2014-09-21 18:08 GMT+02:00 Rémi Cura <remi.c...@gmail.com>:

> Hey, sorry I what I say is obvious for you .
>
> If I understood your problem correctly, it is strictly equivalent to this
> one :
>
> http://postgresql.1045698.n5.nabble.com/Count-of-records-in-a-row-td5775363.html
>
> there is a postgres trick to solve this problem :
> what you want is essentially generate a unique group_id,
> but one that depends of an order of row not defined in the group.
>
> The solution
> is to generate a row number by the order you want , then a row number by
> the group ,
> then a subtraction of the 2 row number gives you an unique id per group.
>
> The cost is that you have to use 2 windows function., hence 2 scans I
> guess.
>

yes, it is little bit similar - I found a pattern described by Andrew is
well too.

regards

Pavel


> Cheers,
> Rémi-C
>
> 2014-09-21 17:51 GMT+02:00 Andrew Gierth <and...@tao11.riddles.org.uk>:
>
>> >>>>> "Pavel" == Pavel Stehule <pavel.steh...@gmail.com> writes:
>>
>>  Pavel> Hi
>>  Pavel> I tried to solve following task:
>>
>>  Pavel> I have a table
>>
>>  Pavel> start, reason, km
>>  Pavel> =============
>>  Pavel>  2014-01-01 08:00:00, private, 10
>>  Pavel>  2014-01-01 09:00:00, commerc, 20
>>  Pavel>  2014-01-01 10:00:00, commerc, 20
>>  Pavel>  2014-01-01 11:00:00, private, 8
>>
>>  Pavel> and I would reduce these rows to
>>
>>  Pavel>  2014-01-01 08:00:00, private, 10
>>  Pavel>  2014-01-01 09:00:00, commerc, 20 + 20 = 40
>>  Pavel>  2014-01-01 11:00:00, private, 8
>>
>>  Pavel> It is relative hard to it now with SQL only.
>>
>> Only relatively. My standard solution is something like this:
>>
>> select start_time, reason, sum(km) as km
>>   from (select max(label_time) over (order by start) as start_time,
>>                reason, km
>>           from (select start, reason, km,
>>                        case when reason
>>                                  is distinct from
>>                                  lag(reason) over (order by start)
>>                             then start
>>                        end as label_time
>>                   from yourtable
>>                ) s2
>>        ) s1
>>  group by start_time, reason
>>  order by start_time;
>>
>> (Your change_number idea is essentially equivalent to doing
>> sum(case when x is distinct from lag(x) over w then 1 end) over w,
>> except that since window functions can't be nested, that expression
>> requires a subquery.)
>>
>> --
>> Andrew (irc:RhodiumToad)
>>
>>
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