您可以使用间隙和孤岛方法并找到连续的运行。这种替代方法很简单：

with data as (
    select *,
        lag(ch_status) over (partition by id order by timestamp) as last_status,
        lag(temperature) over (partition by id order by timestamp) as last_temperature
    from T
)
select id, sum(temperature - last_temperature) as diff
from data
where ch_status = 0 and last_status = 0
group by id;

无论序列是否单调（总是递增），这都将起作用。