Sorry if the title is confusing. Here is the query I have
Select MONTH(DATE(TIMESTAMP)), SUM(FIELD1), SUM(FIELD2) from TABLE WHERE TIMESTAMP BETWEEN '2009-07-26 00:00:00' AND '2010-02-24 23:59:59' GROUP BY MONTH(DATE(TIMESTAMP))
This will let me get the month number out of the query. The problem is that right now it is sorting the months 1,2,3,4.... when it spans two separate years. I need to be able to sort this query by year then month.
If I add "ORDER BY TIMESTAMP" at the end of my query I get this error:
Column TIMESTAMP or expression in SELECT list not valid. SQLCODE=-122
Also I changed the field names for this question to keep it clear the field isn't actually called TIMESTAMP
You need to group by year then month.:
SELECT YEAR(YourField),
Month(YourField),
SUM(Field1),
SUM(Field2)
FROM Table
WHERE...
GROUP BY
YEAR(YourField),
Month(YourField)
ORDER BY
YEAR(YourField),
Month(YourField)
Related
So I am trying and struggling for few days to extend the schema with the custom groupby using something like this
I have a table with few fields like id, country, ip, created_at.
Then I am trying to get them as groups. For example, group the data based on date, hourly of date, or based on country, and based on country with DISTINCT ip.
I am zero with SQLs honestly. But I tried to play around and get what I want. Here's an example.
SELECT Hour(created_at) AS date,
COUNT(*) AS count
FROM session where CAST(created_at AS date) = '2021-04-05'
GROUP BY Hour(created_at)
ORDER BY date;
SELECT country,
count(*) AS count from (SELECT * FROM session where CAST(created_at AS date) <= '2021-05-12' GROUP BY created_at) AS T1
GROUP BY country;
SELECT country, COUNT(*) as count
FROM (SELECT DISTINCT ip, country FROM session) AS T1
GROUP BY country;
SELECT DATE(created_at) AS date,
COUNT(*) AS count
FROM session
GROUP BY DATE(created_at)
ORDER BY date;
Now I am struggling with two things.
How do I make the date as variables? I mean, if I want to group them for a particular date range/ or today's data hourly, or per quarter gap (more of configurable), how do I add the variables in Hasura's Raw SQL?
Also for this approach I have to add schema for each one of them? Like this
CREATE
OR REPLACE VIEW "public"."unique_session_counts_date" AS
SELECT
date(session.created_at) AS date,
count(*) AS count
FROM
session
GROUP BY
(date(session.created_at))
ORDER BY
(date(session.created_at));
Is there a way to make it more generalized? What I mean is, if it
was in Nodejs I could have done something like
return rawQuery(
`
select ${field} x, count(*) y
from ${table}
where website_id=$1
and created_at between $2 and $3
${domainFilter}
${urlFilter}
group by 1
order by 2 desc
`,
params,
);
In this case, based on whatever field and where clause I send, one query would do the trick for me. Can do something similar in hasura?
Thank you so much in advance.
How do I make the date as variables? I mean, if I want to group them for a particular date range/ or today's data hourly, or per quarter gap (more of configurable), how do I add the variables in Hasura's Raw SQL?
My first thought is this. If you're thinking about passing in variables via a GraphQL for example, the GraphQL would look something like:
query MyQuery {
unique_session_counts_date(where: {created_at: {_gte: "<start date here>", _lte: "<end date here>"}}) {
<...any fields, rollups, etc here...>
}
}
The underlying view/query would follow the group by and order by that you've detailed. Then you'd be able to submit a query of the graphql query and just pass in the pertinent parameters like the $1, $2, and $3 in the raqQuery call.
Also for this approach I have to add schema for each one of them?
The schema? The view? I don't think a view specifically would be required, if a multilevel select or similar query can handle it and perform then a view wouldn't particularly be needed.
That's my first stab at the problem. I'm going to try to work through this problem in a few hours via a Twitch stream # HasuraHQ if you can join, happy to walk through it live.
