I have a table with records of orders by customers and a table with dates from Jan 2022 to 10 years. I wanted to get all numbers of customers made everyday for the last 28 days, including those with 0 customers recorded. So I needed to outer join the calendar table to the customer records. However, I cant use outer join correctly.
Here's how I done it:
SELECT order_date as 'date', COUNT(orderstatus) as 'customers'
FROM orders
RIGHT OUTER JOIN calendar ON
calendar.date = orders.order_date
WHERE sellerid = 11
Im getting:
date customers
2022-01-02 9
I wanted to see:
date customers
2022-01-01 0
2022-01-02 9
2022-01-03 0
.
.
.
You would not get the results that you posted in your question unless you group by date, so I guess you missed that part of your code.
You need a WHERE clause to filter the calendar's rows for the last 28 days and you must move the condition sellerid = 11 to the ON clause:
SELECT c.order_date,
COUNT(o.order_date) customers
FROM calendar c LEFT JOIN orders o
ON o.sellerid = 11 AND o.order_date = c.date
WHERE c.date BETWEEN CURRENT_DATE - INTERVAL 28 DAY AND CURRENT_DATE
GROUP BY c.order_date;
Related
I want to count %days when a user was active. A query like this
select
a.id,
a.created_at,
CURRENT_DATE - a.created_at::date as days_since_registration,
NOW() as current_d
from public.accounts a where a.id = 3257
returns
id created_at days_since_registration current_d tot_active
3257 2022-04-01 22:59:00.000 1 2022-04-02 12:00:0.000 +0400 2
The person registered less than 24 hours ago (less than a day ago), but there are two distinct dates between the registration and now. Hence, if a user was active one hour before midnight and one hour after midnight, he is two days active in less than a day (active 200% of days)
What is the right way to count distinct dates and get 2 for a user, who registered at 23:00:00 two hours ago?
WITH cte as (
SELECT 42 as userID,'2022-04-01 23:00:00' as d
union
SELECT 42,'2022-04-02 01:00:00' as d
)
SELECT
userID,
count(d),
max(d)::date-min(d)::date+1 as NrOfDays,
count(d)/(max(d)::date-min(d)::date+1) *100 as PercentageOnline
FROM cte
GROUP BY userID;
output:
userid
count
nrofdays
percentageonline
42
2
2
100
I am working on a query to return the next 7 days worth of data every time an event happens indicated by "where event = 1". The goal is to then group all the data by the user id and perform aggregate functions on this data after the event happens - the event is encoded as binary [0, 1].
So far, I have been attempting to use nested select statements to structure the data how I would like to have it, but using the window functions is starting to restrict me. I am now thinking a self join could be more appropriate but need help in constructing such a query.
The query currently first creates daily aggregate values grouped by user and date (3rd level nested select). Then, the 2nd level sums the data "value_x" to obtain an aggregate value grouped by the user. Then, the 1st level nested select statement uses the lead function to grab the next rows value over and partitioned by each user which acts as selecting the next day's value when event = 1. Lastly, the select statement uses an aggregate function to calculate the average "sum_next_day_value_after_event" grouped by user and where event = 1. Put together, where event = 1, the query returns the avg(value_x) of the next row's total value_x.
However, this doesn't follow my time rule; "where event = 1", return the next 7 days worth of data after the event happens. If there is not 7 days worth of data, then return whatever data is <= 7 days. Yes, I currently only have one lead with the offset as 1, but you could just put 6 more of these functions to grab the next 6 rows. But, the lead function currently just grabs the next row without regard to date. So theoretically, the next row's "value_x" could actually be 15 days from where "event = 1". Also, as can be seen below in the data table, a user may have more than one row per day.
