I'm trying to pull a few SUMs, but I'm getting stuck on how to narrow it down to a specific year.
I have the following code...
SELECT SITE_ID,
Extract(YEAR FROM DATE_ORDERED) YEAR,
Extract(MONTH FROM DATE_ORDERED) MONTH,
SUM(TOTAL_PRICE),
SUM(TOTAL_PRICE),
SUM(TOTAL_SAVINGS)
FROM DB.ACTUAL_SAVINGS_MVIEW
WHERE SITE_ID = 561
GROUP BY SITE_ID,
Extract(YEAR FROM DATE_ORDERED),
Extract(MONTH FROM DATE_ORDERED)
ORDER BY YEAR DESC,
MONTH DESC
This returns all available years, when I'm only looking for 2016.
Any and all help would be greatly appreciated!
How about adding it to your where clause:
SELECT SITE_ID,
Extract(YEAR FROM DATE_ORDERED) YEAR,
Extract(MONTH FROM DATE_ORDERED) MONTH,
SUM(TOTAL_PRICE),
SUM(TOTAL_PRICE),
SUM(TOTAL_SAVINGS)
FROM DB.ACTUAL_SAVINGS_MVIEW
WHERE SITE_ID = 561
AND Extract(YEAR FROM DATE_ORDERED) = 2016
GROUP BY SITE_ID,
Extract(YEAR FROM DATE_ORDERED),
Extract(MONTH FROM DATE_ORDERED)
ORDER BY YEAR DESC,
MONTH DESC
Related
In CockroachDB, I want to have such this query on a specific month for its every day:
select count(*), sum(amount)
from request
where code = 'code_string'
and created_at >= '2022-07-31T20:30:00Z' and created_at < '2022-08-31T20:30:00Z'
the problem is that I want it on my local date. What should I do?
My goal is:
"month, day, count, sum" as result columns for a month.
UPDATE:
I have found a suitable query for this purpose:
select count(amount), sum(amount), extract(month from created_at) as monthTime, extract(day from created_at) as dayTime
from request
where code = 'code_string' and created_at >= '2022-07-31T20:30:00Z' and created_at < '2022-08-31T20:30:00Z'
group by dayTime, monthTime
Thanks to #histocrat for easier answer :) by replacing
extract(month from created_at) as monthTime, extract(day from created_at) as dayTime
by this:
date_part('month', created_at) as monthTime, date_part('day', created_at) as dayTime
To group results by both month and day, you can use the date_part function.
select month, day, count(*), sum(things)
from request
where code = 'code_string'
group by date_part('month', created_at) as month, date_part('day', created_at) as day;
Depending on what type created_at is, you may need to cast or convert it first (for example, group by date_part('month', created_at::timestamptz)).
I need the monthly new user count and the total count of users until that month end in Oracle SQL. Can calculate monthly new users, but struggling to count total for that month.
SELECT COUNT(*) AS REGISTERED_USERS, EXTRACT(MONTH FROM ADD_DATE) AS MN, EXTRACT(YEAR FROM ADD_DATE) AS YR
FROM USERS
WHERE EXTRACT(YEAR FROM ADD_DATE) = 2020
GROUP BY EXTRACT(MONTH FROM ADD_DATE), EXTRACT(YEAR FROM ADD_DATE)
ORDER BY EXTRACT(MONTH FROM ADD_DATE);
You want a running total. You get this with SUM OVER.
