If today is say 15th August 2012 then the query should return the following
15/01/2011,
15/02/2011,
...
...
15/07/2012
15/08/2012
If today is 31st August 2012 then the query would return
31/01/2011,
28/02/2011, <<<<this is the nearest date
...
...
31/07/2012
31/08/2012
We have a vw_DimDate in our Warehouse which should help
edit
It contains the following fields
Currently I'm using the following but it seems rather convoluted! ...
DECLARE #Dt DATETIME = '31 JUL 2012'--GETDATE()
;WITH DateSet_cte(DayMarker)
AS
(
SELECT DayMarker
FROM WHData.dbo.vw_DimDate
WHERE
DayMarker >= CONVERT(DATETIME,CONVERT(CHAR(4),DATEADD(YEAR,-1,#Dt),112) + '0101') AND
DayMarker <=#Dt
)
, MaxDate_cte(MaxDate)
AS
(
SELECT [MaxDate] = MAX(DayMarker)
FROM DateSet_cte
)
SELECT
[Mth] = CONVERT(DATETIME,CONVERT(CHAR(6),a.DayMarker,112) + '01')
, MAX(a.DayMarker) [EquivDate]
FROM DateSet_cte a
WHERE DAY(a.DayMarker) <= (SELECT DAY([MaxDate]) FROM MaxDate_cte)
GROUP BY CONVERT(DATETIME,CONVERT(CHAR(6),a.DayMarker,112) + '01')
;with Numbers as (
select distinct number from master..spt_values where number between 0 and 23
), Today as (
select CONVERT(date,CURRENT_TIMESTAMP) as d
)
select
DATEADD(month,-number,d)
from
Numbers,Today
where DATEPART(year,DATEADD(month,-number,d)) >= DATEPART(year,d) - 1
Seems odd to want a variable number of returned values based on how far through the year we are, but that's what I've implemented.
When you use DATEADD to add months to a value, then it automatically adjusts the day number if it would have produced an out of range date (e.g. 31st February), such that it's the last day of the month. Or, as the documentation puts it:
If datepart is month and the date month has more days than the return month and the date day does not exist in the return month, the last day of the return month is returned.
Of course, if you already have a numbers table in your database, you can eliminate the first CTE. You mentioned that you "have a vw_DimDate in our Warehouse which should help", but since I have no idea on what that (presumably, a) view contains, it wasn't any help.
Related
I know about Snowflake date function to find out day, week, month, year, etc.
I want to have weeks start from Saturday each week to next Saturday.
following gives an idea how to extract, but need something to address my specific case.
How to get week number of month for any given date on snowflake SQL
If four days or more in week period belong to a certain month, I would assign the week to that month; otherwise, to the next month
example:
Week of April 29, 2023 to May 5, 2023 has less then four days in April so want to consider it as May
Week of May 23, 2023 to June 2nd, 2023 has more than four days in May so I would like to consider it as May
I want to assign weeks to a month with more days of one month (four or more days)
Snowflake will allow you to set the first day of the week with a parameter.
https://docs.snowflake.com/en/sql-reference/parameters.html#label-week-start
This will allow you to set the first day of the week at Saturday.
Doing so will result in the WEEK() function counting weeks in a year using saturday as a delimiter between weeks.
Now we just need to find which actual month has the most days for any given week and assign that week to the proper month.
I have an example script below that serves as an example on how to make a custom date dimension table. You can generate the table once and join against it to retrieve your custom date attributes.
/***************************************************************************
A WEEK_START session variable of 0 is the default Snowflake behavior
and has weeks start on Monday and end of Sunday (ISO standard).
https://docs.snowflake.com/en/sql-reference/parameters.html#label-week-start
-- 6 = Saturday is day 1 of of the week
*********************************************************************************************/
alter session set week_start = 6;
/*********************************************************************************************
The parameters below define the temporal boundaries of the calendar table. The values must be
DATE type and can be hardcoded, the result of a query, or a combination of both.
For example, you could set date_start and date_end based on the MIN and MAX date of the table
with the finest date granularity in your data.