Here is the following query I have so far:
select
f.user_id
avg(f.sum_next_day_value_after_event) as sum_next_day_values
from (
select
bld.user_id,
lead(bld.value_x, 1) over(partition by bld.user_id order by bld.daily) as sum_next_day_value_after_event
from (
select
l.user_id,
l.daily,
sum(l.value_x) as sum_daily_value_x
from (
select
user_id, value_x, date_part('day', day_ts) as daily
from table_1
group by date_part('day', day_ts), user_id, value_x) l
group by l.user_id, l.day_ts
order by l.user_id) bld) f
group by f.user_id
Below is a snippet of the data from table_1:
user_id
day_ts
value_x
event
50
4/2/21 07:37
25
0
50
4/2/21 07:42
45
0
50
4/2/21 09:14
67
1
50
4/5/21 10:09
8
0
50
4/5/21 10:24
75
0
50
4/8/21 11:08
34
0
50
4/15/21 13:09
32
1
50
4/16/21 14:23
12
0
50
4/29/21 14:34
90
0
55
4/4/21 15:31
12
0
55
4/5/21 15:23
34
0
55
4/17/21 18:58
32
1
55
4/17/21 19:00
66
1
55
4/18/21 19:57
54
0
55
4/23/21 20:02
34
0
55
4/29/21 20:39
57
0
55
4/30/21 21:46
43
0
Technical details:
PostgreSQL, supported by EDB, version = 14.1
pgAdmin4, version 5.7
Thanks for the help!
"The query currently first creates daily aggregate values"
I don't see any aggregate function in your first query, so that the GROUP BY clause is useless.
select
user_id, value_x, date_part('day', day_ts) as daily
from table_1
group by date_part('day', day_ts), user_id, value_x
could be simplified as
select
user_id, value_x, date_part('day', day_ts) as daily
from table_1
which in turn provides no real added value, so this first query could be removed and the second query would become :
select user_id
, date_part('day', day_ts) as daily
, sum(value_x) as sum_daily_value_x
from table_1
group by user_id, date_part('day', day_ts)
The order by user_id clause can also be removed at this step.
Now if you want to calculate the average value of the sum_daily_value_x in the period of 7 days after the event (I'm referring to the avg() function in your top query), you can use avg() as a window function that you can restrict to the period of 7 days after the event :
select f.user_id
, avg(f.sum_daily_value_x) over (order by f.daily range between current row and '7 days' following) as sum_next_day_values
from (
select user_id
, date_part('day', day_ts) as daily
, sum(value_x) as sum_daily_value_x
from table_1
group by user_id, date_part('day', day_ts)
) AS f
group by f.user_id
The partition by f.user_id clause in the window function is useless because the rows have already been grouped by f.user_id before the window function is applied.
You can replace the avg() window function by any other one, for instance sum() which could better fit with the alias sum_next_day_values
I am trying to get a list of:
all months in a specified year that,
have at least 2 unique rows based on their date
and ignore specific column values
where I got to is:
SELECT DATE_PART('month', "orderDate") AS month, count(*)
FROM public."Orders"
WHERE "companyId" = 00001 AND "orderNumber" != 1 and DATE_PART('year', ("orderDate")) = '2020' AND "orderNumber" != NULL
GROUP BY month
HAVING COUNT ("orderDate") > 2
The HAVING_COUNT sort of works in place of DISTINCT insofar as I can be reasonably sure that condition filters the condition of data required.
However, being able to use DISTINCT based on a given date within a month would return a more reliable result. Is this possible with Postgres?
A sample line of data from the table:
Sample Input
"2018-12-17 20:32:00+00"
"2019-02-26 14:38:00+00"
"2020-07-26 10:19:00+00"
"2020-10-13 19:15:00+00"
"2020-10-26 16:42:00+00"
"2020-10-26 19:41:00+00"
"2020-11-19 20:21:00+00"
"2020-11-19 21:22:00+00"
"2020-11-23 21:10:00+00"
"2021-01-02 12:51:00+00"
without the HAVING_COUNT this produces
month
count
7
1
10
2
11
3
Month 7 can be discarded easily as only 1 record.
Month 10 is the issue: we have two records. But from the data above, those records are from the same day. Similarly, month 11 only has 2 distinct records by day.
The output should therefore be ideally:
month
count
11
2
We have only two distinct dates from the 2020 data, and they are from month 11 (November)
I think you just want to take the distinct count of dates for each month:
SELECT
DATE_PART('month', orderDate) AS month,
COUNT(DISTINCT orderDate::date) AS count
FROM Orders
WHERE
companyId = 1 AND
orderNumber != 1 AND
DATE_PART('year', orderDate) = '2020'
GROUP BY
DATE_PART('month', orderDate)
HAVING
COUNT(DISTINCT orderDate::date) > 2;
I need to find the number of users that were invoiced for an amount greater than 0 in the previous month and were not invoiced in the current month. This calcualtion is to be done for 12 months in a single query. Output should be as below.
