How to natural join the two queries having with clause? - postgresql

I have written two queries that help in finding the minimum and maximum sales quantities for different products. Now, I need to merge these two queries using natural join to output one single table.
Query 1:
with max_quant_table as
(with maxquant_table as
(select distinct month,prod as prod, sum(quant) as quant from sales group by month,prod)
select month as month,prod as MOST_POPULAR_PROD, quant as MOST_POP_TOTAL_Q
from maxquant_table)
select t2.* from
(select month, max(MOST_POP_TOTAL_Q) maxQ FROM max_quant_table group by month order by month asc)
t1 join max_quant_table t2 on t1.month = t2.month and (t2.MOST_POP_TOTAL_Q =maxQ)
Query 2:
with min_quant_table as
(with minquant_table as
(select distinct month,prod as prod, sum(quant) as quant from sales group by month,prod)
select month as month,prod as LEAST_POPULAR_PROD, quant as LEAST_POP_TOTAL_Q
from minquant_table)
select t2.* from
(select month, min(LEAST_POP_TOTAL_Q) minQ FROM min_quant_table group by month order by month asc)
t1 join min_quant_table t2 on t1.month = t2.month and (t2.LEAST_POP_TOTAL_Q = minQ)

You are over complicating things. You don't need to join those two query (and should really stay away from a natural join), you only need to combine them. min() and max() can be used inside the same query, there is no need to run two queries to evaluate both.
You also don't need to nest CTE definitions, you can just write one after the other.
So something like this:
with quant_table as (
select month, prod, sum(quant) as sum_q
from sales
group by month, prod
), min_max as (
select month, max(sum_q) as max_q, min(sum_q) as min_q
from quant_table
group by month
)
select t1.*
from quant_table t1
join min_max t2
on t2.month = t1.month
and t1.sum_q in (t2.min_q, t2.max_q)
order by month, prod;
The condition and t1.sum_q in (t2.min_q, t2.max_q) could also be written as and (t2.max_q = t1.sum_q or t2.min_q = t1.sum_q).
The above can further be simplified by combining group by with window functions and do the calculation of the sum, min and max in a single query:
with min_max as (
select month, prod,
sum(quant) as sum_q,
max(sum(quant)) over (partition by month) as max_q,
min(sum(quant)) over (partition by month) as min_q
from sales
group by month, prod
)
select month, prod, sum_q
from min_max
where sum_q in (max_q, min_q)
order by month, prod;

Related

Is there a SQL code for cumulative count of SaaS customer over months?

I have a table with:
ID (id client), date_start (subscription of SaaS), date_end (could be a date value or be NULL).
So I need a cumulative count of active clients month by month.
any idea on how to write that in Postgres and achieve this result?
Starting from this, but I don't know how to proceed
select
date_trunc('month', c.date_start)::date,
count(*)
from customer
Please check next solution:
select
subscrubed_date,
subscrubed_customers,
unsubscrubed_customers,
coalesce(subscrubed_customers, 0) - coalesce(unsubscrubed_customers, 0) cumulative
from (
select distinct
date_trunc('month', c.date_start)::date subscrubed_date,
sum(1) over (order by date_trunc('month', c.date_start)) subscrubed_customers
from customer c
order by subscrubed_date
) subscribed
left join (
select distinct
date_trunc('month', c.date_end)::date unsubscrubed_date,
sum(1) over (order by date_trunc('month', c.date_end)) unsubscrubed_customers
from customer c
where date_end is not null
order by unsubscrubed_date
) unsubscribed on subscribed.subscrubed_date = unsubscribed.unsubscrubed_date;
share SQL query
You have a table of customers. With a start date and sometimes an end date. As you want to group by date, but there are two dates in the table, you need to split these first.
Then, you may have months where only customers came and others where only customers left. So, you'll want a full outer join of the two sets.
For a cumulative sum (also called a running total), use SUM OVER.
with came as
(
select date_trunc('month', date_start) as month, count(*) as cnt
from customer
group by date_trunc('month', date_start)
)
, went as
(
select date_trunc('month', date_end) as month, count(*) as cnt
from customer
where date_end is not null
group by date_trunc('month', date_end)
)
select
month,
came.cnt as cust_new,
went.cnt as cust_gone,
sum(came.cnt - went.cnt) over (order by month) as cust_active
from came full outer join went using (month)
order by month;

Get an average monthly view of active members (Postgresql)