SELECT yr, mn, new_users, total_users
FROM
(
SELECT
EXTRACT(YEAR FROM add_date) AS yr,
EXTRACT(MONTH FROM add_date) AS mn,
COUNT(*) AS new_users,
SUM(COUNT(*)) OVER (ORDER BY EXTRACT(YEAR FROM add_date), EXTRACT(MONTH FROM add_date)) AS total_users
FROM users
GROUP BY EXTRACT(YEAR FROM add_date), EXTRACT(MONTH FROM add_date)
)
WHERE yr = 2020
ORDER BY yr, mn;
You can use the SUM analytical function with COUNT aggregate function and running the analytical function on add_date by month as follows:
SELECT EXTRACT(YEAR FROM DT) yr, EXTRACT(MONTH FROM DT) mn, new_users, total_users
FROM
(
SELECT
TRUNC(ADD_DATE, 'MON') AS DT,
COUNT(*) AS new_users,
SUM(COUNT(*)) OVER (ORDER BY TRUNC(ADD_DATE, 'MON')) AS total_users
FROM users
GROUP BY TRUNC(ADD_DATE, 'MON')
)
WHERE EXTRACT(YEAR FROM DT) = 2020
ORDER BY YR, MN;
I have a table called 'daily_prices' where I have 'sale_date', 'last_sale_price', 'symbol' as columns.
I need to calculate how many times 'last_sale_price' has gone up compared to previous day's 'last_sale_price' in 10 weeks.
Currently I have my query like this for 2 weeks:
select count(*) as "timesUp", sum(last_sale_price-prev_price) as "dollarsUp", 'wk1' as "week"
from
(
select last_sale_price, LAG(last_sale_price, 1) OVER (ORDER BY sale_date) as prev_price
from daily_prices
where sale_date <= CAST('2020-09-18' AS DATE) AND sale_date >= CAST('2020-09-14' AS DATE)
and symbol='AAPL'
) nest
where last_sale_price > prev_price
UNION
select count(*) as "timesUp", sum(last_sale_price-prev_price) as "dollarsUp", 'wk2' as "week"
from
(
select last_sale_price, LAG(last_sale_price, 1) OVER (ORDER BY sale_date) as prev_price
from daily_prices
where sale_date <= CAST('2020-09-11' AS DATE) AND sale_date >= CAST('2020-09-07' AS DATE)
and symbol='AAPL'
) nest
where last_sale_price > prev_price
I'm using 'UNION' to combine the weekly data. But as the number of weeks increase the query is going to be huge.
Is there a simpler way to write this query?
Any help is much appreciated. Thanks in advance.
you can extract week from sale_date. then apply group by on the upper query
select EXTRACT(year from sale_date) YEAR, EXTRACT('week' FROM sale_date) week, count(*) as "timesUp", sum(last_sale_price-prev_price) as "dollarsUp"
from (
select
sale_date,
last_sale_price,
LAG(last_sale_price, 1) OVER (ORDER BY sale_date) as prev_price
from daily_prices
where symbol='AAPL'
)
where last_sale_price > prev_price
group by EXTRACT(year from sale_date), EXTRACT('week' FROM sale_date)
to extract only weekdays you can add this filter
EXTRACT(dow FROM sale_date) in (1,2,3,4,5)
PS: make sure that monday is first day of the week. In some countries sunday is the first day of the week
You can filter on the last 8 weeks in the where clause, then group by week and do conditional aggregation:
select extract(year from sale_date) yyyy, extract(week from saledate) ww,
sum(last_sale_price - lag_last_sale_price) filter(where lag_last_sale_price > last_sale_price) sum_dollars_up,
count(*) filter(where lag_last_sale_price > last_sale_price) cnt_dollars_up
from (
select dp.*,
lag(last_sale_price) over(partition by extract(year from sale_date), extract(week from saledate) order by sale_date) lag_last_sale_price
from daily_price
where symbol = 'AAPL'
and sale_date >= date_trunc('week', current_date) - '8 week'::interval
) dp
group by 1, 2
Notes:
I am asssuming that you don't want to compare the first price of a week to the last price of the previous week; if you do, then just remove the partition by clause from the over() clause of lag()
this dynamically computes the date as of 8 (entire) weeks ago
if there is no price increase during a whole week, the query still gives you a row, with 0 as sum_dollars_up and cnt_dollars_up
I have SELECT:
SELECT month, year, ROUND(AVG(q_overall) OVER (rows BETWEEN 10000 preceding and current row),2) as avg
FROM (
SELECT EXTRACT(Month FROM date) as month, EXTRACT(Year FROM date) as year, ROUND(AVG(q_overall),1) as q_overall
FROM fb_parsed
WHERE business_id = 1
GROUP BY year, month
ORDER BY year, month) a
output:
month year avg
-----------------
12 2012 5
1 2013 4.5
2 2013 4.1
4 2013 4.8
5 2013 4.7
And I have to append this table with missing values (in this example with 3-rd month in 2013 year). The avg must be same as in previous row, that means I need to append this table with:
3 2013 4.1
Can I do this with SELF JOINS and generate_series, or with some UNION select?
