This query returns the number of alarms created by day between a specific date range.
SELECT CAST(created_at AS DATE) AS date, SUM(1) AS count
FROM ew_alarms
LEFT JOIN site ON site.id = ew_alarms.site_id
AND ew_alarms.created_at BETWEEN '12/22/2020' AND '01/22/2021' AND (CAST(EXTRACT(HOUR FROM ew_alarms.created_at) AS INT) BETWEEN 0 AND 23.99)
GROUP BY CAST(created_at AS DATE)
ORDER BY date DESC
Result: screenshot
What the best way to fill in the missing dates (1/16, 1/17, 1/18, etc)? Due to no alarms created on those days these results throw off the daily average I'm ultimately trying to achieve.
Would it be a generate_series query?
Yes, use generate_series(). I would suggest:
SELECT gs.date, COUNT(s.site_id) AS count
FROM GENERATE_SERIES('2020-12-22'::date, '2021-01-22'::date, INTERVAL '1 DAY') gs(dte) LEFT JOIN
ew_alarms a
ON ew.created_at >= gs.dte AND
ew.created_at < gs.dte + INTERVAL '1 DAY' LEFT JOIN
site s
ON s.id = a.site_id
GROUP BY gs.dte
ORDER BY date DESC;
I don't know what the hour comparison is supposed to be doing. The hour is always going to be between 0 and 23, so I removed that logic.
Note: Presumably, you want to count something from either site or ew_alarms. That is expected with LEFT JOINs so 0 can be returned.
Related
What I mean by this is if I have a table called payments with a created_at column and user_id column I want to select the count of purchases aggregated weekly (can be any interval I want) but only selecting first time purchases e.g. if a user purchased for the first time in week 1 it would be counted but if he purchased again in week 2 he would not be counted.
created_at
user_id
timestamp
1
timestamp
1
This is the query I came up with. The issue is if the user purchases multiple times they are all included. How can I improve this?
WITH dates AS
(
SELECT *
FROM generate_series(
'2022-07-22T15:30:06.687Z'::DATE,
'2022-11-21T17:04:59.457Z'::DATE,
'1 week'
) date
)
SELECT
dates.date::DATE AS date,
COALESCE(COUNT(DISTINCT(user_id)), 0) AS registrations
FROM
dates
LEFT JOIN
payment ON created_at::DATE BETWEEN dates.date AND dates.date::date + '1 ${dateUnit}'::INTERVAL
GROUP BY
dates.date
ORDER BY
dates.date DESC;
You want to count only first purchases. So get those first purchases in the first step and work with these.
WITH dates AS
(
SELECT *
FROM generate_series(
'2022-07-22T15:30:06.687Z'::DATE,
'2022-11-21T17:04:59.457Z'::DATE,
'1 week'
) date
)
, first_purchases AS
(
SELECT user_id, MIN(created_at:DATE) AS purchase_date
FROM payment
GROUP BY user_id
)
SELECT
d.date,
COALESCE(COUNT(p.purchase_date), 0) AS registrations
FROM
dates d
LEFT JOIN
first_purchases p ON p.purchase_date >= d.date
AND p.purchase_date < d.date + '1 ${dateUnit}'::INTERVAL
GROUP BY
d.date
ORDER BY
d.date DESC;
In PostgreSQL, I have a table that logs activity for all users, with an account ID and a timestamp field:
SELECT account_id, created FROM activity_log;
A single account_id can appear many times in a day, or not at all.
I would like a chart showing the number of "active users" each day, where "active users"
means "users who have done any activity within the previous X days".
If X is 1, then we can just truncate timestamp to 'day' and aggregate:
SELECT date_trunc('day', created) AS date, count(DISTINCT account_id)
FROM activity_log
GROUP BY date_trunc('day', created) ORDER BY date;
If X is exactly 7, then we could truncate to 'week' and aggregate - although this gives
me only one data point for a week, when I actually want one data point per day.
But I need to solve for the general case of different X, and give a distinct data point for each day.
One method is to generate the dates and then count using left join and group by or similar logic. The following uses a lateral join:
select gs.dte, al.num_accounts
from generate_series('2021-01-01'::date, '2021-01-31'::date, interval '1 day'
) gs(dte) left join lateral
(select count(distinct al.account_id) as num_accounts
from activity_log al
where al.created >= gs.dte - (<n - 1>) * interval '1 day' and
al.created < gs.dte + interval '1 day'
) al
on 1=1
order by gs.dte;
<n - 1> is one less than the number of days. So for one week, it would be 6.
If your goal is to get day wise distinct account_id for last X days you can use below query. Instead of 7 you can use any number as you wise:
SELECT date_trunc('day', created) AS date, count(DISTINCT account_id)
FROM activity_log
where date_trunc('day', created)>=date_trunc('day',CURRENT_DATE) +interval '-7' day
GROUP BY date_trunc('day', created)
ORDER BY date
(If there is no activity in any given date then the date will not be in the output.)
I am trying to get all the paid contracts from my contracts table and group them by month. I can get the data but for months where there is no new paid contract I want to get a zero instead of missing month. I have tried coalesce and generate_series but I cannot seem to get the missing row.
Here is my query:
with months as (
select generate_series(
'2019-01-01', current_date, interval '1 month'
) as series )
select date(months.series) as day, SUM(contracts.price) from months
left JOIN contracts on date(date_trunc('month', contracts.to)) = months.series
where contracts.tier='paid' and contracts.trial=false and (contracts.to is not NULL) group by day;
I want the results to look like:
|Contract Value| Month|
| 20 | 01-2020|
| 10 | 02-2020|
| 0 | 03-2020|
I can get the rows where there is a contract but cannot get the zero row.
Postgres Version 10.9
I think that you want:
with months as (
select generate_series('2019-01-01', current_date, interval '1 month' ) as series
)
select m.series as day, coalesce(sum(c.price), 0) sum_price
from months m
left join contracts c
on c.to >= m.series
and c.to < m.series + interval '1' month
and co.tier = 'paid'
and not c.trial
group by m.series;
That is:
you want the condition on the left joined table in the on clause of the join rather than in the where clause, otherwise they become mandatory, and evict rows where the left join came back empty
the filter on the date can be optimized to avoid using date functions; this makes the query SARGeable, ie the database may take advantage of an index on the date column
table aliases make the query easier to read and write
You need to move conditions to the on clause:
with months as (
select generate_series( '2019-01-01'::date, current_date, interval '1 month') as series
)
select dm.series as day, coalesce(sum(c.price), 0)
from months m left join
contracts c
on c.to >= m.series and
c.to < m.series + interval '1 month' and
c.tier = 'paid' and
c.trial = false
group by day;
Note some changes to the query:
The conditions on c that were in the where clause are in the on clause.
The date comparison uses simple data comparisons, rather than truncating to the month. This helps the optimizer and makes it easier to use an index.
Table aliases make the query easier to write and to read.
There is no need to convert day to a date. It already is.
to is a bad choice for a column name because it is reserved. However, I did not change it.
I am trying to get data to populate a multi-line graph. The table jobs has the columns id, created_at, and partner_id. I would like to display the sum of jobs for each partner_id each day. My current query has 2 problems. 1) It is missing a lot of jobs. 2) It only contains an entry for a given day if there was a row on that day. My current query is where start is an integer denoting how many days back we are looking for data:
SELECT d.date, count(j.id), j.partner_id FROM (
select to_char(date_trunc('day', (current_date - offs)), 'YYYY-MM-DD')
AS date
FROM generate_series(0, #{start}, 1)
AS offs
) d
JOIN (
SELECT jobs.id, jobs.created_at, jobs.partner_id FROM jobs
WHERE jobs.created_at > now() - INTERVAL '#{start} days'
) j
ON (d.date=to_char(date_trunc('day', j.created_at), 'YYYY-MM-DD'))
GROUP BY d.date, j.partner_id
ORDER BY j.partner_id, d.date;
This returns records like the following:
[{"date"=>"2019-06-21", "count"=>3, "partner_id"=>"099"},
{"date"=>"2019-06-22", "count"=>1, "partner_id"=>"099"},
{"date"=>"2019-06-21", "count"=>3, "partner_id"=>"075"},
{"date"=>"2019-06-23", "count"=>1, "partner_id"=>"099"}]
what I want is something like this:
[{"date"=>"2019-06-21", "count"=>3, "partner_id"=>"099"},
{"date"=>"2019-06-22", "count"=>1, "partner_id"=>"099"},
{"date"=>"2019-06-21", "count"=>3, "partner_id"=>"075"},
{"date"=>"2019-06-22", "count"=>0, "partner_id"=>"075"},
{"date"=>"2019-06-23", "count"=>0, "partner_id"=>"075"},
{"date"=>"2019-06-23", "count"=>1, "partner_id"=>"099"}]
So that for every day in the query I have an entry for every partner even if that count is 0. How can I adjust the query to populate data even when the count is 0?
