I have this table (products with information about start and end publish
):
SKU start_time end_time
id1 21.01.2020 14:10:00 22.01.2020 16:18:05
id1 23.01.2020 16:18:05 24.01.2020 19:03:14
id2 21.01.2020 16:18:05 21.01.2020 18:33:50
id3 25.01.2020 18:33:50 25.01.2020 19:03:14
and expect active products by days in two variants (not include comments in braces):
date active_sku active sku_end_of_day
21.01.2020 2 (id1,id2) 1 (id1)
22.01.2020 1 (id1) 0
23.01.2020 1 (id1) 1 (id1)
24.01.2020 1 (id1) 0
25.01.2020 1 (id3) 0
Below is for BigQuery Standard SQL
Assuming that start_time and end_time are timestamps data type - consider below
select date,
count(distinct SKU) as active_sku,
count(distinct if(offset = 0, null, SKU)) as active_sku_end_of_date
from `project.dataset.table`,
unnest(array_reverse(generate_date_array(date(start_time),date(end_time)))) date with offset
group by date
# order by date
if applied to sample data in your question - output is
In case if start_time and end_time are strings - you should use parse_timestamp() function to parse timestamp from string - as in below example
select date,
count(distinct SKU) as active_sku,
count(distinct if(offset = 0, null, SKU)) as active_sku_end_of_date
from `project.dataset.table`,
unnest(array_reverse(generate_date_array(date(parse_timestamp('%d.%m.%Y %T', start_time)),date(parse_timestamp('%d.%m.%Y %T', end_time))))) date with offset
group by date
# order by date
Related
I have 3 tables, a user table, an admin table, and a cust table. Both admin and cust tables are foreign keyed to the user_account table. Basically, every user has a user record, and the type of user they are is determined by if they have a record in the admin or the cust table.
user admin cust
user_id user_id | admin_id user_id | cust_id
--------- ---------|---------- ---------|---------
1 1 | a 2 | dd
2 4 | b 3 | ff
3
4
Then I have a login_history table that records the user_id and login timestamp every time a user logs into the app
login_history
user_id | login_on
---------|---------------------
1 | 2022-01-01 13:22:43
1 | 2022-01-02 16:16:27
3 | 2022-01-05 21:17:52
2 | 2022-01-11 11:12:26
3 | 2022-01-12 03:34:47
I would like to create a view that would contain all dates for the first day of each week in the year starting from jan 1st, and a count column that contains the count of unique admin users that logged in that week and a count of unique cust users that logged in that week. So the resulting view should contain the following 53 records, one for each week.
login_counts_view
week_start_date | admin_count | cust_count
-----------------|-------------|------------
2022-01-01 | 1 | 1
2022-01-08 | 0 | 2
2022-01-15 | 0 | 0
.
.
.
2022-12-31 | 0 | 0
Note that the first week (2022-01-01) only has 1 count for admin_count even though the admin with user_id 1 logged in twice that week.
Below is the current query I have for the view. However, the tables are pretty large and it takes over 10 seconds to retrieve all records from the view, mainly because of the left joined date comparisons.
CREATE VIEW login_counts_view AS
SELECT
week_start_dates.week_start_date::text AS week_start_date,
count(distinct a.user_id) AS admin_count,
count(distinct c.user_id) AS cust_count
FROM (
SELECT
to_char(i::date, 'YYYY-MM-DD') AS week_start_date
FROM
generate_series(date_trunc('year', NOW()), to_char(NOW(), 'YYYY-12-31')::date, '1 week') i
) week_start_dates
LEFT JOIN login_history l ON l.login_on::date BETWEEN week_start_dates.week_start_date::date AND (week_start_dates.week_start_date::date + INTERVAL '6 day')::date
LEFT JOIN admin a ON a.user_id = l.user_id
LEFT JOIN cust c ON c.user_id = l.user_id
GROUP BY week_start_date;
Does anyone have any tips as to how to make this query execute more efficiently?
Idea
Compute the pseudo-week of each login date: partition the year into 7-day slices and number them consecutively. The pseudo-week of a given date would be the ordinal number of the slice it falls into.
Then operate the joins on integers representing the pseudo-weeks instead of date values and comparisons.
Implementation
A view to implement this follows:
CREATE VIEW login_counts_view_fast AS
WITH RECURSIVE Numbers(i) AS ( SELECT 0 UNION ALL SELECT i + 1 FROM Numbers WHERE i < 52 )
SELECT CAST ( date_trunc('year', NOW()) AS DATE) + 7 * n.i week_start_date
, count(distinct lw.admin_id) admin_count
, count(distinct lw.cust_id) cust_count
FROM (
SELECT i FROM Numbers
) n
LEFT JOIN (
SELECT admin_id
, cust_id
, base
, pit
, pit-base delta
, (pit-base) / (3600 * 24 * 7) week
FROM (
SELECT a.user_id admin_id
, c.user_id cust_id
, CAST ( EXTRACT ( EPOCH FROM l.login_on ) AS INTEGER ) pit
, CAST ( EXTRACT ( EPOCH FROM date_trunc('year', NOW()) ) AS INTEGER ) base
FROM login_history l
LEFT JOIN admin a ON a.user_id = l.user_id
LEFT JOIN cust c ON c.user_id = l.user_id
) le
) lw
ON lw.week = n.i
GROUP BY n.i
;
Some remarks:
The epoch values are the number of seconds elapsed since an absolute base datetime (specifically 1/1/1970 0h00).
