Question

我有以下在SQL 2012中运行的T-SQL

SELECT machine_id, speed_gps_kph, odometer_total, event_timestamp, 
   1.0 * (speed_gps_kph - LAG(speed_gps_kph, 1) OVER (PARTITION BY machine_id ORDER BY event_timestamp)) /
   datediff(ss, LAG(event_timestamp, 1) OVER (PARTITION BY machine_id ORDER BY event_timestamp), event_timestamp)
FROM   Simple_speed

然而，正如错误所示，我有一个除以零的问题。

在这个问题上有很多帖子和答案，但我很难接受工作。

Answer 1

这是一种使用NULLIF(...,0)来控制除零错误的简单方法，而无需重复自己：

SELECT machine_id, speed_gps_kph, odometer_total, event_timestamp, 
   1.0 * (speed_gps_kph - LAG(speed_gps_kph, 1) OVER (PARTITION BY machine_id ORDER BY event_timestamp)) 
   / NULLIF(datediff(ss, LAG(event_timestamp, 1) OVER (PARTITION BY machine_id ORDER BY event_timestamp), event_timestamp),0)
FROM Simple_speed

对于分母为零的行，表达式现在将返回NULL。

遇到零除错误。 T-SQL

1 个答案: