I am trying to write a piece for a query that grabs the closest, past June 1st. For example, today is 10/2/2018. If I run the query today, I need it to use the date 6/1/2018. If I run it on 5/29/2019, it still needs to grab 6/1/2018. If I run it on 6/2/2019, it should then grab 6/1/2019. If I run it on 6/2/2022, it should then grab 6/1/2022 and so on.
I believe I need to start with something like this:
SELECT CASE WHEN EXTRACT(MONTH FROM NOW())>=6 THEN 'CURRENT' ELSE 'RF LAST' END AS X
--If month is greater than or equal to 6, you are in the CURRENT YEAR (7/1/CURRENT YEAR)
--If month is less than 6, then reference back to the last year (YEAR MINUS ONE)
And I believe I need to truncate the date then perform an operation. I am unsure of which approach to take (if I should be adding a year to a timestamp such as '6/1/1900', or if I should try to disassemble the date parts to perform an operation. I keep getting errors in my attempts such as "operator does not exist". Things I have tried include:
SELECT (CURRENT_DATE- (CURRENT_DATE-INTERVAL '7 months'))
--This does not work as it just gives me a count of days.
SELECT (DATE_TRUNC('month',NOW())+TIMESTAMP'1900-01-01 00:00:00')
--Variations of this just don't work and generally error out.
Use a case expression to determine if you need to use the current year, or, the previous year (months 1 to 5)
case when extract(month from current_date) >= 6 then 0 else -1 end
then add that to the year extracted from current_date, e.g. using to_date()
select to_Date('06' || (extract(year from current_date)::int + case when extract(month from current_date) >= 6 then 0 else -1 end)::varchar, 'mmYYYY');
You could also use make_date(year int, month int, day int) in postgres 9.4+
select make_date(extract(year from current_date) + case when extract(month from current_date) >= 6 then 0 else -1 end, 6, 1) ;
If month lower than 6, trunc year and minus 6 months.
Else trunc year and add 6 months.
set datestyle to SQL,MDY;
select
case when (extract( month from (date::date)))<6 then date_trunc('year',date)-'6 month'::interval
else date_trunc('year',date)+'6 months'::interval
end as closest_prev_june,
another_column,
another_column2
from mytable;
But format is default and supposed you have a column that named date.
If you want to do this with now(), change date columns with now()
function.
I want to just select data for the current week. So...
If the current date is a Monday just select Monday
If the current date is a Tuesday select Monday and Tuesday's data
If the current date is Wednesday select Monday, Tuesday and Wednesday
...and so on. I want it to reset on Sunday and I believe it's some kind of "where" clause just don't know what. As you can see below I'm just counting the number of pieces into the oven and want it to accumulate as the week goes on and then reset on Sunday.
select
count(*) as PiecesIntoOven
from ovenfeederfloat
where...??
Thanks for the help.
If you're looking to do this in Sql Server, see below. Essentially this converts the current date to its numeric (0-6) value, then finds the 0th date for that week and uses it to set the lower bound of the where clause.
select sum(numberofpieces)
from Test
where dateofwork <= getdate()
and dateofwork >= (DATEADD(DAY, DATEPART(WEEKDAY,getdate()) * -1, getdate()) + 1)
Note that the '0' value is impacted by DATEFIRST. https://stackoverflow.com/a/1113891/4824030
I'm not certain how to do this in Oracle. Something like the below should work, but it's being finicky in sqlfiddle.
select sum(numberofpieces)
from Test
where dateofwork <= current_timestamp
and dateofwork >= (((to_char(level+trunc(current_timestamp,'D'),'Day') * -1) + current_timestamp) + 1)
For previous day I used use the below expression .
DATE_INSERTED >=DATEADD(day, DATEDIFF(day,0,GETDATE())-1,0)
AND DATE_INSERTED < DATEADD(day, DATEDIFF(day,0,GETDATE()),0)
How to get rows from yesterday 10Am to today 10AM
-- yesterday at midnight:
DECLARE #yesterday DATETIME = DATEADD(DAY,DATEDIFF(DAY,1,GETDATE()),0);
SELECT
...
WHERE DATE_INSERTED >= DATEADD(HOUR, 10, #yesterday) -- 10 AM yesterday
AND DATE_INSERTED < DATEADD(HOUR, 34, #yesterday); -- 10 AM today
Instead of using zeroes, use some date(time)s that have the desirable properties:
DATE_INSERTED >=
DATEADD(day, DATEDIFF(day,'20010102',GETDATE()),'2001-01-01T10:00:00')
AND DATE_INSERTED <
DATEADD(day, DATEDIFF(day,'20010102',GETDATE()),'2001-01-02T10:00:00')
I.e. if you add the total number of days that have occurred since 2nd January 2001 onto 10:00am on the 1st January 2001, you'll always obtain a value which is "yesterday at 10am". The second one is almost identical.
How can i calculate the last working five days which is monday to Friday. my current script gets the last monday's date, but i cannot get the last friday's date. Please help
declare #StartDate datetime
declare #EndDate datetime
--Calculate date range for report
select #EndDate = Cast(convert(char(10), getdate(), 101)+' 00:00:00' as datetime)
select #StartDate = DateAdd(d, -7, #EndDate)
select #EndDate = Cast(convert(char(10), getdate(), 101)+' 23:59:59' as datetime)
select #StartDate startdate
select #EndDate enddate
Datepart offers you an easy way to get the week day:
SET DATEFIRST 1 -- monday is first day of the week
SELECT DATEPART(weekday, '20110725')
-- result is 1
See T-SQL Date functions and SET DATEFIRST for more information.
Using the weekday, you can work out how many days ago last monday and friday are and use 'AddDate' (like you are doing now) to calculate those.
Note that you should really use DATEDIFF for this type of date range selection. If you select everything up to 23:59:59 there's always a chance that some records are left out. For example 23:59:59.001 is out of range but it's still on the same day. With DATEDIFF you can test whether it's on the same day, discarding the time part. No need to bother with casting to string, adding time and casting back.
The answer is more complex than people are assuming. What you need is to go 5 days back and find the first monday before that, and the first friday after that. You can eather use ##datefirst, a calculation or 'set firstdate 1' for that. I prefer not using the last one, because it can't be done in functions. As you can see i used the calculation, ##datefirst is just as good.
Assuming you want the last group of monday to friday that is in the past. This query will get that. If you trust your current monday, you can just add 5 days and subtract 1 minute (I wouldn't trust it, it only returns last monday if you run the query on a monday).
In my sql, I am not aiming for simplicity, I am aiming for effectivity.
DECLARE #getdate datetime = dateadd(day, cast(getdate() as int), 0)
-- the 'Declare' can also be written like this thanks to #Andriy M
--DECLARE #getdate = CAST(GETDATE() - 0.5 AS int)
SELECT #getdate - 5 - CAST(#getdate- 5 as int) % 7 monday,
dateadd(minute, -1, #getdate) - CAST(#getdate- 5 as int) % 7 friday
Result:
Monday Friday
2011-07-18 0:00:00 2011-07-22 23:59:00
*first solution was a day off #AndriyM pointed it out, it has been solved.
Answer to #Andriy M
For some reason it acted different than I expected. I can't explain it but try this
select cast(dateadd(day, cast(getdate() as int) - .5, 0) as datetime),
cast(dateadd(day, cast(getdate() as int), 0) as datetime),
cast(dateadd(day, cast(getdate() as int) + .5, 0) as datetime)
in the morning the last 2 fields has same value, in the evening the first 2 fields has the same value. I am as surprised as you are, I wish I could explain it. It was tested here
https://data.stackexchange.com/stackoverflow/query/new
These two questions:
Find last sunday
How to get last day of last week in sql?
might be of some help.
Although, if you already know how to find the last Monday, you can easily find the corresponding Friday by adding 4 days to the Monday date using the DATEADD() function. For example:
SELECT #EndDate = DATEADD(DAY, 4, #StartDate)
It's all relative to TODAY's date, right?
So find out what today is DATEPART(weekday, getdate())
And then turn that into an adjustment variable -- e.g.:
declare #adjustment int
set #adjustment = CASE DATEPART(weekday, getdate()) WHEN 'MONDAY' THEN 0 WHEN 'TUESDAY' THEN 1, ...END
So then the Monday you want would be today - 7 - #adjustment
...and the Friday you want would be Monday + 5
Declare #myMonday smalldatetime,
#myFriday smalldatetime
set #myMonday = getdate() - 7 - #adjustment
set #myFriday = #myMonday + 5
DECLARE #my int
DECLARE #myDeduct int
DECLARE #day INT
DECLARE #mydate DATETIME
SET #mydate = '2012-08-01'
SET #myDeduct = 0
SET DateFirst 1 -- Set it monday=1 (value)
--Saturday and Sunday on the first and last day of a month will Deduct 1
IF (DATEPART(weekday,(DATEADD(dd,-(DAY(#mydate)-1),#mydate))) > 5)
SET #myDeduct = #myDeduct + 1
IF (DATEPART(weekday,(DATEADD(dd,-(DAY(DATEADD(mm,1,#mydate))),DATEADD(mm,1,#mydate)))) > 5)
SET #myDeduct = #myDeduct + 1
SET #my = day(DATEADD(dd,-(DAY(DATEADD(mm,1,#mydate))),DATEADD(mm,1,#mydate)))
select (((#my/7) * 5 + (#my%7)) - #myDeduct) as Working_Day_per_month
I'll throw my hat in the ring too. :-)
DECLARE #Date datetime = '01/11/2015'
DECLARE #StartDate datetime = DATEADD(d, (1 - (DATEDIFF(d, CAST('1899.12.31' AS datetime), #Date - 6) % 7)), #Date - 6) -- MONDAY
DECLARE #EndDate datetime = DATEADD(d, (5 - (DATEDIFF(d, CAST('1899.12.31' AS datetime), #Date - 6) % 7)), #Date - 6) -- FRIDAY
SELECT '#Date' as Variable ,CONVERT(date, #Date) as DateValue ,DATENAME(dw, #Date) as DayOfTheWeek
UNION SELECT '#StartDate' as Variable ,CONVERT(date, #StartDate) as DateValue ,DATENAME(dw, #StartDate) as DayOfTheWeek
UNION SELECT '#EndDate' as Variable ,CONVERT(date, #EndDate) as DateValue ,DATENAME(dw, #EndDate) as DayOfTheWeek
-- Variable DateValue DayOfTheWeek
-- ---------- ---------- ------------
-- #Date 2015-01-11 Sunday
-- #StartDate 2015-01-05 Monday
-- #EndDate 2015-01-09 Friday
BONUS: Here you can generate a quick table of the 5 weekdays using the same technique.
SELECT DATENAME(dw, DATEADD(d, TT.DaysToAdd, DATEADD(d, (1 - (DATEDIFF(d, CAST('1899.12.31' AS datetime), #Date - 6) % 7)), #Date - 6))) as DayOfTheWeek
, DATEADD(d, TT.DaysToAdd, DATEADD(d, (1 - (DATEDIFF(d, CAST('1899.12.31' AS datetime), #Date - 6) % 7)), #Date - 6)) as DateValue
FROM (SELECT ROW_NUMBER() OVER (ORDER BY (SELECT NULL)) - 1 as DaysToAdd FROM (VALUES(0),(0),(0),(0),(0)) a(n)) as TT
-- DayOfTheWeek DateValue
-- ------------------------------ -----------------------
-- Monday 2015-01-05 00:00:00.000
-- Tuesday 2015-01-06 00:00:00.000
-- Wednesday 2015-01-07 00:00:00.000
-- Thursday 2015-01-08 00:00:00.000
-- Friday 2015-01-09 00:00:00.000
Here's an explanation:
1) First we need to know what date to begin our evaluation from. For this example, we chose to use Sunday, January 11th, 2015.
DECLARE #Date2 datetime = '01/11/2015'
1b) Here's a nice to have bonus technique of getting the name for the day of the week, given a date value
SELECT #Date2 as DateValue, DATENAME(dw, #Date2) as DayOfTheWeek
2) Next we need to know deterministically (based on the US calendar) what the given day of the week is numerically between 1 and 7
NOTE: 1899.12.31 is the first Sunday before 1900.01.01, which is the MINIMUM value for the SmallDateTime data type.
NOTE: Yes, you could use DATEPART(dw, #Date) like this more simply, but it is not deterministic given that certain server environments could have different configurations
RESULTS: 1 = Sunday | 2 = Monday | 3 = Tuesday | 4 = Wednesday | 5 = Thursday | 6 = Friday | 7 = Saturday
SELECT ((DATEDIFF(d, CAST('1899.12.31' AS datetime), #Date2) % 7) + 1) [DayOfWeek Deterministic (Based on US)]
3) Now, given any date, you should have a deterministic way of determining the Monday for that given week
SELECT DATEADD(d, (1 - (DATEDIFF(d, CAST('1899.12.31' AS datetime), #Date2) % 7)), #Date2) as [Monday Day of the Week - Deterministic (Based on US)]
4) Now, given any date, you should have a deterministic way of determining the Friday for that given week
SELECT DATEADD(d, (5 - (DATEDIFF(d, CAST('1899.12.31' AS datetime), #Date2) % 7)), #Date2) as [Monday Day of the Week - Deterministic (Based on US)]
5) The last date manipulation technique we need is to know how to get into a week that has the first full week of weekdays happening before it. For instance, if we are on a Sunday and we subtract 1 day from it, then we get to Saturday, which puts us in the previous week, which is the first full week of weekdays. Alternatively, if we also subtracted 1 day from Monday, it would only get us to Sunday, which is not the previous week, so subtracting 1 day is not enough. On the flip side, if we were on a Saturday and subtracted 7 days, it would take us past the previous full week of weekday, into the week before it, which is too far. Here's a run down of the analysis to figure out what the magic numbers is that you can subtract by that will work with any day of the week. As you can see below, the magic number is 6.