CASTS are necessary to convert doubles to integers and timestamps to dates as mandated by the signatures of postgresql date functions and in order to enforce integer arithmetics.
The recursive subquery is a generator of consecutive integers. It could possibly be replaced by a generate_series call (untested)
Evaluation
See it in action in this db fiddle
The query plan indicates savings of 50-70% in execution time.
Here is the schema I'm working with
-- Table Definition ----------------------------------------------
CREATE TABLE transactions (
id BIGSERIAL PRIMARY KEY,
date date,
amount double precision,
category character varying,
full_category character varying,
transaction_id character varying,
created_at timestamp(6) without time zone NOT NULL,
updated_at timestamp(6) without time zone NOT NULL
);
-- Indices -------------------------------------------------------
CREATE UNIQUE INDEX transactions_pkey ON transactions(id int8_ops);
I would like to group the data with the following columns:
Category, January Total, February Total, March Total, and so on for every month.
This is as far as I've got:
SELECT
category, sum(amount) as january_total
from transactions
where category NOT IN ('Transfer', 'Payment', 'Deposit', 'Income')
AND date >= '2021-01-01' AND date < '2021-02-01'
group by category
Order by january_total asc
How do I add a column for every month to this output?
Here is the solution I came up with:
SELECT
category,
SUM(CASE WHEN date >= '2021-01-01' AND date <'2021-02-01' THEN amount ELSE 0.00 END) AS january,
SUM(CASE WHEN date >= '2021-02-01' AND date <'2021-03-01' THEN amount ELSE 0.00 END) AS february,
SUM(CASE WHEN date >= '2021-03-01' AND date <'2021-04-01' THEN amount ELSE 0.00 END) AS march,
SUM(CASE WHEN date >= '2021-04-01' AND date <'2021-05-01' THEN amount ELSE 0.00 END) AS april
from transactions
where category NOT IN ('Transfer', 'Payment', 'Deposit', 'Income')
GROUP BY category
Order by category
I have a table called Position, in this table, I have the following, dates are inclusive (yyyy-mm-dd), below is a simplified view of the employment dates
id, person_id, start_date, end_date , title
1 , 1 , 2001-12-01, 2002-01-31, 'admin'
2 , 1 , 2002-02-11, 2002-03-31, 'admin'
3 , 1 , 2002-02-15, 2002-05-31, 'sales'
4 , 1 , 2002-06-15, 2002-12-31, 'ops'
I'd like to be able to calculate the gaps in employment, assuming some of the dates overlap to produce the following output for the person with id=1
person_id, start_date, end_date , last_position_id, gap_in_days
1 , 2002-02-01, 2002-02-10, 1 , 10
1 , 2002-06-01, 2002-06-14, 3 , 14
I have looked at numerous solutions, UNIONS, Materialized views, tables with generated calendar date ranges, etc. I really am not sure what is the best way to do this. Is there a single query where I can get this done?
step-by-step demo:db<>fiddle
You just need the lead() window function. With this you are able to get a value (start_date in this case) to the current row.
SELECT
person_id,
end_date + 1 AS start_date,
lead - 1 AS end_date,
id AS last_position_id,
lead - (end_date + 1) AS gap_in_days
FROM (
SELECT
*,
lead(start_date) OVER (PARTITION BY person_id ORDER BY start_date)
FROM
positions
) s
WHERE lead - (end_date + 1) > 0
After getting the next start_date you are able to compare it with the current end_date. If they differ, you have a gap. These positive values can be filtered within the WHERE clause.
(if 2 positions overlap, the diff is negative. So it can be ignored.)
first you need to find what dates overlaps Determine Whether Two Date Ranges Overlap
then merge those ranges as a single one and keep the last id
finally calculate the ranges of days between one end_date and the next start_date - 1
SQL DEMO
with find_overlap as (
SELECT t1."id" as t1_id, t1."person_id", t1."start_date", t1."end_date",
t2."id" as t2_id, t2."start_date" as t2_start_date, t2."end_date" as t2_end_date
FROM Table1 t1
LEFT JOIN Table1 t2
ON t1."person_id" = t2."person_id"
AND t1."start_date" <= t2."end_date"
AND t1."end_date" >= t2."start_date"
AND t1.id < t2.id
), merge_overlap as (
SELECT
person_id,
start_date,
COALESCE(t2_end_date, end_date) as end_date,
COALESCE(t2_id, t1_id) as last_position_id
FROM find_overlap
WHERE t1_id NOT IN (SELECT t2_id FROM find_overlap WHERE t2_ID IS NOT NULL)
), cte as (
SELECT *,
LEAD(start_date) OVER (partition by person_id order by start_date) next_start
FROM merge_overlap
)
SELECT *,
DATE_PART('day',
(next_start::timestamp - INTERVAL '1 DAY') - end_date::timestamp
) as days
FROM cte
WHERE next_start IS NOT NULL
OUTPUT
| person_id | start_date | end_date | last_position_id | next_start | days |
|-----------|------------|------------|------------------|------------|------|
| 1 | 2001-12-01 | 2002-01-31 | 1 | 2002-02-11 | 10 |
| 1 | 2002-02-11 | 2002-05-31 | 3 | 2002-06-15 | 14 |
I’m having trouble getting my head around this.