You can simplify your select. It doesn't need a subquery:
SELECT EXTRACT(Month FROM date) as month,
EXTRACT(Year FROM date) as year,
ROUND(AVG(q_overall), 1) as q_overall,
ROUND(AVG(AVG(q_overall)) OVER (rows BETWEEN 10000 preceding and current row), 2)
FROM fb_parsed
WHERE business_id = 1
GROUP BY year, month;
The windows function needs an order by. I assume you really intend:
SELECT EXTRACT(Month FROM date) as month,
EXTRACT(Year FROM date) as year,
ROUND(AVG(q_overall), 1) as q_overall,
ROUND(AVG(AVG(q_overall)) OVER (ORDER BY year, month)), 2)
FROM fb_parsed
WHERE business_id = 1
GROUP BY year, month;
Then, to fill in the values you can use generate_series():
SELECT EXTRACT(Month FROM ym.date) as month,
EXTRACT(Year FROM ym.date) as year,
ROUND(AVG(AVG(q_overall)) OVER (ORDER BY year, month)), 2)
FROM (SELECT generate_series(date_trunc('month', min(date)),
date_trunc('month', max(date)),
interval '1 month') as date
FROM fb_parsed
) ym LEFT JOIN
fb_parsed p
ON EXTRACT(year FROM ym.date) = EXTRACT(year FROM p.date) AND
EXTRACT(month FROM ym.date) = EXTRACT(month FROM p.date) AND
p.business_id = 1
GROUP BY year, month;
I think this will do what you want.
Final query:
SELECT EXTRACT(Month FROM ym.date) as month,
EXTRACT(Year FROM ym.date) as year,
ROUND(AVG(AVG(q_overall)) OVER (ORDER BY EXTRACT(Year FROM ym.date), EXTRACT(Month FROM ym.date)), 2)
FROM
(SELECT generate_series(date_trunc('month', min(date)),
date_trunc('month', max(date)),
interval '1 month') as date
FROM fb_parsed WHERE business_id = 1 AND site = 'facebook')
ym LEFT JOIN
fb_parsed p
ON EXTRACT(year FROM ym.date) = EXTRACT(year FROM p.date) AND
EXTRACT(month FROM ym.date) = EXTRACT(month FROM p.date) AND
p.business_id = 1 AND site = 'facebook'
GROUP BY year, month;
Can I do this with SELF JOINS and generate_series?
Yep, you're close, but your current query does a Cumulative Average. The tricky part is the fill the gaps with the previous value (If PostgreSQL supported the IGNORE NULLS option of LAST_VALUE this would be easier...)
SELECT month,
year,
MAX(q_overall) -- assign the value to all rows within the same group
OVER (PARTITION BY grp)
FROM
(
SELECT all_months.month, all_months.year, p.q_overall,
-- assign a new group number whenever there's a value in q_overall
SUM(CASE WHEN q_overall IS NULL THEN 0 ELSE 1 END)
OVER (ORDER BY all_months.month, all_months.year
ROWS UNBOUNDED PRECEDING) AS grp
FROM
( -- create all months with min and max date
SELECT generate_series(date_trunc('month', min(date)),
date_trunc('month', max(date)),
interval '1 month') as date
FROM fb_parsed
) AS all_months
LEFT JOIN
( -- do the average per month calculation
SELECT EXTRACT(Month FROM date) as month,
EXTRACT(Year FROM date) as year,
ROUND(AVG(q_overall),1) as q_overall
FROM fb_parsed
WHERE business_id = 1
GROUP BY year, month
) AS p
ON EXTRACT(year FROM ym.date) = all_months.month
AND EXTRACT(month FROM ym.date) = all_months.year
) AS dt
Edit:
Oops, this was overly complicated, the question asked for a Cumulative Average and then NULLs will not change the result and there's no need to fill the gaps
I have a following problem with my SQL query. I was managed to successfully execute about 15 of them but this one makes me sick. It's even hard to translate but probably you will understand.
