sql - 如何在 BigQuery 中获取连续时间戳之间的差异

Question

我有一个看起来像这样的表:

ID    DateTime
1     5-1-16 12:25:13
1     5-1-16 12:28:46
2     5-1-16 12:25:18
2     5-1-16 12:29:34

我想找出每个 ID 的每个连续时间戳之间的秒数差异。有没有办法在 BigQuery 中执行此操作？我有几千条记录。我知道我需要先将时间与日期分开。

Answer 1

试试下面

SELECT
  ID, 
  TIMESTAMP_TO_SEC(TIMESTAMP(DateTime))-TIMESTAMP_TO_SEC(TIMESTAMP(prev_DateTime)) AS diff,
FROM (
  SELECT 
    ID,
    DateTime,
    LAG(DateTime) OVER(PARTITION BY ID ORDER BY DateTime) AS prev_DateTime
  FROM 
    (SELECT 1 AS ID, '2016-05-01 12:25:13' AS DateTime),
    (SELECT 1 AS ID, '2016-05-01 12:28:46' AS DateTime),
    (SELECT 2 AS ID, '2016-05-01 12:25:18' AS DateTime),
    (SELECT 2 AS ID, '2016-05-01 12:29:34' AS DateTime)
)

added

 SELECT
  ID, 
  TIMESTAMP_TO_SEC(TIMESTAMP(DateTime))-TIMESTAMP_TO_SEC(TIMESTAMP(prev_DateTime)) AS diff,
FROM (
  SELECT 
    ID,
    DateTime,
    LAG(DateTime) OVER(PARTITION BY ID ORDER BY DateTime) AS prev_DateTime
  FROM YourTable
)

我注意到 - 看起来您的 DateTime 字段格式不正常 - '5-1-16 12:29:34'
如果在执行上述查询时这对您来说是个问题 - 您可以尝试下面的一个
请注意:对于此查询，您需要 enable Standard SQL

SELECT
  ID, 
  UNIX_SECONDS(DateTime) - UNIX_SECONDS(prev_DateTime) AS diff
FROM (
  SELECT 
    ID,
    DateTime,
    LAG(DateTime) OVER(PARTITION BY ID ORDER BY DateTime) AS prev_DateTime
  FROM (
    SELECT
      ID, 
      PARSE_TIMESTAMP("%m-%d-%y %H:%M:%S", DateTime) AS DateTime
    FROM YourTable  
  )
)