Grouping in Mysql - mysql

Question

i need to get the top touristCount in each month like January Zambia has 4 touristCount i need to select only Zambia for January and so on
user
`useri_id` | `username` | `email` | `nationality`
1 Joseph `` US
2 Abraham. `` UK
3 g.wood '' Zambia
4 Messi. '' France
5 Ronaldo. '' Namibia
6 Pogba. '' Holand.
bookings
booking_id | user_id | booking_date | tour_id
1 1 2022-01-01 1
2 1 2022-01-01 6
3 1 2022-05-01 2
4 3 2022-01-01 5
5 2 2022-04-01 5
6 2 2022-11-01 7
7 3 2022-12-01 2
8 6 2022-01-01 1
this is what i have tried
SELECT s.nationality AS Nationality,
COUNT(b.tourist_id) AS touristsCount,
MONTH(STR_TO_DATE(b.booked_date, '%d-%m-%Y')) AS `MonthNumber`
FROM bookings b, users s
WHERE s.user_id = b.tourist_id
AND YEAR(STR_TO_DATE(b.booked_date, '%d-%m-%Y')) = '2022'
GROUP BY Nationality,MonthNumber
order BY MonthNumber ASC
LIMIT 100
i need the results to be like
nationality | TouritIdCount | MonthNumber
US 2 01
UK 1 04
US 1 05
UK 1 11
ZAMBIA 1 12

Answer 1

Try this :
SELECT nationality, COUNT(booking_id) AS TouristIdCount, MONTH(booking_date) AS MonthNumber
FROM users u
JOIN bookings b ON u.user_id = b.user_id
WHERE YEAR(booking_date) = 2022
GROUP BY nationality, MonthNumber
ORDER BY TouristIdCount DESC, MonthNumber ASC

Answer 2

you can use
having COUNT(b.tourist_id) >= 2

Answer 3

You want to count bookings per month and tourist's nationality and then show only the top nationality (or nationalities) per month.
There are two very similar approaches:
Rank the nationalities' booking counts per month with RANK and only show the best ranked rows.
Select the top booking count per month and only show rows matching their top count.
The following query uses the second method. It shows one row per month and top booking nationality. Often there may be excatly one row for a month showing the one top booking nationality, but there may also be months where nationalities tie and share the same top booking count, in which case we see more than one row for a month.
select year, month, nationality, booking_count
from
(
select
year(b.booking_date) as year,
month(b.booking_date) as month,
u.nationality,
count(*) as booking_count,
max(count(*)) over (partition by year(b.booking_date), month(b.booking_date)) as months_max_booking_count
from bookings b
join users u on u.user_id = b.tourist_id
group by year(b.booking_date), month(b.booking_date), u.nationality
) ranked
where booking_count = months_max_booking_count
order by year, month, nationality;
As your own sample data doesn't contain any edge cases, here is some other sample data along with my query's result and an explanation. (In other words, this is what you should have shown in your request ideally.)
users
user_id
username
email
nationality
1
Joseph
joseph#mail.us
US
2
Mary
mary#mail.us
US
3
Abraham
abraham#mail.uk
UK
bookings
booking_id
user_id
booking_date
tour_id
1
1
2022-01-11
1
2
2
2022-01-11
1
3
3
2022-01-11
1
4
3
2022-01-22
2
5
1
2022-05-01
3
6
2
2022-05-01
3
7
1
2022-05-12
4
8
2
2022-05-12
4
9
3
2022-05-14
5
10
3
2022-05-20
6
11
3
2022-05-27
7
result
year
month
nationality
booking_count
2022
1
UK
2
2022
1
US
2
2022
5
US
4
In January there were two tours, but we are not interested in tours. We see four bookings, two by the Americans, two by the Britsh person. This is a tie, and we show two rows, one for UK and one for US with two bookings each.
In May there were five tours, but again, we are not interested in tours. There are seven bookings, four by the Americans, three by the Britsh person. So we only show US as the top country with four bookings here.