELECT count(*) INTO adults 
FROM 
client_room, clients
WHERE 
client_id = clients.id and client_room.room_id = r_id 
and (age >18 
or age = 0);

END

我试过这个方法，似乎很管用：

BEGIN

 DECLARE adults INT;

    SELECT COUNT(*) INTO adults
    FROM clients
    WHERE clients.room_id = r_id
     and (age >= 18 or age = 0);

    RETURN IF( adults>0, adults, (SELECT count(*) FROM client_room, clients
         WHERE client_id = clients.id
         and client_room.room_id = r_id)
             );

   END