SQL184 某宝店铺连续2天及以上购物的用户及其对应的天数
题目描述
11月结束后,小牛同学需要对其在某宝的网店就11月份用户交易情况和产品情况进行分析以更好的经营小店。
11月份销售数据表sales_tb如下(其中,sales_date表示销售日期,user_id指用户编号,item_id指货号,sales_num表示销售数量,sales_price表示结算金额):
sales_date | user_id | item_id | sales_num | sales_price |
---|---|---|---|---|
2021-11-01 | 1 | A001 | 1 | 90 |
2021-11-01 | 2 | A002 | 2 | 220 |
2021-11-01 | 2 | B001 | 1 | 120 |
2021-11-02 | 3 | C001 | 2 | 500 |
2021-11-02 | 4 | B001 | 1 | 120 |
2021-11-03 | 5 | C001 | 1 | 240 |
2021-11-03 | 6 | C002 | 1 | 270 |
2021-11-04 | 7 | A003 | 1 | 180 |
2021-11-04 | 8 | B002 | 1 | 140 |
2021-11-04 | 9 | B001 | 1 | 125 |
2021-11-05 | 10 | B003 | 1 | 120 |
2021-11-05 | 10 | B004 | 1 | 150 |
2021-11-05 | 10 | A003 | 1 | 180 |
2021-11-06 | 11 | B003 | 1 | 120 |
2021-11-06 | 10 | B004 | 1 | 150 |
请你统计连续2天及以上在该店铺购物的用户及其对应的次数(若有多个用户,按user_id升序排序),以上例子的输出结果如下:
user_id | days_count |
---|---|
10 | 2 |
SQL Schema
sql
drop table if exists sales_tb;
CREATE TABLE sales_tb(
sales_date date NOT NULL,
user_id int(10) NOT NULL,
item_id char(10) NOT NULL,
sales_num int(10) NOT NULL,
sales_price int(10) NOT NULL
);
INSERT INTO sales_tb VALUES('2021-11-1', 1, 'A001', 1, 90);
INSERT INTO sales_tb VALUES('2021-11-1', 2, 'A002', 2, 220);
INSERT INTO sales_tb VALUES('2021-11-1', 2, 'B001', 1, 120);
INSERT INTO sales_tb VALUES('2021-11-2', 3, 'C001', 2, 500);
INSERT INTO sales_tb VALUES('2021-11-2', 4, 'B001', 1, 120);
INSERT INTO sales_tb VALUES('2021-11-3', 5, 'C001', 1, 240);
INSERT INTO sales_tb VALUES('2021-11-3', 6, 'C002', 1, 270);
INSERT INTO sales_tb VALUES('2021-11-4', 7, 'A003', 1, 180);
INSERT INTO sales_tb VALUES('2021-11-4', 8, 'B002', 1, 140);
INSERT INTO sales_tb VALUES('2021-11-4', 9, 'B001', 1, 125);
INSERT INTO sales_tb VALUES('2021-11-5', 10, 'B003', 1, 120);
INSERT INTO sales_tb VALUES('2021-11-5', 10, 'B004', 1, 150);
INSERT INTO sales_tb VALUES('2021-11-5', 10, 'A003', 1, 180);
INSERT INTO sales_tb VALUES('2021-11-6', 11, 'B003', 1, 120);
INSERT INTO sales_tb VALUES('2021-11-6', 10, 'B004', 1, 150);
答案
sql
SELECT user_id, COUNT(*) AS `days_count`
FROM (SELECT user_id,
DATE_SUB(sales_date, INTERVAL ROW_NUMBER() OVER (PARTITION BY user_id ORDER BY sales_date) DAY) AS `diff`
FROM (SELECT DISTINCT user_id, sales_date FROM sales_tb) st) t
GROUP BY user_id, diff
HAVING COUNT(*) >= 2;