Month Count
01/07/2019 50
01/08/2019 34
01/09/2019 23
01/10/2019 98
01/11/2019 10
01/12/2019 5
01/01/2020 32
01/02/2020 65
01/03/2020 23
01/04/2020 12
01/05/2020 64
01/06/2020 54
01/07/2020 78
I am able to get the value only for one month. I want to get it for all months in a single query.
This is my current query:
SELECT COUNT(DISTINCT TWO_MONTHS_AGO.USER_ID), TWO_MONTHS_AGO.MONTH AS INVOICE_MONTH
FROM (
SELECT USER_ID, LAST_DAY(invoice_ct_dt)) AS MONTH
FROM table a AS ID
WHERE invoice_amt > 0
AND LAST_DAY(invoice_ct_dt)) = ADD_MONTHS(LAST_DAY(CURRENT_DATE - 1), - 2)
GROUP BY user_id
) AS TWO_MONTHS_AGO
LEFT JOIN (
SELECT user_id,LAST_DAY(invoice_ct_dt)) AS MONTH
FROM table a AS ID
AND LAST_DAY(invoice_ct_dt)) = ADD_MONTHS(LAST_DAY(CURRENT_DATE - 1), - 1)
GROUP BY USER_ID
) AS ONE_MONTH_AGO ON TWO_MONTHS_AGO.USER_ID = ONE_MONTH_AGO.USER_ID
WHERE ONE_MONTH_AGO.USER_ID IS NULL
GROUP BY INVOICE_MONTH;
Thank you in advance.
Lona
Probably lots of different approaches but the way I would do it is as follows:
Summarise data by user and month for the last 13 months (you need 12 months plus the previous month to that first month
Compare "this" month (that has data) to "next" month and select records where there is no "next" month data
Summarise this dataset by month and distinct userid
For example, assuming a table created as follows:
create table INVOICE_DATA (
USERID varchar(4),
INVOICE_DT date,
INVOICE_AMT NUMBER(10,2)
);
the following query should give you what you want - you may need to adjust it depending on whether you are including this month, or only up to the end of last month, in your calculation, etc.:
--Summarise data by user and month
WITH MONTH_SUMMARY AS
(
SELECT USERID
,TO_CHAR(INVOICE_DT,'YYYY-MM') "INVOICE_MONTH"
,TO_CHAR(ADD_MONTHS(INVOICE_DT,1),'YYYY-MM') "NEXT_MONTH"
,SUM(INVOICE_AMT) "MONTHLY_TOTAL"
FROM INVOICE_DATA
WHERE INVOICE_DT >= TRUNC(ADD_MONTHS(current_date(),-13),'MONTH') -- Last 13 months of data
GROUP BY 1,2,3
),
--Get data for users with invoices in this month but not the next month
USER_DATA AS
(
SELECT USERID, INVOICE_MONTH, MONTHLY_TOTAL
FROM MONTH_SUMMARY MS_THIS
WHERE NOT EXISTS
(
SELECT USERID
FROM MONTH_SUMMARY MS_NEXT
WHERE
MS_THIS.USERID = MS_NEXT.USERID AND
MS_THIS.NEXT_MONTH = MS_NEXT.INVOICE_MONTH
)
AND MS_THIS.INVOICE_MONTH < TO_CHAR(current_date(),'YYYY-MM') -- Don't include this month as obviously no next month to compare to
)
SELECT INVOICE_MONTH, COUNT(DISTINCT USERID) "USER_COUNT"
FROM USER_DATA
GROUP BY INVOICE_MONTH
ORDER BY INVOICE_MONTH
;
I want to count number of month for particular user from subscription table. for example user_id = 1 occur 10 times in subscription table like in January it appears 2 time and in February = 0 and again in march = 1 like that
user_id type started_on ended_on
2 P 2009-10-21 2010-03-18
2 F 2010-03-18 2010-03-20
2 P 2010-03-20 2012-05-19
2 F 2012-05-19 till now
This is pretty basic SQL, I reccomend you read some manual before asking it here. About the Aggregate functions for example. But there you go:
If you want one user:
SELECT
count(distinct month)
FROM
subscription
WHERE
user_id=your_user_id_number
If you want every user's:
SELECT
count(distinct month),
user_id
FROM
subscription
GROUP BY
user_id
Edit:
Ok, so you want the month difference between two date columns, here is how you do it with age():
SELECT user_id, extract(YEAR from age(coalesce(ended_on,current_date),started_on)) * 12 + extract(MONTH FROM age(coalesce(ended_on,current_date),started_on))
FROM subscription