每个连续行 BIGQUERY SQL 之间的时间戳差异答案

【问题标题】：Timestamp difference between every consecutive row BIGQUERY SQL每个连续行 BIGQUERY SQL 之间的时间戳差异
【发布时间】：2020-10-21 19:52:33
【问题描述】：

我在 SQL BQ 中有一个带有 ID 和 DateTime (TIMESTAMP) 列的表。我想计算每个连续行之间的时间戳差异，比如说秒数，并创建一个具有计算时间差异的新列。

表：

ID     DateTime
a      2019-10-15 10:00:19 UTC
a      2019-10-15 10:00:29 UTC
a      2019-10-15 10:00:39 UTC
a      2019-10-15 10:00:49 UTC
a      2019-10-15 10:00:59 UTC

the desired result would look like this:

ID     DateTime                    TimeDiff
a      2019-10-15 10:00:19 UTC      null
a      2019-10-15 10:00:29 UTC       10
a      2019-10-15 10:00:39 UTC       10
a      2019-10-15 10:00:49 UTC       10
a      2019-10-15 10:00:59 UTC       10

到目前为止，我已经尝试了这些选项但没有成功：

select ID, DateTime,
(LAG(DateTime) OVER (PARTITION BY ID ORDER BY DateTime ASC) - DateTime) AS TimeDiff
from `xxx.yyy.table` 
order by DateTime

和

select ID, DateTime,
timestamp_diff(lag(DateTime, 1) OVER (ORDER BY DateTime)) as TimeDiff
from `xxx.yyy.table`
order by DateTime

和

select ID, DateTime,
LAG(DateTime) OVER (PARTITION BY FieldID ORDER BY DateTime ASC) AS timeDiff
from `xxx.yyy.table` 
order by DateTime

【问题讨论】：

标签： sql google-bigquery

【解决方案1】：

LAG() 是从上一行获取值的正确函数。你只需要正确使用TIMESTAMP_DIFF()：

select ID, DateTime,
       timestamp_diff(DateTime,
                      lag(DateTime, 1) OVER (ORDER BY DateTime), 
                      second
                     ) as TimeDiff
from `xxx.yyy.table`
order by DateTime;

请注意，您似乎想要这个 per id。如果是这样，你也应该PARTITION BY：

       timestamp_diff(DateTime,
                      lag(DateTime, 1) OVER (PARTITION BY id ORDER BY DateTime), 
                      second
                     ) as TimeDiff

【讨论】：