I have a table that looks like that
foo_date | bar
---------------
2018-01-01 | bar
2018-01-09 | bar
2018-01-10 | bar
2018-01-20 | bar
And I would like to build a request that retrieves,for each week, the row which occurs first in the week.
Cheers
You can simply do:
select distinct on (datetrunc('week', foo_date)) t.*
from t
order by datetrunc('week', foo_date), foo_date;
demo:db<>fiddle
SELECT DISTINCT ON (year, week)
foodate, bar
FROM (
SELECT
foodate,
bar,
EXTRACT('isoyear' FROM foodate) as year,
EXTRACT('week' FROM foodate) as week
FROM dates
ORDER BY foodate
)s
EXTRACT('week'...) gives the week. So two date in the same week give the same output at this column.
DISTINCT ON (week) gives the first (ordered!) row for each week.
Postgres Date functions
Notice the definition of the week:
The number of the ISO 8601 week-numbering week of the year. By
definition, ISO weeks start on Mondays and the first week of a year
contains January 4 of that year. In other words, the first Thursday of
a year is in week 1 of that year.
Edit: If you have data from more then a year of course you should add the year as well. Other wise you get the first row of all first weeks of all years for example.
Related
I am new using SQL and I need to compare current year sales with the same weekday last year.
In the image below for example 06/01/2001 was on a Tuesday so I need to retrieve the same weekday last year with was on 05/01/2020 "Tuesday"
Thanks in advance
Data with columns
If you have data every day, then you can use lag(). The same weekday last year would be 52 weeks ago (at least that is a reasonable definition). So, using lag():
select t.*, lag(total_sales, 52*7) over (order by current_year)
from t;
I have a week number of year in ISO format and I want to get start date of this week in Hive. First date of week is Monday.
Example: year 2020 week 50 - start date should be 2020-12-07
Try the code below, where year and week are the corresponding column names of your table.
select date(
from_unixtime(
unix_timestamp(concat(year,'-',week,'-','1'), 'yyyy-w-u')
)
) from <your_table_name>;
How to get data of last 6 quarter in Oracle including current quarter. I mean if I run the query today so data between 01-JAN-2018 to 30-JUN-2019 should come in the query.
You could do something like this:
SELECT
*
FROM
DUAL
WHERE
DATE_FIELD >= (SYSDATE - (30*(3*6)))
What this query is doing is taking the current date (SYSDATE), and grabbing all values from the previous 6 quarters. The rationale is:
30 = days in a month | 3 = months in a quarter | 6 = quarters specified by OP
You can use add_months and trunc functions for date value with Q(quarter) argument
select t.*
from tab t
where insert_date between
trunc(add_months(sysdate,-3*5),'Q')
and trunc(add_months(sysdate,3),'Q')-1;
Demo for verification
for the starting date, -3*(6-1) = -3*5 considered, starting from 5 quarter back to be able to count 6 quarter including the current quarter. 3 is
obvious as being the number of months in each quarter.
I have a date in sql which will always fall on a Monday and I'm subtracting 13 weeks to get a weekly report. I am trying to get the same 13 week report but for last year's figures as well.
At the moment, I'm using the following:
calendar_date >= TRUNC(sysdate) - 91
which is working fine.
I need the same for last year.
However, when I split this into calendar weeks, there will also be a partially complete week as it will include 1 or 2 days from the previous week. I need only whole weeks.
e.g. the dates that will be returned for last year will be 14-Feb-2015 to 16-May-2015. I need it to start on the Monday and be 16-Feb-2015. This will change each week as I am only interested in complete weeks...
I would do this:
Get the date by substracting 91 days as you're already doing.
Get the number of the day of the week with TO_CHAR(DATE,'d')
Add the number of days until the next monday to the date.
Something like this:
SELECT TO_DATE(TO_DATE('16/05/2015','DD/MM/YYYY'),'DD/MM/YYYY')-91 + MOD(7 - TO_NUMBER(TO_CHAR(TO_DATE(TO_DATE('16/05/2015','DD/MM/YYYY'),'DD/MM/RRRR')-91,'d'))+1,7) d
FROM dual
next_day - returns date of first weekday named by char.
with dates as (select to_date('16/05/2015','DD/MM/YYYY') d from dual)
select
trunc(next_day( trunc(d-91) - interval '1' second,'MONDAY'))
from dates;
I want to get next monday from calculated date. In situation when calculated date is monday i have to move back to previous week ( -1 second).
Getting the month for the current date is, obviously, straight forward, but I'm needing to get the month name with a different end date.
I need to get the month name with the start date of the month being the first Thursday after the first Wednesday of the month and the end date of the month being the first Wednesday of the following month. It's for an accounting thing, so I'm not going to argue with the spec!
e.g. for 2014, January would run from 9th Jan - 5th Feb, February would run from 6th February - 5th March, March would run from 6th March - 2nd April.
I would suggest that you create a table with your 'accounting months' in it, having a start date, end date and month name columns.
You could then query this to find the row where your date is between the start and end dates and return the month name. Putting this into a scalar function would then allow it to be reusable and relatively easily updated for next years months as well.
I think, as per Paddy's answer, a lookup table is the simplest thing to do. Here's one way to generate the rows for it:
; With Numbers(n) as (
select 4 union all select 5
), Months as (
select CONVERT(date,'20010104') as StartDt,CONVERT(date,'20010207') as EndDt,
DATENAME(month,'20010103') as Month
union all
select DATEADD(week,n1.n,StartDt),DATEADD(week,n2.n,EndDt),
DATENAME(month,DATEADD(week,n1.n,StartDt))
from Months,Numbers n1,Numbers n2 --Old-skool join, just for once
where DATEPART(day,DATEADD(week,n1.n,StartDt)) between 2 and 8 and
DATEPART(day,DATEADD(week,n2.n,EndDt)) between 1 and 7 and
StartDt < '21000101'
)
select * from Months option (maxrecursion 0)
(CW since this is effectively just an extension to Paddy's answer but I don't want to edit their answer, nor is it suitable for a comment