I have a quest that is about doing a statistic of the sales per day in the last 30 day...i've found a way to only show the last month:
SELECT *
FROM purchase
WHERE date >= date('01-05-2021', current_date - interval '1 month')
and date < date('01-05-2021', current_date)
the columns in purchase are just id, value, date, cashier and store id what do you think is the best way to do this?
i have this and i don't know way it is not working...i'm new in postgresql so please don't be offended by this
Group by date and use sum to find the total.
select date,sum(value)
from purchase
where date between current_date - interval '1 month' and current_date - 1
group by date
Related
I need to pull out first date and last date of the month from the given from_date and to_date as input, For Example:-
I have my psql output table as the following:
Year
Term
Start Date
End Date
2022
Odd
01-02-2022
30-04-2022
2022
Even
01-07-2022
30-09-2022
I need the output as the following:-
Year
Term
Start Date
End Date
2022
Odd
01-02-2022
28-02-2022
2022
Odd
01-03-2022
31-03-2022
2022
Odd
01-04-2022
30-04-2022
2022
Even
01-07-2022
30-07-2022
2022
Even
01-08-2022
31-08-2022
2022
Even
01-09-2022
30-09-2022
I need the ouput in Postgresql, Pls help
Thanks
Your issue boils down to given a period with start and end dates, determine the first and last dates for each month in that period. In Postgres given a date you can determine the first (with date_trunc function) and last of the a month with the expressions:
-- for a given date
date_trunc('month', given_date) -- returns first day of month
date_trunc('month', given_date + interval '1 month' - interval '1 day') -- returns last day of month
Use the first expression above, with generate_series with dates, to create the first of each month in the period. The use the second expression to generate the end of each month. (see demo)
with range_dates (year, term, gsom) as
( select year
, term
, generate_series( date_trunc('month', od.start_date)::date
, date_trunc('month', od.end_date )::date
, interval '1 month'
)::date
from output_data od
)
select year
, term
, gsom start_date
, (gsom + interval '1 month' - interval '1 day')::date end_date
from range_dates
order by term desc, start_date;
I work with a hotel client where they have a BigQuery database which has hotel booking data. I've shared the relevant columns in the image below which list the names of each hotel, the arrival date of the guest, the departure date, and the revenue generated from the each booking:
My problem statement is that I have to showcase how many rooms have been booked, and how much revenue has been made for each hotel every month where my final grid would look similar to this:
The important points to remember are:
the depart_dt - arrival_dt are the number of nights that the guest is staying
the Rez_rate_total / (depart_dt - arrival_dt) is the revenue made per night
My problem here is trying to figure out how to change the start date and end date columns into groups of months. The challenge comes when a guest arrives in one month and leaves in the next month. For example, Row 5 in the original data has the guest coming in on 18th July and leaving on 1st Aug - so 13 days of his stay and 13 days of revenue has to be included in July and 1 day has to be included in August.
I haven't used SQL in a while so this is as far as I got:
WITH
temp_table AS (
SELECT
hotel_long_nm,
arrival_dt,
depart_dt,
DATE_DIFF(depart_dt, arrival_dt, day) AS room_nights,
rez_rate_total
FROM
`DATABASE.analytics.bookings` )
SELECT
*
FROM
temp_table
Any help would be greatly appreciated!
Consider the following approach:
with bookings as (
select hotel_long_nm, date(arrival_dt) as arrival_dt, date(depart_dt) as depart_dt, rez_rate_total from project.dataset.bookings
),
tmp as (
-- expose the dates in the reservation (excluding last day of reservation)
select *, generate_date_array(arrival_dt,date_sub(depart_dt, interval 1 day)) as stay_dates from bookings
),
calc as (
-- unnest and calculate the daily rate
select
hotel_long_nm,
stay_dt,
1 as stay_nights,
rez_rate_total/array_length(stay_dates) as rez_rate_daily
from tmp
left join unnest(stay_dates) as stay_dt
),
agg as (
-- aggregate to the year-month level
select
date_trunc(stay_dt, month) as year_month,
hotel_long_nm,
sum(stay_nights) as room_nights,
round(sum(rez_rate_daily),2) as rez_rate_total
from calc
group by 1,2
)
select * from agg
order by hotel_long_nm, year_month
You can consider this approach, following this logic.
Validate if both dates are in the same month
If are not in the same month, i get the final date of the month of
arrival date and subtract both dates
I get the first date of the month of the depart date and subtract
and subtract both dates
In this code you can see an example:
SELECT
/*arrival date*/
CURRENT_DATE() AS the_arival,
/*depart_dt*/
DATE_ADD(CURRENT_DATE(), INTERVAL 30 DAY) AS the_depart,
/*total of night between arrival date and depart date*/
DATE_DIFF(DATE_ADD(CURRENT_DATE(), INTERVAL 30 DAY) , CURRENT_DATE(), DAY) AS total_room_nights,
/* validate if the dates are in the same month or different month if equal 0 same month if >0 another month */
DATE_DIFF(DATE_ADD(CURRENT_DATE(), INTERVAL 30 DAY) , CURRENT_DATE(), MONTH) AS Same_Month,/*1 no and 0 yes/
/*in this case are in different month*/
/*I get the final date of the arrival month and subtract with the arrival date*/
DATE_DIFF(DATE_SUB(DATE_TRUNC(DATE_ADD(DATE_ADD(CURRENT_DATE(), INTERVAL 30 DAY), INTERVAL 1 MONTH), MONTH), INTERVAL 1 DAY),DATE_ADD(CURRENT_DATE(), INTERVAL 30 DAY), DAY) as total_room_nights_first_mont,
/*I get the initial date of the depart month and subtract with the depart date i add +1 because is the night between last day of the mont and first day of the next month*/
DATE_DIFF(DATE_ADD(CURRENT_DATE(), INTERVAL 30 DAY),DATE_TRUNC(DATE_ADD(CURRENT_DATE(), INTERVAL 30 DAY), MONTH), DAY)+1 as total_room_nights_second_month
You can see more information about the date function.Click Here.