I am working with members data. I have the responsible Coach, the coachee entry, exit status and date. Because some coachees might graduate/leave during a month I want to calculate a daily number and then get a monthly average of active members for each coach. That means that I need to take in the account all coachees from previous months, that are still active that current month. This is my data:
I am thinking of creating a variable first where I can get the daily active member count for each coach. This is my first approach:
with all_years as (
select y.year, m.month, d.day
from generate_series(2019, 2022) as y(year)
cross join generate_series(1, 12) as m(month)
cross join generate_series(1, 31) as d(day) --<<*not sure how to adjust for days with less than 31 days??*
select ay.*, coach, coachee, entry_status, entry_date, exit_reason, exit_date, sum(count) over (partition by ay.coach order by ay.year, ay.month, ay.day)
from all_years ay
left join table t
on --.... *not sure what I can join on in this case*;
I am open to an easier approach, this logic is just an idea.
You can cross join the list of distinct coaches with all dates to generat combinations, then bring the table with a left join:
select d.dt, c.coach, count(t.coach) no_coachees
from (select distinct coach from mytable) c
cross join generate_series('2019-01-01'::date, '2022-12-31'::date, '1 day':: interval) d(dt)
left join mytable t on t.coach = c.coach and t.entry_date <= d.dt and t.exit_date > d.dt
group by d.dt, c.coach
Then you can use another level of aggregation to get the monthly average:
select date_trunc('month', d.dt) d_month, coach, avg(no_coachees) avg_coaches
from (
select d.dt, c.coach, count(t.coach) no_coachees
from (select distinct coach from mytable) c
cross join generate_series('2019-01-01'::date, '2022-12-31'::date, '1 day':: interval) d(dt)
left join mytable t on t.coach = c.coach and t.entry_date <= d.dt and t.exit_date > d.dt
group by d.dt, c.coach
) t
group by date_trunc('month', d.dt), coach

I have two queries and I want to natural join them so that the table for each quantity is displayed separately

The two queries display max and min quantities along with their products and month, I want to join them to display a single table with each entity(except month) having a seperate column.
Query 1(max):
with max_quant_table as
(with maxquant_table as
(select month,
prod as prod,
sum(quant) as quant
from sales
group by month,prod)
select month as month,
prod as MOST_POPULAR_PROD,
quant as MOST_POP_TOTAL_Q
from maxquant_table)
select t2.*
from (select month,
max(MOST_POP_TOTAL_Q) maxQ
FROM max_quant_table
group by month
order by month asc)t1
join max_quant_table t2
on t1.month = t2.month and (t2.MOST_POP_TOTAL_Q = maxQ)
Query 2(min):
with min_quant_table as
(with minquant_table as
(select month,
prod as prod,
sum(quant) as quant
from sales
group by month, prod)
select month as month,
prod as LEAST_POPULAR_PROD,
quant as LEAST_POP_TOTAL_Q
from minquant_table)
select t2.*
from (select month,
min(LEAST_POP_TOTAL_Q) minQ
FROM min_quant_table
group by month
order by month asc) t1
join min_quant_table t2
on t1.month = t2.month and (t2.LEAST_POP_TOTAL_Q = minQ)

Join on generate_series and count

I'm trying to find the # users who did action A or action B on a monthly basis.
Table: User
- id
- "creationDate"
Table: action_A
- user_id (= user.id)
- "creationDate"
Table: action_B
- user_id (= user.id)
- "creationDate"
The general idea of what I was trying to do was that I'd find the list of users who did action A in Month X and the list of users who did action B in Month X, then count how many ids are there for every month based on a generate_series of monthly dates.
I tried the following, however, the query times out when running and I'm not sure if there's any way to optimize it (or if it is even correct).
SELECT monthseries."Month", count(*)
FROM
(SELECT to_char(DAY::date, 'YYYY-MM') AS "Month"
FROM generate_series('2014-01-01'::date, CURRENT_DATE, '1 month') DAY) monthseries
LEFT JOIN
(SELECT to_char("creationDate", 'YYYY-MM') AS "Month",
id
FROM action_A) did_action_A ON monthseries."Month" = did_action_A."Month"
LEFT JOIN
(SELECT to_char("creationDate", 'YYYY-MM') AS "Month",
id
FROM action_B) did_action_B ON monthseries."Month" = did_action_B."Month"
GROUP BY monthseries."Month"
Any comments/ help would be immensely helpful!
If you want to count distinct users:
select to_char(month, 'YYYY-MM') as "Month", count(*)
from
generate_series(
'2014-01-01'::date, current_date, '1 month'
) monthseries (month)
left join (
(
select distinct date_trunc('month', "creationDate") as month, id
from action_a
) a
full outer join (
select distinct date_trunc('month', "creationDate") as month, id
from action_b
) b using (month, id)
) s using (month)
group by 1
order by 1

multiple extract() with WHERE clause possible?