Use a LEFT JOIN. You also don't need so many subqueries and there is no need to translate to a date to a string and then back to a date:
SELECT d.date, count(j.id), j.partner_id
FROM (SELECT to_char(dte, 'YYYY-MM-DD') AS date , dte
FROM generate_series(current_date - {start} * interval '1 day', current_date, interval '1 day') gs(dte)
) d LEFT JOIN
jobs j
ON DATE_TRUNC('day', j.created_at) = d.dte
GROUP BY d.date, j.partner_id
ORDER BY j.partner_id, d.date;
I am trying to calculate the median response time for conversations on each date for the last X days.
I use the following query below, but for some reason, it will generate multiple rows with the same date.
with grouping as (
SELECT a.id, d.date, extract(epoch from (first_response_at - started_at)) as response_time
FROM (
select to_char(date_trunc('day', (current_date - offs)), 'YYYY-MM-DD') AS date
FROM generate_series(0, 2) AS offs
) d
LEFT OUTER JOIN apps a on true
LEFT OUTER JOIN conversations c ON (d.date=to_char(date_trunc('day'::varchar, c.started_at), 'YYYY-MM-DD')) and a.id = c.app_id
and c.app_id = a.id and c.first_response_at > (current_date - (2 || ' days')::interval)::date
)
select
*
from grouping
where grouping.id = 'ASnYW1-RgCl0I'
Any ideas?
First a number of issues with your query, assuming there aren't any parts you haven't shown us:
You don't need a CTE for this query.
From table apps you only use column id whose value is the same as c.app_id. You can remove the table apps and select c.app_id for the same result.
When you use to_char() you do not first have to date_trunc() to a date, the to_char() function handles that.
generate_series() also works with timestamps. Just enter day values with an interval and cast the end result to date before using it.
So, removing all the flotsam we end up with this which does exactly the same as the query in your question but now we can at least see what is going on.
SELECT c.app_id, to_date(d.date, 'YYYY-MM-DD') AS date,
extract(epoch from (first_response_at - started_at)) AS response_time
FROM generate_series(CURRENT_DATE - 2, CURRENT_DATE, interval '1 day') d(date)
LEFT JOIN conversations c ON d.date::date = c.started_at::date
AND c.app_id = 'ASnYW1-RgCl0I'
AND c.first_response_at > CURRENT_DATE - 2;
You don't calculate the median response time anywhere, so that is a big problem you need to solve. This only requires data from table conversations and would look somewhat like this to calculate the median response time for the past 2 days:
SELECT app_id, started_at::date AS start_date,
percentile_disc(0.5) WITHIN GROUP (ORDER BY first_response_at - started_at) AS median_response
FROM conversations
WHERE app_id = 'ASnYW1-RgCl0I'
AND first_response_at > CURRENT_DATE - 2
GROUP BY 2;
When we fold the two queries, and put the parameters handily in a single place, this is the final result:
SELECT p.id, to_date(d.date, 'YYYY-MM-DD') AS date,
extract(epoch from (c.median_response)) AS response_time
FROM (VALUES ('ASnYW1-RgCl0I', 2)) p(id, days)
JOIN generate_series(CURRENT_DATE - p.days, CURRENT_DATE, interval '1 day') d(date) ON true
LEFT JOIN LATERAL (
SELECT started_at::date AS start_date,
percentile_disc(0.5) WITHIN GROUP (ORDER BY first_response_at - started_at) AS median_response
FROM conversations
WHERE app_id = p.id
AND first_response_at > CURRENT_DATE - p.days
GROUP BY 2) c ON d.date::date = c.start_date;
If you want to change the id of the app or the number of days to look back, you only have to change the VALUES clause accordingly. You can also wrap the whole thing in a SQL function and convert the VALUES clause into two parameters.