怎样对MYSQL中的两列时间做减法？

--这样是显示的结果是差值总的秒数

select sum(unix_timestamp(B)-unix_timestamp(A)) as 秒数

from tb

--这样是显示成 xxx:xx:xx 的形式

select sec_to_time(sum(unix_timestamp(B)-unix_timestamp(A))) as ti

from tb

--

select count(*) from handup where unix_timestamp(hinserttime) >(select unix_timestamp(max(hinserttime)) from handup) - 3600

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