Show years and months numbers and their sum of costs of ‘borrowed-time’ in those months where monthly sum of costs was lesser then the biggest one in February and March in 2006.
And this is what I have so far (one of the version of the query because I tried many of them)
SELECT EXTRACT(MONTH FROM DATA_WYP), SUM(KOSZT)
FROM WYPOZYCZENIA
WHERE KOSZT<(SELECT SUM(KOSZT)
FROM WYPOZYCZENIA
WHERE SUM(KOSZT)<(SELECT MAX(SUM(KOSZT))
FROM WYPOZYCZENIA
WHERE EXTRACT(MONTH FROM DATA_WYP)='2' OR
EXTRACT(MONTH FROM DATA_WYP)='3'
GROUP BY EXTRACT(MONTH FROM DATA_WYP)))
GROUP BY EXTRACT(MONTH FROM DATA_WYP);
The problem is that I cannot equal SUM(KOSZT), tried to save them using AS but it doesn't work either.
Please help me because it already ruined my day.
Thanks in advance.
To filter by the results of an aggregate function in a SQL query, place the comparisons in a HAVING statement.
Are you looking for something like this?
SELECT YEAR(data_wyp) year,
MONTH(data_wyp) month,
SUM(koszt) koszt
FROM wypozyczenia
GROUP BY year, month
HAVING koszt < (SELECT SUM(koszt) koszt
FROM wypozyczenia
WHERE MONTH(data_wyp) IN (2, 3)
AND YEAR(data_wyp) = 2006
GROUP BY MONTH(data_wyp)
ORDER BY koszt DESC LIMIT 1)
SQLFiddle (MySql)
I'm assuming Oracle or DB2 because of the keywords in your sample query.
Using a CTE to find the max for the 2 months and using HAVING for the condition simplifies the expression somewhat, this should (more or less) do what you need;
WITH cte AS (
SELECT SUM(KOSZT) mx FROM WYPOZYCZENIA
WHERE EXTRACT(MONTH FROM DATA_WYP) IN ('2','3')
AND EXTRACT(YEAR FROM DATA_WYP) = '2006'
GROUP BY EXTRACT(MONTH FROM DATA_WYP)
)
SELECT EXTRACT(YEAR FROM DATA_WYP) Year, EXTRACT(MONTH FROM DATA_WYP) Month,
SUM(KOSZT) KOSZT
FROM WYPOZYCZENIA
GROUP BY EXTRACT(Year FROM DATA_WYP), EXTRACT(Month FROM DATA_WYP)
HAVING SUM(KOSZT) < (SELECT MAX(mx) FROM cte)
An SQLfiddle to test with.
Try this out-:
SELECT EXTRACT(YEAR FROM DATA_WYP)
EXTRACT(MONTH FROM DATA_WYP),
SUM(KOSZT)
FROM WYPOZYCZENIA
GROUP BY EXTRACT(YEAR FROM DATA_WYP)
EXTRACT(MONTH FROM DATA_WYP)
HAVING Sum(KOSZT)<(SELECT MAX(SUM(KOSZT))
FROM WYPOZYCZENIA
WHERE EXTRACT(YEAR FROM DATA_WYP)=2006
EXTRACT(MONTH FROM DATA_WYP) IN ('2','3')
GROUP BY EXTRACT(MONTH FROM DATA_WYP))
I hope this solves your problem.