【字节】
3月：order表如下，找出连续登录7天的用户
SELECT id, COUNT(*)
FROM(
SELECT *, date0 – rk
FROM(
SELECT *, ROW_NUMBER() OVER(PARTITION BY id, ORDER BY date0) AS rk
FROM(
SELECT id, DISTINCT DATE(date) AS date0
FROM order
)a)b)c
GROUP BY id
HAVING COUNT(*) > 7
【LLS】
4月：查employee 表中第二高薪水
+----+--------+
| Id | Salary |
+----+--------+
| 1 | 100 |
| 2 | 200 |
| 3 | 300 |
+----+--------+
返回：
+---------------------+
| SecondHighestSalary |
+---------------------+
| 200 |
+-------------------
