两个时间戳相减得到interval day to second。如果要保留亚秒级精度，则需要提取间隔的每个部分并进行数学运算。

select t.*,
    round(
          extract(day    from duration)  * 24
        + extract(hour   from duration) 
        + extract(minute from duration) / 60
        + extract(second from duration) / 60 / 60,
        3
    ) as rounded
from (
    select lot_id, 
        max(inspection_time) as max_time, 
        min(inspection_time) as min_time,  
        max(inspection_time) - min(inspection_time) as duration
    from sensormaintab.sensor_inspections 
    where lot_id = 1392353501 and inspection_id = 49 
    group by lot_id
) t

使用子查询可以避免重复max(...) - min(...)表达式（我们也可以在横向连接中这样做）。

如果你不受小数秒的困扰，那么把时间戳转换成日期，然后减去一天的一小部分，然后乘以24：

SELECT LOT_ID, 
    MAX(INSPECTION_TIME), 
    MIN(INSPECTION_TIME), 
    (CAST(MAX(INSPECTION_TIME) AS DATE) - CAST(MIN(INSPECTION_TIME) AS DATE)) * 24
      AS Duration
FROM SENSORMAINTAB.SENSOR_INSPECTIONS
WHERE LOT_ID = 1392353501 AND INSPECTION_ID = 49
GROUP BY LOT_ID

如果需要小数秒，则将时间戳截短到分钟的开始，并找到差值，然后包括秒差：

SELECT LOT_ID, 
    MAX(INSPECTION_TIME), 
    MIN(INSPECTION_TIME), 
    (
      TRUNC(MAX(INSPECTION_TIME), 'MI')
      - TRUNC(MIN(INSPECTION_TIME), 'MI')
    ) * 24
    + (
      EXTRACT(SECOND FROM MAX(INSPECTION_TIME))
      - EXTRACT(SECOND FROM MIN(INSPECTION_TIME))
    ) / 60 / 60
      AS Duration
FROM SENSORMAINTAB.SENSOR_INSPECTIONS
WHERE LOT_ID = 1392353501 AND INSPECTION_ID = 49
GROUP BY LOT_ID