leetcode刷mysql遇到的一些小tips

題目是刪除重複郵箱，抄的解答

#因為mysql不允許同時對乙個表delete和select聯合操作select, 所以會select乙個虛擬的表出來

delete
from person where id 
notin
(select need.id from
(select
min(id)
as id from person group
by email)
as need
);

datediff和timestampdiff是後減前

select timestampdiff(

day,
'2019-05-20'
,'2019-05-21');
# 1

elect x.request_at day

,round
(sum(if
(status
!='completed',1
,0))
/count(*
),2)
'cancellation rate'
from
(select t.
*from trips  t, users  u1, users u2
where t.client_id=u1.users_id and t.driver_id=u2.users_id
and u1.banned=
'no'
and u2.banned=
'no'
and request_at between
'2013-10-01'
and'2013-10-03'
) xgroup
by x.request_at
order
by x.request_at;

# write your mysql query statement below

select a.
*from stadium as a
inner
join stadium as b
on a.id = b.id -
1inner
join stadium as c
on b.id = c.id -
1where a.people >
100and b.people >
100and c.people >
100union
select a.
*from stadium as a
inner
join stadium as b
on a.id = b.id +
1inner
join stadium as c
on b.id = c.id +
1where a.people >
100and b.people >
100and c.people >
100;

比如, id為: 1, 2, 3, 4的滿足, 我們討論id 為2的這一天, id為2 滿足是因為, 1, 2, 3; 也可以是因為2, 3, 4

select

distinct a.
*from stadium a,stadium b,stadium c
where a.people>=
100and b.people>=
100and c.people>=
100and
((a.id = b.id-
1and b.id = c.id -1)
or##a(a.id = b.id-
1and a.id = c.id +1)
or##c(a.id = b.id+
1and b.id = c.id +1)
##c)
order
by a.id

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