Easy

select firstName ,lastName ,CASE WHEN addressId is NULL then NULL ELSE city END as city ,CASE WHEN addressId is NULL then NULL ELSE state END as state from Person A left join Address B ON A.personId = B.personId;

select a.Name as Employee from Employee a where a.Salary > (select b.Salary from Employee b where Id = a.managerId);

select email as Email from Persons group by email having count(id) >= 2;

select c.name as Customers from Customers c left join Orders o on c.id = o.customerId where o.id is null;

select customers.name as 'Customers' from customers where customers.id not in ( select customerid from orders );

delete a from person a, person b where a.id > b.id # id는 순서값을 가진 칼럼명을 사용한다. and a.email = b.email;

select a.id from weather a, weather b where datediff(a.recordDate, b.recordDate) = 1 and a.temperature > b.temperature;

select player_id ,min(event_date) as first_login from activity group by player_id;

select name from customer where referee_id <> '2' or referee_id is null

select customer_number from orders group by customer_number having count(*) >= all(select count(*) from orders group by customer_number);

select class from courses group by class having count(distinct student) >= 5;

select name from salesperson where sales_id not in (select sales_id from orders where com_id = (select com_id from company where name = 'RED'))

SELECT s.name FROM salesperson s WHERE s.sales_id NOT IN (SELECT o.sales_id FROM orders o LEFT JOIN company c ON o.com_id = c.com_id WHERE c.name = 'RED')

select * from cinema where id % 2 = 1 and description <> 'boring' order by rating desc;

update salary set sex = case when sex = 'f' then 'm' else 'f' end

select actor_id, director_id from actordirector group by actor_id, director_id having count(*) >= 3

select distinct b.product_id, a.product_name from product a inner join sales b on a.product_id = b.product_id where b.product_id not in ( select distinct product_id from sales where sale_date not between '2019-01-01' and '2019-03-31' );

select product_id, product_name from sales join product using(product_id) # 필드 이름이 같기 때문에 group by product_id having min(sale_date) >= '2019-01-01' and max(sale_date) <= '2019-03-31'

select activity_date as day, count(distinct user_id) as active_users from activity where activity_date >= date_sub('2019-07-27',interval 29 day) and activity_date <= '2019-07-27' group by activity_date having count(distinct user_id) > 0

SELECT activity_date AS day , COUNT(DISTINCT user_id) AS active_users FROM Activity WHERE (activity_date > "2019-06-27" AND activity_date <= "2019-07-27") GROUP BY activity_date having count(distinct user_id) > 0;

select distinct author_id as id from views group by author_id having count(if(author_id = viewer_id,1,null)) >= 1 order by id;

select id ,sum(case when month='Jan' then revenue else null end) as Jan_Revenue ,sum(case when month='Feb' then revenue else null end) as Feb_Revenue ,sum(case when month='Mar' then revenue else null end) as Mar_Revenue ,sum(case when month='Apr' then revenue else null end) as Apr_Revenue ,sum(case when month='May' then revenue else null end) as May_Revenue ,sum(case when month='Jun' then revenue else null end) as Jun_Revenue ,sum(case when month='Jul' then revenue else null end) as Jul_Revenue ,sum(case when month='Aug' then revenue else null end) as Aug_Revenue ,sum(case when month='Sep' then revenue else null end) as Sep_Revenue ,sum(case when month='Oct' then revenue else null end) as Oct_Revenue ,sum(case when month='Nov' then revenue else null end) as Nov_Revenue ,sum(case when month='Dec' then revenue else null end) as Dec_Revenue from department group by id;

select u.name as name ,ifnull(sum(r.distance),0) as travelled_distance from users u left join rides r on u.id = r.user_id group by u.id, u.name order by travelled_distance desc, name;

select sell_date ,count(distinct product) as num_sold ,group_concat(distinct product order by product) as products from activities group by sell_date order by sell_date;

select patient_id ,patient_name ,conditions from patients where conditions like 'DIAB1%' OR conditions like '%DIAB1%';

select customer_id ,count(*) as count_no_trans from visits where visit_id not in (select visit_id from transactions) group by customer_id;

select u.name ,sum(t.amount) as balance from users u left join transactions t on u.account = t.account group by u.account having sum(t.amount) > 10000;

select user_id ,concat(upper(left(name,1)),lower(substr(name,2))) as name from users order by user_id;

select date_id ,make_name ,count(distinct lead_id) as unique_leads ,count(distinct partner_id) as unique_partners from dailysales group by date_id, make_name;

select user_id ,count(distinct follower_id) as followers_count from followers group by user_id order by user_id;

select event_day as day ,emp_id ,sum(out_time - in_time) as total_time from employees group by emp_id, event_day;

select product_id from products where low_fats = 'Y' and recyclable = 'Y';

select product_id ,'store1' as store ,store1 as price from products where store1 is not null union all select product_id ,'store2' as store ,store2 as price from products where store2 is not null union all select product_id ,'store3' as store ,store3 as price from products where store3 is not null

select employee_id ,case when name not like 'M%' and employee_id % 2 = 1 then salary else 0 end as bonus from employees order by employee_id;

select user_id ,max(time_stamp) as last_stamp from logins where year(time_stamp) = '2020' group by user_id;

select case when a.salary is null then e_id when a.name is null then s_id end as employee_id from( select e.employee_id as e_id, e.name as name, s.employee_id as s_id, s.salary as salary from employees e left join salaries s on e.employee_id = s.employee_id union select e.employee_id as e_id, e.name as name, s.employee_id as s_id, s.salary as salary from employees e right join salaries s on s.employee_id = e.employee_id ) a where a.salary is null or a.name is null order by employee_id;