谷歌大查询行之间的时差

Question

我目前正在尝试计算google big query中行之间的时间戳差异附加的是我用来测试代码的示例表。

我正在使用此代码

SELECT 
  A.row,
  A.issue.updated_at,
 (B.issue.updated_at - A.issue.updated_at) AS timedifference
FROM [icxmedia-servers:icx_metrics.gh_zh_data_production] A
INNER JOIN [icxmedia-servers:icx_metrics.gh_zh_data_production] B
  ON B.row = (A.row + 1)
WHERE issue.number==6 and issue.name=="archer"
ORDER BY A.requestid ASC

参考此问题Calculate the time difference between of two rows

Answer 1

而不是JOIN，这更自然地使用分析函数表达。 analytic functions with standard SQL in BigQuery的文档解释了分析函数的工作原理以及语法。例如，如果您希望在列x确定订单的y值中产生连续差异，则可以执行以下操作：

WITH T AS (
  SELECT
    x,
    y
  FROM UNNEST([9, 3, 4, 7]) AS x WITH OFFSET y)
SELECT
  x,
  x - LAG(x) OVER (ORDER BY y) AS x_diff
FROM T;

请注意，要在BigQuery中运行此功能，您需要取消选中＆＃34;使用旧版SQL＆＃34;框下＆＃34;显示选项＆＃34;启用标准SQL。 WITH T子句只是为示例设置了一些数据。

对于您的具体情况，您可能需要查询，例如：

SELECT
  row,
  issue.updated_at,
  issue.updated_at - LAG(issue.updated_at) OVER (ORDER BY issue.updated_at) AS timedifference
FROM `icxmedia-servers.icx_metrics.gh_zh_data_production`
WHERE issue.number = 6
  AND issue.name = "archer"
ORDER BY requestid ASC;

如果您想确定updated_at仅在一个问题编号之外的差异，您也可以使用PARTITION BY子句。例如：

SELECT
  row,
  issue.name,
  issue.number,
  issue.updated_at,
  issue.updated_at - LAG(issue.updated_at) OVER (
      PARTITION BY issue.number
      ORDER BY issue.updated_at) AS timedifference
FROM `icxmedia-servers.icx_metrics.gh_zh_data_production`
ORDER BY requestid ASC;