计算48小时内出现的时间记录数-SQL

Question

我们如何选择48小时内出现多次的记录计数？

例如

ID DATE 
1  9/24/2018
1  9/23/2018
1  9/20/2018
2  9/20/2018

ID 1在48小时内的出现次数超过一次

请让我知道如何编写一个sql查询来做到这一点

Answer 1

有很多方法，但我将从使用LAG()和日期比较开始。假设您的DATE列是日期数据类型？

WITH
  entity_summary AS
(
  SELECT
    ID,
    CASE
      WHEN LAG("DATE") OVER (PARTITION BY ID ORDER BY "DATE") >= "DATE" - INTERVAL '2' DAY
      THEN 1
      ELSE 0
    END
      AS occurence_within_2_day
  FROM
    Table1
)
SELECT
  ID,
  SUM(occurence_within_2_day)
FROM
  entity_summary
GROUP BY
  ID
HAVING
  SUM(occurence_within_2_day) >= 1