[SQL]문법정리4(groupby,orderby)

SELECT *
FROM users u
group by name ; 이** abc3b3@hanmail.com

select name,
count(*)
from users
group by name ;

SELECT payment_method ,
COUNT(*)
FROM orders o
group by payment_method;

SELECT payment_method ,
COUNT()
FROM orders o
WHERE course_title ='웹개발 종합반'
group by payment_method;
SELECT payment_method ,
COUNT()
FROM orders o
WHERE course_title ='웹개발 종합반'
group by payment_method
order by COUNT(*) ASC ;

SELECT payment_method ,
COUNT()
FROM orders o
WHERE course_title ='웹개발 종합반'
group by payment_method
order by COUNT() ;

SELECT payment_method ,
COUNT()
FROM orders o
WHERE course_title ='웹개발 종합반'
group by payment_method
order by COUNT() DESC ;

SELECT *
FROM users u
order by email ;

SELECT *
FROM users u
order by name DESC ;

SELECT *
FROM users u
order by created_at ASC ;

SELECT payment_method ,
COUNT(*)
FROM orders o
group by payment_method ;

SELECT payment_method ,
COUNT(*)
FROM orders o
WHERE course_title ='앱개발 종합반'
group by payment_method;

SELECT name ,
email,
COUNT(*)
FROM users u
WHERE email LIKE '%gmail.com'
GROUP BY name;

SELECT course_id ,
ROUND(AVG(likes),1)
FROM checkins c
group by course_id ;