两个日期字段之间的时差

Question

我有一个日期字段名称＆＃34; dts＆＃34;字符串格式。我想根据他们的时差（以小时为单位）找出所有记录。运行事件时间应大于或等于eat事件。

输出应为：

Answer 1

将时间戳转换为秒，然后减去，将结果除以3600得到小时，用例+计数按范围计算，如下所示：

select count(case when diff_hrs >24 then 1 end) as more_24,
       count(case when diff_hrs <=24 then 1 end) as less_than_24,
       ...
       count(case when diff_hrs >=2 and diff_hrs <=3 then 1 end) as hrs_2_to_3,
       ...
from
(
select
abs(unix_timestamp(dts) - unix_timestamp(dts-eat)))/60/60 as diff_hrs
from table
)s;