LeetCode 1565 Interview SQL Question with Detailed Explanation | Practice SQL

[@dornalanavya1722]

great explanation sir thank you so much

[@aakashbhandari9761]

Thanks a lot for this playlist❤

[@the_fury07]

we also can use date_format(order_date, yyyy-mm) to group by

[@sukumar-m4t]

LeetCode prime Q/A

[@mlvprasadofficial]

select left(order_date, 7) as month,
count( order_id ) as order_count,
count(distinct customer_id) as customer_count
from Orders
where invoice>20
group by left(order_date, 7)

[@mlvprasadofficial]

30

[@prathamsarraf6999]

CREATE TABLE Orders (
order_id INT,
order_date DATE,
customer_id INT,
invoice INT
);
INSERT INTO Orders (order_id, order_date, customer_id, invoice) VALUES
(1, '2020-09-15', 1, 30),
(2, '2020-09-17', 2, 90),
(3, '2020-10-06', 3, 20),
(4, '2020-10-20', 3, 21),
(5, '2020-11-10', 1, 10),
(6, '2020-11-21', 2, 15),
(7, '2020-12-01', 4, 55),
(8, '2020-12-03', 4, 77),
(9, '2021-01-07', 3, 31),
(10, '2021-01-15', 2, 20);
SELECT SUBSTRING(order_date,1,7) AS month ,COUNT(*) AS order_count,COUNT(DISTINCT customer_id) AS customer_count
FROM Orders
WHERE invoice>20
GROUP BY SUBSTRING(order_date,1,7)

[@sammyd4042]

Thanks bro