I want to query my postgresql data and group the results by week, so I'm using the following query:
select
date_trunc ('week', date_column) as week,
sum (orders) as orders_count
from database
group by week
But it uses Monday as start day of the week, while I want my weeks to be like 'Saturday -> Friday'. How can I acheive this?
Just subtract two days, and you land on saturday:
select
date_trunc('week', date_column)::date - 2 as week,
sum (orders) as orders_count
from the_table
group by week
You can just offset by two days, as follows:
select
date_trunc ('week', date_column + interval '2 days') - interval '2 days' as week,
sum (orders) as orders_count
from database
group by week
The below query returns a distinct count of 'members' for a given month and brand (see image below).
select to_char(transaction_date, 'YYYY-MM') as month, brand,
count(distinct UNIQUE_MEM_ID) as distinct_count
from source.table
group by to_char(transaction_date, 'YYYY-MM'), brand;
The data is collected with a 15 day lag after the month closes (meaning September 2016 MONTHLY data won't be 100% until October 15). I am only concerned with monthly data.
The query I would like to build: Until the 15th of this month (October), last month's data (September) should reflect August's data. The current partial month (October) should default to the prior month and thus also to the above logic.
After the 15th of this month, last month's data (September) is now 100% and thus September should reflect September (and October will reflect September until November 15th, and so on).
The current partial month will always = the prior month. The complexity of the query is how to calc prior month.
This query will be ran on a rolling basis so needs to be dynamic.
To be clear, I am trying to build a query where distinct_count for the prior month (until end of current month + 15 days) should reflect (current month - 2) value (for each respective brand). After 15 days of the close of the month, prior month = (current month - 1).
Partial current month defaults to prior month's data. The 15 day value should be variable/modifiable.
First, simplify the query to:
select to_char(transaction_date, 'YYYY-MM') as month, brand,
count(distinct members) as distinct_count
from source.table
group by members, to_char(transaction_date, 'YYYY-MM'), brand;
Then, you are going to have a problem. The problem is that one row (say from Aug 20th) needs to go into two groups. A simple group by won't handle this. So, let's use union all. I think the result is something like this:
select date_trunc('month', transaction_date) as month, brand,
count(distinct members) as distinct_count
from source.table
where (date_trunc('month', transaction_date) < date_trunc('month' current_date) - interval '1 month') or
(day(current_date) > 15 and date_trunc('month', transaction_date) = date_trunc('month' current_date) - interval '1 month')
group by date_trunc('month', transaction_date), brand
union all
select date_trunc('month' current_date) - interval '1 month' as month, brand,
count(distinct members) as distinct_count
from source.table
where (day(current_date) < 15 and date_trunc('month', transaction_date) = date_trunc('month' current_date) - interval '1 month')
group by brand;
Since you already have a working query, I concentrate on the subselect. The condition you can use here is CASE, especially "Searched CASE"
case
when extract(day from current_date) < 15 then
extract(month from current_date - interval '2 months')
else
extract(month from current_date - interval '1 month')
end case
This may be used as part of a where clause, for example.
Here is some sudo code to get the begin date and the end date for your interval.
Begin date:
date DATE_TRUNC('month', CURRENT_DATE - integer 15) - interval '1 month'
This will return the current month only after the 15th day, from there you can subtract a full month to get your starting point.
End Date:
To calculate this, grab the begin date, plus a month, minus a day.
If the source table is partitioned by transaction_date, this syntax (not masking transaction_date with expression) enables partitions eliminatation.
select to_char(transaction_date, 'YYYY-MM') as month
,count (distinct members) as distinct_count
,brand as brand
FROM source.table
where transaction_date between date_trunc('month', current_date) - case when extract (day from current_date) >= 15 then 1 else 2 end * interval '1' month
and date_trunc('month', current_date) - case when extract (day from current_date) >= 15 then 0 else 1 end * interval '1' month - interval '1' day
group by to_char(transaction_date, 'YYYY-MM')
,brand
;
Hello i have some record:
date Money
1/6/2014 100
12/6/2014 2200
13/6/2014 500
1/3/2014 100
2/5/2014 2200
30/5/2014 500
30/6/2014 100
23/6/2014 2200
31/6/2014 500
Well, i have one param, its date for example i need all record of 30 of july but i need the sum of record if very easy that
select sum(money) from info where date <= param
group by month
But now i need all record group by day of month but i need every fifteen for every month to lower the parameter in my result i need for example
param = 30/6/2014
The result I hope to get:
15/6/2014 sum(money)
15/5/2014 sum(money)
15/4/2014 sum(money)
I need the record for every fifteen of month
The question is how to arrange it to days up to the 15th are in one month and days from the 16th onward are in the next month. You can do this by subtracting 15 days. This puts all days from 1-15 in the previous month. Then, add a month. Here is an approach in Postgres:
select to_char(date - 15 * interval '1 day' + interval '1 month', 'YYYY-MM') as mon,
sum(money)
from info
where date <= param
group by to_char(date - 15 * interval '1 day' + interval '1 month', 'YYYY-MM')
order by 1;
EDIT:
If you want month-to-the-15th, then you can do:
select to_char(date, 'YYYY-MM') as mon,
sum(money)
from info
where extract(day from date) <= param
group by to_char(date, 'YYYY-MM')
order by 1;
Or, if it is only for one month:
select sum(money)
from info
where to_char(date, 'YYYY-MM') = to_char(param, 'YYYY-MM') and
date <= param;