*********************************************************************************************/
SET date_start = TO_DATE('2022-12-18');
SET date_end = current_date(); --TIP: for the current date use current_date();
--This sets the num_days parameter to the number of days between start and end
--this value is used for the generator
set num_days = (select datediff(day, $date_start, $date_end+1));
--CTE to hold generated date range
create or replace transient table calendar as
with gen_cte as (
select
dateadd(day,'-' || row_number() over (order by null),
dateadd(day, '+1', $date_end)
) as date_key
from table (generator(rowcount => ($num_days)))
order by 1)
-- calendar table expressions
, step_1 as (
select
date_key,
, dayofmonth(date_key) as day_of_month
, week(date_key) as week_num --*see comments
--, dayofweekiso(date_key) as day_of_week_iso,
, dayofweek(date_key) as day_of_week
, dayname(date_key) as day_name
, month(date_key) as month_num
--, weekiso(date_key) as week_iso_num, --*see comments
, year(date_key) as year_
, year_ || '-' ||week_num::string as year_week_key
, count(date_key) over (partition by year_week_key, month_num) as days_of_week_in_month
--ceil(dayofmonth(date_key) / 7) as day_instance_in_month --used to identify 'floating' events such as "fourth thursday of november"
FROM gen_cte)
-- calculate the max number of days in each month for any week in year
, step_2 as (
select
year_week_key
, month_num
, max(step_1.days_of_week_in_month) as max_days_of_week_in_month
from step_1
group by year_week_key, month_num)
-- for any week with 2 actual month values, assign the month with the most number of days
, step_3 as (
select
year_week_key
, month_num
, row_number() over (partition by year_week_key order by max_days_of_week_in_month desc ) as month_rank
from step_2
qualify month_rank = 1
)
select
s1.date_key
, s1.day_of_month
, s1.week_num
, s1.day_of_week
, s1.day_name
, s3.month_num as assigned_month_num
, s1.month_num as actual_month_num
, s1.year_
from step_1 s1
left join step_3 s3
on s1.year_week_key = s3.year_week_key
;
-- select from your new date dimension table
select * from calendar;
I currently have a database which has a start date and and end date when when a Car is being held for Maintenance. I am able to figure the difference between these dates using datediff but the problem is that the value when across multiple months is linked either to the month of the start date or the end date. I am looking to spread the difference between the months so in my analysis I can see how long the cars have been held each month in days.
Example:
StartDate: '2022-04-28 06:33:34.000'
EndDate: '2022-06-20 14:09:45.000'
Days Difference: 53 days 7 Hours 36 Minutes 11 Seconds
What I need to do is spread the 54 rounded up as
April: 3
May: 31
June: 20
I currently calculating the difference either as day percentage or in days using this logic.
'CAST(CAST(DATEDIFF(s, STARTDATE, ENDDATE)AS float)/86400 AS DECIMAL(16,3)) AS CAR_TOTAL_DAYS_PERC' returns 53.317
'DATEDIFF(s, STARTDATE, ENDDATE) / 86400 AS CAR_TOTAL_DAYS' returns 53
Any assistance would be greatly appreciated.
You may want to implement this as a stored procedure, something like GetDateSpreadString
--Variables for start and end date
DECLARE #startDate AS DATETIME
DECLARE #endDate AS DATETIME
SET #startDate='2022-04-28 06:33:34.000'
SET #endDate ='2022-06-20 14:09:45.000';
--CTE to expand the count of date numbers to be used for DATEADD
WITH numbers
as
(
Select 1 as value
Union ALL
Select value + 1 from numbers
where value + 1 <= DATEDIFF( dd,#startDate, #endDate)
),
--CTE to group individual days covered by spread in each of the months.
MonthsDays AS (
Select EOMONTH(DATEADD(d,value, #startDate)) as MonthSpread, DATEPART(dd,DATEADD(d,value, #startDate)) as DaySpread
From numbers
)
--USE FOR XML PATH() and STUFF() to make it a single string result
SELECT STUFF(
(
Select CONCAT(' ',COUNT(daySpread),','), CONCAT( DATENAME(month,MonthSpread),',')
FROM MonthsDays
GROUP BY MonthSpread
FOR XML PATH('')
),1,1,'') AS Result
Current output with no DAYS_LEFT_IN_QUARTERI am new to using Snowflake and was tasked to create a Calendar Dimension table that would aid in reporting weekly / monthly /quarterly reports. I am confused on how to return days remaining in the FISCAL_QUARTER. Q1 spans from Feb - Apr.
Attached below is the code I have been writing to generate the dates projecting 14 years in the future.
