hive 组内任意两次登录时间差

根据引用\[2\]中的代码，可以使用Hive来计算组内任意两次登录时间差。首先，需要使用窗口函数和子查询来为每个用户的登录记录进行编号，并计算出登录时间的差值。然后，使用HAVING子句来筛选出时间差大于等于指定值的记录。

以下是一个示例代码：

SELECT uid, dt_diff, DATEDIFF(MAX(dt), MIN(dt)) AS time_diff
FROM (
  SELECT uid, dt, ROW_NUMBER() OVER (PARTITION BY uid ORDER BY dt) AS rnk,
         DATEDIFF(dt, LAG(dt) OVER (PARTITION BY uid ORDER BY dt)) AS dt_diff
  FROM your_table
  WHERE login_status = 1
) t
GROUP BY uid, dt_diff
HAVING time_diff >= your_specified_value;

请注意，你需要将代码中的"your_table"替换为你实际使用的表名，"your_specified_value"替换为你指定的时间差值。

引用\[2\]提供了一个连续登录的示例，你可以根据你的实际需求进行修改和适应。
#### 引用[.reference_title]
- *1* *3* [hiveSql 计算两段时间范围交集的连续每月天数](https://blog.csdn.net/i_mycode/article/details/125834337)[target="_blank" data-report-click={"spm":"1018.2226.3001.9630","extra":{"utm_source":"vip_chatgpt_common_search_pc_result","utm_medium":"distribute.pc_search_result.none-task-cask-2~all~insert_cask~default-1-null.142^v91^insertT0,239^v3^insert_chatgpt"}} ] [.reference_item]
- *2* [hive sql 经典题目 连续登陆｜间隔连续登陆｜行列转换｜累加｜topN | 炸裂](https://blog.csdn.net/u010569893/article/details/128918864)[target="_blank" data-report-click={"spm":"1018.2226.3001.9630","extra":{"utm_source":"vip_chatgpt_common_search_pc_result","utm_medium":"distribute.pc_search_result.none-task-cask-2~all~insert_cask~default-1-null.142^v91^insertT0,239^v3^insert_chatgpt"}} ] [.reference_item]
[ .reference_list ]