So far I have come up with the below:
WHERE (extract(month FROM orders)) =
(SELECT min(extract(month from orderdate))
FROM orders)
However, that will consequently return zero to many rows, and in my case, many, because many orders exist within that same earliest (minimum) month, i.e. 4th February, 9th February, 15th Feb, ...
I know that a WHERE clause can contain multiple columns, so why wouldn't the below work?
WHERE (extract(day FROM orderdate)), (extract(month FROM orderdate)) =
(SELECT min(extract(day from orderdate)), min(extract(month FROM orderdate))
FROM orders)
I simply get: SQL Error: ORA-00920: invalid relational operator
Any help would be great, thank you!
Sample data:
02-Feb-2012
14-Feb-2012
22-Dec-2012
09-Feb-2013
18-Jul-2013
01-Jan-2014
Output:
02-Feb-2012
14-Feb-2012
Desired output:
02-Feb-2012
I recreated your table and found out you just messed up the brackets a bit. The following works for me:
where
(extract(day from OrderDate),extract(month from OrderDate))
=
(select
min(extract(day from OrderDate)),
min(extract(month from OrderDate))
from orders
)
Use something like this:
with cte1 as (
select
extract(month from OrderDate) date_month,
extract(day from OrderDate) date_day,
OrderNo
from tablename
), cte2 as (
select min(date_month) min_date_month, min(date_day) min_date_day
from cte1
)
select cte1.*
from cte1
where (date_month, date_day) = (select min_date_month, min_date_day from cte2)
A common table expression enables you to restructure your data and then use this data to do your select. The first cte-block (cte1) selects the month and the day for each of your table rows. Cte2 then selects min(month) and min(date). The last select then combines both ctes to select all rows from cte1 that have the desired month and day.
There is probably a shorter solution to that, however I like common table expressions as they are almost all the time better to understand than the "optimal, shortest" query.
If that is really what you want, as bizarre as it seems, then as a different approach you could forget the extracts and the subquery against the table to get the minimums, and use an analytic approach instead:
select orderdate
from (
select o.*,
row_number() over (order by to_char(orderdate, 'MMDD')) as rn
from orders o
)
where rn = 1;
ORDERDATE
---------
01-JAN-14
The row_number() effectively adds a pseudo-column to every row in your original table, based on the month and day in the order date. The rn values are unique, so there will be one row marked as 1, which will be from the earliest day in the earliest month. If you have multiple orders with the same day/month, say 01-Jan-2013 and 01-Jan-2014, then you'll still only get exactly one with rn = 1, but which is picked is indeterminate. You'd need to add further order by conditions to make it deterministic, but I have no idea what you might want.
That is done in the inner query; the outer query then filters so that only the records marked with rn = 1 is returned; so you get exactly one row back from the overall query.
This also avoids the situation where the earliest day number is not in the earliest month number - say if you only had 01-Jan-2014 and 02-Feb-2014; comparing the day and month separately would look for 01-Feb-2014, which doesn't exist.
SQL Fiddle (with Thomas Tschernich's anwer thrown in too, giving the same result for this data).
To join the result against your invoice table, you don't need to join to the orders table again - especially not with a cross join, which is skewing your results. You can do the join (at least) two ways:
SELECT
o.orderno,
to_char(o.orderdate, 'DD-MM-YYYY'),
i.invno
FROM
(
SELECT o.*,
row_number() over (order by to_char(orderdate, 'MMDD')) as rn
FROM orders o
) o, invoices i
WHERE i.invno = o.invno
AND rn = 1;
Or:
SELECT
o.orderno,
to_char(o.orderdate, 'DD-MM-YYYY'),
i.invno
FROM
(
SELECT orderno, orderdate, invno
FROM
(
SELECT o.*,
row_number() over (order by to_char(orderdate, 'MMDD')) as rn
FROM orders o
)
WHERE rn = 1
) o, invoices i
WHERE i.invno = o.invno;
The first looks like it does more work but the execution plans are the same.
SQL Fiddle with your pastebin-supplied query that gets two rows back, and these two that get one.