I have this query
SELECT DISTINCT ON (tours.departure.departure_month)
tours.departure.departure_month
FROM tours.departure
But I want to order the distinct months by month name. I've tried this from a similar question to_date(tours.departure.departure_month, 'Month'),
but I cannot get it to work with DISTINCT ON.
What is the column type of departure_month is it date or month names, if its date type column you can try the following:
SELECT DISTINCT ON (tours.departure.departure_month)
tours.departure.departure_month
FROM tours.departure
ORDER BY month(tours.departure.departure_month) DESC;
I have a rather simple query:
SELECT table.foo, array_agg([ARRAY[EXTRACT(epoch FROM table.date), table.bar]) AS array
FROM table
GROUP BY table.foo,
ORDER BY table.date ASC;
When I run this query I get an error:
ERROR: column "table.date" must appear in the GROUP BY clause or be used in an aggregate function
I don't quite understand why that is happening because date appears in aggregate function. Is there any way to achieve that grouping?
you cant order by not existing column. If you want to order values in aggregation, use:
SELECT table.foo, array_agg([ARRAY[EXTRACT(epoch FROM table.date), table.bar] ORDER BY table.date ASC) AS array
FROM table
GROUP BY table.foo;
Im using redshift now.
then Id like to run query like
SELECT to_char(created_at, 'HH24') AS hour , to_char(created_at, 'YYYY-MM-DD HH24') AS tmp FROM log GROUP BY tmp;
this returns error, when I do it in mysql, it seems to be good.
this error is
ERROR: column "log.created_at" must appear in the GROUP BY clause or be used in an aggregate function
when I changed group by clause like "group by created_at", it returns results, but it has duplicated list.
Is is due to redshift?
If you're using a GROUP BY clause, any column in your query must either appear in the clause or you have to specify how you want it to be aggregated.
In your case, you seem to be trying to aggregate your log entries by hour. I suggest using the postgres date manipulation functions, for example:
SELECT created_at::date AS date,
extract('HOUR' FROM created_at) as hour
FROM log
GROUP BY date, hour;
I'm getting an error running this query
SELECT date(updated_at), count(updated_at) as total_count
FROM "persons"
WHERE ("persons"."updated_at" BETWEEN '2012-10-17 00:00:00.000000' AND '2012-11-07 12:25:04.082224')
GROUP BY date(updated_at)
ORDER BY persons.updated_at DESC
I get the error ERROR: column "persons.updated_at" must appear in the GROUP BY clause or be used in an aggregate function LINE 5: ORDER BY persons.updated_at DESC
This works if I remove the date( function from the group by call, however I'm using the date function because i want to group by date, not datetime
any ideas
At the moment it is unclear what you want Postgres to return. You say it should order by persons.updated_at but you do not retrieve that field from the database.
I think, what you want to do is:
SELECT date(updated_at), count(updated_at) as total_count
FROM "persons"
WHERE ("persons"."updated_at" BETWEEN '2012-10-17 00:00:00.000000' AND '2012-11-07 12:25:04.082224')
GROUP BY date(updated_at)
ORDER BY count(updated_at) DESC -- this line changed!
Now you are explicitly telling the DB to sort by the resulting value from the COUNT-aggregate. You could also use: ORDER BY 2 DESC, effectively telling the database to sort by the second column in the resultset. However I highly prefer explicitly stating the column for clarity.
Note that I'm currently unable to test this query, but I do think this should work.
the problem is that, because you are grouping by date(updated_at), the value for updated_at may not be unique, different values of updated_at can return the same value for date(updated_at). You need to tell the database which of the possible values it should use, or alternately use the value returned by the group by, probably one of
SELECT date(updated_at) FROM persons GROUP BY date(updated_at)
ORDER BY date(updated_at)
or
SELECT date(updated_at) FROM persons GROUP BY date(updated_at)
ORDER BY min(updated_at)