SQL选择每天的最后一个时间戳

Question

I am trying to reduce a list of timestamps to only the last one from each day.

I can use the following SQL to pull in the timestamps that I'm looking for but I'm also converting these to a formatted date, which I don't want to do.

select
     from_unixtime(intTimeStamp) as day
from
    (
        select distinct intTimeStamp from TableA order by intTimeStamp desc
    ) as t2
group by
    date(day)
order by
    intTimeStamp desc

This is the output:

2014-08-29 13:21:01
2014-08-28 13:21:01
2014-08-27 13:21:01
2014-08-26 13:21:01
2014-08-25 13:21:21
2014-08-22 10:11:01
2014-08-21 13:21:01
2014-08-20 13:21:01
[remaining omitted]

This is as expected but I need them in their original unixtime format.

Simply removing the from_unixtime() function yields very odd behaviour, however. It only returns 6 rows , as opposed to the expected 168 . I noticed that only one of the six rows returned is even the last timestamp for its day. It's skipping 95% of the days with timestamps available but I'm not sure why.

select
     intTimeStamp as day
from
    (
        select distinct intTimeStamp from TableA order by intTimeStamp desc
    ) as t2
group by
    date(intTimeStamp)
order by
    intTimeStamp desc

This is the output:

1409343661
1409170501
1406060101
1401221701
1400271301
1400012101

EXPLAIN Output for both:

1   PRIMARY <derived2>  ALL                 13338   Using temporary; Using filesort
2   DERIVED TableA  range       PRIMARY 4       10  Using index for group-by; Using temporary; Using filesort

How can I retrieve only the last timestamp from each day, as in the first query, but not have to convert them from a unix timestamp (or at least have them output as one)?

Thanks.

Answer 1

You can do this with a simple group by :

select max(intTimeStamp) as lasttimestamp
from TableA 
group by date(from_unixtime(intTimeStamp))
order by lasttimestamp desc;