--Set the start date and number of years to produce
SET START_DATE = '2012-01-01';
SET NUMBER_DAYS = (SELECT TRUNC(14 * 365));
--Set parameters to force ISO
ALTER SESSION SET WEEK_START = 1, WEEK_OF_YEAR_POLICY = 1;
WITH CTE_MY_DATE AS (
SELECT DATEADD(DAY, SEQ4(), $START_DATE) AS MY_DATE
FROM TABLE(GENERATOR(ROWCOUNT=>$NUMBER_DAYS)) -- Number of days after reference date in previous line
)
SELECT
MY_DATE::date
,YEAR(MY_DATE) AS YEAR
,MONTH(MY_DATE) AS MONTH
,MONTHNAME(MY_DATE) AS MONTH_ABBREVIATION
,DAY(MY_DATE)
,DAYOFWEEK(MY_DATE)
,WEEKOFYEAR(MY_DATE)
,DAYOFYEAR(MY_DATE)
,YEAR(ADD_MONTHS(DATE_TRUNC('month', MY_DATE),11)) AS FISCAL_YEAR
,CONCAT('Q', QUARTER(ADD_MONTHS(DATE_TRUNC('month', MY_DATE),11))) AS FISCAL_QUARTER
,MONTH(ADD_MONTHS(DATE_TRUNC('month', MY_DATE),11)) AS FISCAL_MONTH
FROM CTE_MY_DATE
;
firstly your generator will get gaps, as SEQx() function are allowed to have gaps, so you need to use SEQx() as the OVER BY of a ROW_NUMBER like so:
WITH cte_my_date AS (
SELECT DATEADD(DAY, ROW_NUMBER() OVER(ORDER BY SEQ4()), $START_DATE) AS my_date
FROM TABLE(GENERATOR(ROWCOUNT=>$NUMBER_DAYS)) -- Number of days after reference date in previous line
)
and days left in quarter, is the day truncated to quarter, +1 quarter, date-diff in days to day:
,DATEDIFF('days', my_date, DATEADD('quarter', 1, DATE_TRUNC('quarter', my_date))) AS days_left_in_quarter
How's this? You can copy/paste the code straight into snowflake to test.
Using last_day() tends to make it look a little tidier :-)
WITH CTE_MY_DATE AS (
SELECT DATEADD(DAY, SEQ4(), current_date()) AS MY_DATE
FROM TABLE(GENERATOR(ROWCOUNT=>300)))
SELECT
MY_DATE::date
,YEAR(last_day(my_date,year)) AS FISCAL_YEAR
,concat('Q',quarter(my_date)) AS FISCAL_QUARTER
,datediff(d, my_date, last_day(my_date,quarter)) AS
DAYS_LEFT_IN_QUARTER
FROM CTE_MY_DATE
I need a suggestion around one of the columns that I'm creating in the Date dimension in SQL Server, basically rolling weeks..
I have a table dimDate in my datawarehouse.
I want to create a column in the dimdate table which will have week number in any year and each week should have 7 days.
For eg: In year 2015 there are 53 weeks but the 53rd week has only 5 days (because the week starts on Sunday in SQL Server I guess).
I want to include 2 more days from 2016 (1st and 2nd Jan in 2016) to complete the 53rd week with 7 days and also the the 1st week in 2016 should start on 3rd of Jan 2016, so on and so forth.
If there are any suggestions that will be great to start with.
Assuming that you already have weeks populated (but not extended into the next year), and making some assumptions about columns names
This query finds the last week in a year (which would almost always always be 53 but don't count on it:) and the date that it ends on
SELECT YearNo, MAX(Week) As Week, MAX(DateKey) As DateKey
FROM dimDate
GROUP BY YearNo
This query finds all weeks that are shorter than 7 days, and how many extra days are required to make them 7 days.
SELECT
YearNo,
Week,
7-COUNT(DISTINCT DateKey) As ExtraDaysRequired
FROM dimDate
GROUP BY YearNo, Week
HAVING COUNT(DISTINCT DateKey) < 7
This might always be the last week of the year but lets not make assumptions.
Lets combine these to find all final weeks that have less than 7 days, as well as add the number of days required:
SELECT
Under7Days.YearNo, Under7Days.Week, Under7Days.ExtraDaysRequired,
FinalWeeks.DateKey StartDate,
DATEADD(d,Under7Days.ExtraDaysRequired,FinalWeeks.DateKey) EndDate
FROM
(
SELECT YearNo, MAX(Week) As Week, MAX(DateKey) As DateKey
FROM dimDate
GROUP BY YearNo
) As FinalWeeks
INNER JOIN
(
SELECT YearNo, Week, 7-COUNT(DISTINCT DateKey) As ExtraDaysRequired
FROM dimDate
GROUP BY YearNo, Week
HAVING COUNT(DISTINCT DateKey) < 7
) As Under7Days
ON FinalWeeks.Week = Under7Days.Week
AND FinalWeeks.YearNo = Under7Days.YearNo
So we have a query that identifies the start date and end date and week number that it needs to be updated to. So now we run an update:
UPDATE TGT
SET Week = SRC.Week
FROM dimDate TGT
INNER JOIN
(
SELECT
Under7Days.YearNo, Under7Days.Week, Under7Days.ExtraDaysRequired,
FinalWeeks.DateKey StartDate,
DATEADD(d,Under7Days.ExtraDaysRequired,FinalWeeks.DateKey) EndDate
FROM
(
SELECT YearNo, MAX(Week) As Week, MAX(DateKey) As DateKey
FROM dimDate
GROUP BY YearNo
) As FinalWeeks
INNER JOIN
(
SELECT YearNo, Week, 7-COUNT(DISTINCT DateKey) As ExtraDaysRequired
FROM dimDate
GROUP BY YearNo, Week
HAVING COUNT(DISTINCT DateKey) < 7
) As Under7Days
ON FinalWeeks.Week = Under7Days.Week
AND FinalWeeks.YearNo = Under7Days.YearNo
) SRC
ON TGT.DateID BETWEEN SRC.StartDate AND SRC.EndDate
Looks complicated? There's half a dozen ways to write the same thing but this approach is step-by-step. You could probably write a windowing function to do the same thing but I leave that as an exercise for someone else.
