Обсуждение: group by hour + distinct

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group by hour + distinct

От
Michele Petrazzo - Unipex
Дата:
Hi list,
Into a table like this:
id_user | my_date

and some data inside
1 | 2010/11/25 00:01:00
1 | 2010/11/25 00:02:00
1 | 2010/11/25 01:01:00
2 | 2010/11/25 02:01:00
3 | 2010/11/25 02:01:00
3 | 2010/11/25 02:06:00
1 | 2010/11/25 03:01:00

I'm looking for a query that say me, hour per hour, how many unique
id_user are inside that range.
With the simple data above, I'm looking for:
hour | count
0 | 1
1 | 1
2 | 2
3 | 1

Like now, with my tests, I achieve only a
hour | count
0 | 2
1 | 1
2 | 3
3 | 1

My real query and data:

SELECT count(id_user) from some_table where my_date >= '2010/11/25 
00:00:00' and my_date < '2010/11/25 01:00:00'; count
-------    90
(1 row)

SELECT distinct(id_user) from some_table where my_date >= '2010/11/25 
00:00:00' and my_date < '2010/11/25 01:00:00' order by id_user;

Give me 69 rows, that are the real unique id_user that I have and I'm 
looking for.

One of a query that I use without success:

SELECT count(distinct some_table.my_date), EXTRACT(hour from my_date) as 
h from some_table where my_date >= '2010/11/25' and my_date < 
'2010/11/26' group by h order by h; count | h
-------+----    90 |  0    63 |  1
... and so on

Someone?
Thanks,
Michele


Re: group by hour + distinct

От
"Oliveiros d'Azevedo Cristina"
Дата:
Howdy, Michelle,

If you write something like this,

SELECT  hour , COUNT(id_user) as count
FROM
(
SELECT EXTRACT(hour from my_date) as hour, id_user FROM michelle
GROUP BY EXTRACT(hour from my_date),id_user
)subquery
GROUP BY hour

for each hour it will count the number of distinct user_id's there are . If 
I understood correctly what you need...

Can you please test it and see if it is OK for your needs?
With me, it worked on the sample data you provided

Best,
Oliver

----- Original Message ----- 
From: "Michele Petrazzo - Unipex" <michele.petrazzo@unipex.it>
To: <pgsql-sql@postgresql.org>
Sent: Friday, November 26, 2010 11:39 AM
Subject: [SQL] group by hour + distinct


> Hi list,
> Into a table like this:
> id_user | my_date
>
> and some data inside
> 1 | 2010/11/25 00:01:00
> 1 | 2010/11/25 00:02:00
> 1 | 2010/11/25 01:01:00
> 2 | 2010/11/25 02:01:00
> 3 | 2010/11/25 02:01:00
> 3 | 2010/11/25 02:06:00
> 1 | 2010/11/25 03:01:00
>
> I'm looking for a query that say me, hour per hour, how many unique
> id_user are inside that range.
> With the simple data above, I'm looking for:
> hour | count
> 0 | 1
> 1 | 1
> 2 | 2
> 3 | 1
>
> Like now, with my tests, I achieve only a
> hour | count
> 0 | 2
> 1 | 1
> 2 | 3
> 3 | 1
>
> My real query and data:
>
> SELECT count(id_user) from some_table where my_date >= '2010/11/25 
> 00:00:00' and my_date < '2010/11/25 01:00:00';
>  count
> -------
>     90
> (1 row)
>
> SELECT distinct(id_user) from some_table where my_date >= '2010/11/25 
> 00:00:00' and my_date < '2010/11/25 01:00:00' order by id_user;
>
> Give me 69 rows, that are the real unique id_user that I have and I'm 
> looking for.
>
> One of a query that I use without success:
>
> SELECT count(distinct some_table.my_date), EXTRACT(hour from my_date) as h 
> from some_table where my_date >= '2010/11/25' and my_date < '2010/11/26' 
> group by h order by h;
>  count | h
> -------+----
>     90 |  0
>     63 |  1
> ... and so on
>
> Someone?
> Thanks,
> Michele
>
> -- 
> Sent via pgsql-sql mailing list (pgsql-sql@postgresql.org)
> To make changes to your subscription:
> http://www.postgresql.org/mailpref/pgsql-sql 



Re: group by hour + distinct

От
Michele Petrazzo - Unipex
Дата:
Oliveiros d'Azevedo Cristina ha scritto:
> Howdy, Michelle,
>

Ciao

> for each hour it will count the number of distinct user_id's there
> are . If I understood correctly what you need...
>
> Can you please test it and see if it is OK for your needs? With me,
> it worked on the sample data you provided
>

Yes! Simply perfect!

> Best, Oliver
>

Thanks,
Michele

P.s. Have you some references about the "subquery" keyword? I found only
the word subquery as "use" (for example: select a from b where id in 
(select id from table)), but not as sql command.



Re: group by hour + distinct

От
"Oliveiros d'Azevedo Cristina"
Дата:
>
> Ciao
>
* Olá!

>> for each hour it will count the number of distinct user_id's there
>> are . If I understood correctly what you need...
>>
>> Can you please test it and see if it is OK for your needs? With me,
>> it worked on the sample data you provided
>>
>
> Yes! Simply perfect!
>

* Great to hear it worked !

>>
>
> Thanks,
> Michele
* Don't mention it ;-)

>
> P.s. Have you some references about the "subquery" keyword? I found only
> the word subquery as "use" (for example: select a from b where id in 
> (select id from table)), but not as sql command.
>
>

* I don't know subquery as an SQL keyword too. In the query I gave you the 
"subquery" thing was just an alias. Such kind of things are mandatory. If I 
try to omit it, I get this error

ERROR:  subquery in FROM must have an alias
HINT:  For example, FROM (SELECT ...) [AS] foo.

********** Erro **********

ERROR: subquery in FROM must have an alias
Estado de SQL:42601
Sugestão:For example, FROM (SELECT ...) [AS] foo.

So, even though the "subquery" alias was not used anywhere else, I believe 
it had to be there, otherwise the SQL parser would complain


Best,
Oliver



Re: group by hour + distinct

От
Lew
Дата:
Michele Petrazzo - Unipex wrote:
> P.s. Have you some references about the "subquery" keyword? I found only
> the word subquery as "use" (for example: select a from b where id in
> (select id from table)), but not as sql [sic] command.

"subquery" is not an SQL keyword.
<http://www.postgresql.org/docs/9.0/interactive/sql-keywords-appendix.html>

Nor is it a command all by itself.

Reading the documentation might help you:
<http://www.postgresql.org/docs/9.0/interactive/queries-table-expressions.html#QUERIES-SUBQUERIES>
<http://www.postgresql.org/docs/9.0/interactive/sql-expressions.html#SQL-SYNTAX-SCALAR-SUBQUERIES>
<http://www.postgresql.org/docs/9.0/interactive/functions-subquery.html>

Or try:
<http://lmgtfy.com/?q=SQL+subquery>

-- 
Lew