I’m looking for a single query, if possible, running PostgreSQL 9.6.6 under pgAdmin3 v1.22.1
I have a table with a date and a row for each event on the date:
Date Events
2018-12-10 1
2018-12-10 1
2018-12-10 0
2018-12-09 1
2018-12-08 0
2018-12-07 1
2018-12-06 1
2018-12-06 1
2018-12-06 1
2018-12-05 1
2018-12-04 1
2018-12-03 0
I’m looking for the longest sequence of dates without a break. In this case, 2018-12-08 and 2018-12-03 are the only dates with no events, there are two dates with events between 2018-12-08 and today, and four between 2018-12-8 and 2018-12-07 - so I would like the answer of 4.
I know I can group them together with something like:
Select Date, count(Date) from Table group by Date order by Date Desc
To get just the most recent sequence, I’ve got something like this- the subquery returns the most recent date with no events, and the outer query counts the dates after that date:
select count(distinct date) from Table
where date>
( select date from Table
group by date
having count (case when Events is not null then 1 else null end) = 0
order by date desc
fetch first row only)
But now I need the longest streak, not just the most recent streak.
Thank you!
Your instinct is a good one in looking at the rows with zero events and working off them. We can use a subquery with a window function to get the "gaps" between zero event days, and then in a query outside it take the record we want, like so:
select *
from (
select date as day_after_streak
, lag(date) over(order by date asc) as previous_zero_date
, date - lag(date) over(order by date asc) as difference
, date_part('days', date - lag(date) over(order by date asc) ) - 1 as streak_in_days
from dates
group by date
having sum(events) = 0 ) t
where t.streak_in_days is not null
order by t.streak_in_days desc
limit 1
user timespent(in sec) date(in timestamp)
u1 10 t1(2015-08-15)
u1 20 t2(2015-08-19)
u1 15 t3(2015-08-28)
u1 16 t4(2015-09-06)
Above is the format of my table, which represents timespent by user on a course and it is ordered by timestamp. I want to get sum of timespent by a particular user, say u1 weekly in the format :
start_date end_date sum
2015-08-15 2015-08-21 30
2015-08-22 2015-08-28 15
2015-08-29 2015-09-04 0
2015-09-05 2015-09-11 16
The difficulty lies in the fact that the seven-day periods that you want to get are not regular weeks starting with Monday.
You can not therefore use standard functions to get the week number based on the date, and have to use your own weeks generator using generate_series().
Example data:
create table sessions (user_name text, time_spent int, session_date timestamp);
insert into sessions values
('u1', 10, '2015-08-15'),
('u1', 20, '2015-08-19'),
('u1', 15, '2015-08-28'),
('u1', 16, '2015-09-06');
The query for an arbitrary chosen period from 2015-08-15 to 2015-09-06:
with weeks as (
select d::date start_date, d::date+ 6 end_date
from generate_series('2015-08-15', '2015-09-06', '7d'::interval) d
)
select w.start_date, w.end_date, coalesce(sum(time_spent), 0) total
from weeks w
left join (
select start_date, end_date, coalesce(time_spent, 0) time_spent
from weeks
join sessions
on session_date between start_date and end_date
where user_name = 'u1'
) s
on w.start_date = s.start_date and w.end_date = s.end_date
group by 1, 2
order by 1;
start_date | end_date | total
------------+------------+-------
2015-08-15 | 2015-08-21 | 30
2015-08-22 | 2015-08-28 | 15
2015-08-29 | 2015-09-04 | 0
2015-09-05 | 2015-09-11 | 16
(4 rows)
select
ui,
date_trunc('week', the_date)::date as start_date,
date_trunc('week', the_date)::date + 6 as end_date,
sum(timespent) as "sum"
from t
group by 1, 2, 3
order by 1,2
Something like this (assuming that by timestamp you mean the data type timestamp).
In order to make the 1st day of the week to be Sunday, I added and extra day to "date" in the group by.
select (start_date - date_part('dow', start_date) * interval '1 day')::date start_date,
(start_date + (6 - date_part('dow', start_date)) * interval '1 day')::date end_date,
total_time_spent
from (
select min("date") start_date, sum(timespent) total_time_spent
from mytable
where user=u1
group by date_part('year', "date"), date_part('week', "date" + interval '1 day')) "tmp"
order by start_date
This is a more generic approach, for any date interval.