I have a table that contains multiple records for each day of the month, over a number of years. Can someone help me out in writing a query that will only return the last day of each month.
SQL Server (other DBMS will work the same or very similarly):
SELECT
*
FROM
YourTable
WHERE
DateField IN (
SELECT MAX(DateField)
FROM YourTable
GROUP BY MONTH(DateField), YEAR(DateField)
)
An index on DateField is helpful here.
PS: If your DateField contains time values, the above will give you the very last record of every month, not the last day's worth of records. In this case use a method to reduce a datetime to its date value before doing the comparison, for example this one.
The easiest way I could find to identify if a date field in the table is the end of the month, is simply adding one day and checking if that day is 1.
where DAY(DATEADD(day, 1, AsOfDate)) = 1
If you use that as your condition (assuming AsOfDate is the date field you are looking for), then it will only returns records where AsOfDate is the last day of the month.
Use the EOMONTH() function if it's available to you (E.g. SQL Server). It returns the last date in a month given a date.
select distinct
Date
from DateTable
Where Date = EOMONTH(Date)
Or, you can use some date math.
select distinct
Date
from DateTable
where Date = DATEADD(MONTH, DATEDIFF(MONTH, -1, Date)-1, -1)
In SQL Server, this is how I usually get to the last day of the month relative to an arbitrary point in time:
select dateadd(day,-day(dateadd(month,1,current_timestamp)) , dateadd(month,1,current_timestamp) )
In a nutshell:
From your reference point-in-time,
Add 1 month,
Then, from the resulting value, subtract its day-of-the-month in days.
Voila! You've the the last day of the month containing your reference point in time.
Getting the 1st day of the month is simpler:
select dateadd(day,-(day(current_timestamp)-1),current_timestamp)
From your reference point-in-time,
subtract (in days), 1 less than the current day-of-the-month component.
Stripping off/normalizing the extraneous time component is left as an exercise for the reader.
A simple way to get the last day of month is to get the first day of the next month and subtract 1.
This should work on Oracle DB
select distinct last_day(trunc(sysdate - rownum)) dt
from dual
connect by rownum < 430
order by 1
I did the following and it worked out great. I also wanted the Maximum Date for the Current Month. Here is what I my output is. Notice the last date for July which is 24th. I pulled it on 7/24/2017, hence the result
Year Month KPI_Date
2017 4 2017-04-28
2017 5 2017-05-31
2017 6 2017-06-30
2017 7 2017-07-24
SELECT B.Year ,
B.Month ,
MAX(DateField) KPI_Date
FROM Table A
INNER JOIN ( SELECT DISTINCT
YEAR(EOMONTH(DateField)) year ,
MONTH(EOMONTH(DateField)) month
FROM Table
) B ON YEAR(A.DateField) = B.year
AND MONTH(A.DateField) = B.Month
GROUP BY B.Year ,
B.Month
SELECT * FROM YourTableName WHERE anyfilter
AND "DATE" IN (SELECT MAX(NameofDATE_Column) FROM YourTableName WHERE
anyfilter GROUP BY
TO_CHAR(NameofDATE_Column,'MONTH'),TO_CHAR(NameofDATE_Column,'YYYY'));
Note: this answer does apply for Oracle DB
Here's how I just solved this. day_date is the date field, calendar is the table that holds the dates.
SELECT cast(datepart(year, day_date) AS VARCHAR)
+ '-'
+ cast(datepart(month, day_date) AS VARCHAR)
+ '-'
+ cast(max(DATEPART(day, day_date)) AS VARCHAR) 'DATE'
FROM calendar
GROUP BY datepart(year, day_date)
,datepart(month, day_date)
ORDER BY 1