-- DAYS TO SUBTRACT
-- Day of the Week - 0 - 1 - 2 - 3 - 4 - 5 - 6 - 7
-- =============== ==== ==== ==== ==== ==== ==== ==== ====
-- Sunday Bad Good Good Good Good Good Good Good
-- Monday Bad Bad Good Good Good Good Good Good
-- Tuesday Bad Bad Bad Good Good Good Good Good
-- Wednesday Bad Bad Bad Bad Good Good Good Good
-- Thursday Bad Bad Bad Bad Bad Good Good Good
-- Friday Bad Bad Bad Bad Bad Bad Good Good
-- Saturday Good Good Good Good Good Good Good Bad
BONUS) If you want to have all the weekdays in little table, then you would want to also use a quick zero based "tally table". There are many ways to do this, so pick your flavor. Here are few.
SELECT * FROM (SELECT 0 as DaysToAdd UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4) as TT
SELECT * FROM (SELECT TOP 5 ROW_NUMBER() OVER (ORDER BY (SELECT NULL)) - 1 as DaysToAdd FROM sys.all_columns a CROSS JOIN sys.all_columns b) as TT
SELECT * FROM (SELECT ROW_NUMBER() OVER (ORDER BY (SELECT NULL)) - 1 as DaysToAdd FROM (VALUES(0),(0),(0),(0),(0)) a(n)) as TT
I work at a college and our student management systems academic year start date is determined by the Monday on or before the 1st of August. I need to match this in my query, is there a way to easily get the date of the Monday on or before this date.
The accepted answer didn't work for me because I needed both a Sunday week and a Monday week in the same query. This works across different "datefirst" settings:
SELECT DATEADD(d, -((DATEPART(WEEKDAY, '20110515') - DATEPART(dw, '19000101') + 7) % 7), '20110515')
"DATEPART(dw, '19000101')" will determine your "datefirst" setting since 1900-01-01 was on a Monday. If you want a Tuesday based week, you can change 19000101 to 19000102.
BTW, '20110515' is the only date format that works across all SQL Server culture settings. Dates like '2011-05-06' will get mis-interpreted in certain countries. (credit to Itzik Ben-Gan for pointing this out)
set datefirst 1; -- Make Monday the first day of the week
select dateadd(dd, -1*(datepart(dw, '2009-08-01')-1), '2009-08-01')
Returns July 27th, 2009, which is the Monday on or before August 1. Change it to 2005 when Aug 1 was a Monday and the query will return 08-01
You could use datepart to get the weekday, and then do a little math to back into your monday. This example is using the US default of datefirst 7 (in which Monday is day 2 of the week). Adjust the days to add to be which day of the week Monday is for your locale.
select dateadd(dd, -datepart(dw, '2009-08-01') + 2, '2009-08-01')
very hacky
DECLARE #weekday int
SELECT #weekday = DATEPART(WEEKDAY, '1-Aug-2009')
SELECT CASE
WHEN #weekday = 1 THEN '1-Aug-2009' ELSE DATEADD ( dd,(#weekday-2)*-1, '1-Aug-2009')
END
This is a generic algorithm that will return the first Monday of any month (#inputdate):
DATEADD(wk, DATEDIFF(wk, 0, dateadd(dd, 6 - datepart(day, #inputDate), #inputDate)), 0)
It is a common method for getting the first monday of the month in SQL Server. This link explains how the above calculation works along with many other date calculations.
Here is how the above algorithm could be used to get the Monday on or before the 1st day of a month:
-- Set month to get Monday before or at 1st of month.
DECLARE #inputDate DATETIME
SET #inputDate = '2009-08-01'
-- Get first Monday of month.
DECLARE #firstMonday DATETIME
SET #firstMonday = DATEADD(wk, DATEDIFF(wk, 0, dateadd(dd, 6 - datepart(day, #inputDate), #inputDate)), 0)
-- Determine date for first Monday on or before 1st of month.
DECLARE #startDate DATETIME
SET #startDate = #firstMonday
IF #firstMonday > #inputDate
SET #startDate = DATEADD(wk, -1, #firstMonday)
SELECT #startDate