选择给定数据的最新值并缺少记录

Question

... where "missing records" are identical to the last recorded value, hence no record.

This may be subjective, but I'm hoping there's a standardised way of doing this.

So, let's say I have a bunch of analytics in a MySQL table. There is some missing information, but as mentioned above, that's because their previous value is the same as the current value.

table "table":

id    value      datetime
1     5          1285891200    // Today
1     4          1285804800    // Yesterday
2     18         1285804800    // Yesterday
2     16         1285771094    // The day before yesterday

As you can see, I don't have a value for today for id 2.

If I wanted to pull the "most recent value" from this table (that is, 1's "today", and 2's "yesterday", how do I do that? I've achieved it by running the following query:

SELECT id, value FROM (SELECT * FROM table ORDER BY datetime DESC) as bleh GROUP BY id

Which utilizes a subquery to order the data first, and then I rely on "GROUP BY" to pick the first value (which, since it is ordered, is the most recent) from each id. However, I don't know if shoving a subquery in there is the best way to get the most recent value.

How would you do it?

The desired table:

id    value      datetime
1     5          1285891200    // Today
2     18         1285804800    // Yesterday

Thanks...

Answer 1

Gotta love MySQL for allowing an order by in a subquery. That's not allowed by the SQL standard :)

You could rewrite the query in a standards complaint way like:

select  *
from    YourTable a
where   not exists
        (
        select  *
        from    YourTable b
        where   a.id = b.id
        and     a.datetime < b.datetime
        )

In case there are duplicates that you can't split apart in the subquery, you can group by and then pick an arbitrary value:

select  a.id
,       max(a.value)
,       max(a.datetime)
from    YourTable a
where   not exists
        (
        select  *
        from    YourTable b
        where   a.id = b.id
        and     a.datetime < b.datetime
        )
group by
        a.id

This chooses the maximum a.value sharing the latest datetime . Now datetime is the same for all duplicate rows, but standard SQL doesn't know that, so you have to specify a way to pick from the equal days. Here, I'm using max , but min or even avg would work just as well.