I need a calculated month value within DATEFROMPARTS function. The month has to be seven month prior to CURRENT_TIMESTAMP month.
This is what I tried:
DATEFROMPARTS(Year(CURRENT_TIMESTAMP), Month(CURRENT_TIMESTAMP)-7, 1) as SevenMoAgo;
I will eventually use this in the following expression where '12-01-2018' is:
where RECORDED_SERVICE_STARTTIME > ='12-01-2018'
I later used
declare #CurMo AS INT;
declare #MonPri7 AS INT;
set #CurMo = Month(CURRENT_TIMESTAMP);
set #MonPri7 = (#CurMo -7);
Datefromparts(Year(CURRENT_TIMESTAMP), #MonPri7, 1) as SevenMoAgo;
This also did not work.
I get the following error message:
"Cannot construct data type date, some of the arguments have values which are not valid."
For the second code I get:
Msg 102, Level 15, State 1, Line 8
Incorrect syntax near 'Datefromparts'.
Try this...
SELECT DATEADD(MONTH, DATEDIFF(MONTH, 0, CURRENT_TIMESTAMP) - 7, 0)
Let me explain. First off, we need to understand that SQL Server interprets 0 as 1900-01-01 as shown by the following DATEPART functions.
SELECT DATEPART(YEAR, 0) AS Year
, DATEPART(MONTH, 0) AS Month
, DATEPART(DAY, 0) AS Day;
Which returns...
Year Month Day
----------- ----------- -----------
1900 1 1
Therefore, my SQL could be rewritten as...
SELECT DATEADD(MONTH, DATEDIFF(MONTH, '1900-01-01', CURRENT_TIMESTAMP) - 7, '1900-01-01')
Now perhaps it is a little easier to see what is going on here. The DATEDIFF function returns the number number of months between 1900-01-01 and today (CURRENT_TIMESTAMP) which is 1434.
SELECT DATEADD(MONTH, 1434 - 7, '1900-01-01')
Then we subtract 7 from 1434 which is 1427 and add that many months back to 1900-01-01.
SELECT DATEADD(MONTH, 1427, '1900-01-01')
Which yields 2018-12-01.
The reason is #MonPri7 is equal to ZERO when you say (#CurMo -7)
There are many different ways to calculate it, but if you want to fix your logic, you should use this:
declare #CurMo AS INT;
declare #MonPri7 AS INT;
set #CurMo = Month(CURRENT_TIMESTAMP);
set #MonPri7 = (#CurMo -7);
declare #Y int = Year(CURRENT_TIMESTAMP) -- <-- This is new variable
-- if 7 months ago is ZERO then you should go back to prev year December
if #MonPri7 = 0
begin
set #MonPri7 = 12
set #Y = #Y - 1
end
Edit:
declare #SevenMonthsAgo datetime;
select #SevenMonthsAgo = Datefromparts(#Y, #MonPri7, 1);
SELECT yourfields
FROM yourtable
where RECORDED_SERVICE_STARTTIME > = '01-01-2019' and
RECORDED_SERVICE_STARTTIME > = #SevenMonthsAgo
Related
I am trying to find the total number of members in a given period. Say I have the following data:
member_id start_date end_date
1 9/1/2013 12/31/2013
2 10/1/2013 11/12/2013
3 12/1/2013 12/31/2013
4 5/1/2012 8/5/2013
5 9/1/2013 12/31/2013
6 7/1/2013 12/31/2013
7 6/6/2012 12/5/2013
8 10/1/2013 12/31/2013
9 7/8/2013 12/31/2013
10 1/1/2012 11/5/2013
In SQL I need to create a report that will list out the number of members in each month of the year. In this case something like the following:
Date Members Per Month
Jan-12 1
Feb-12 1
Mar-12 1
Apr-12 1
May-12 2
Jun-12 3
Jul-12 3
Aug-12 3
Sep-12 3
Oct-12 3
Nov-12 3
Dec-12 3
Jan-13 3
Feb-13 3
Mar-13 3
Apr-13 3
May-13 3
Jun-13 3
Jul-13 5
Aug-13 4
Sep-13 6
Oct-13 8
Nov-13 6
Dec-13 6
So there is only 1 member from Jan-12 (member id 10) until May-12 when member id 4 joins making the count 2 and so on.
The date range can be all over so I can't specify the specific dates but it is by month, meaning that even if someone ends 12-1 it is considered active for the month for Dec.
I was able to create the following stored procedure that was able to accomplish what I needed:
USE [ValueBasedSandbox]
GO
/****** Object: StoredProcedure [dbo].[sp_member_count_per_month] Script Date: 01/08/2015 12:02:37 ******/
SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER ON
GO
-- =============================================
-- Create date: 2015-08-01
-- Description: Find the counts per a given date passed in
-- =============================================
CREATE PROCEDURE [dbo].[sp_member_count_per_month]
-- Add the parameters for the stored procedure here
#YEAR int
, #ENDYEAR int
AS
DECLARE #FIRSTDAYMONTH DATETIME
DECLARE #LASTDAYMONTH DATETIME
DECLARE #MONTH INT = 1;
--Drop the temporary holding table if exists
IF OBJECT_ID('tempdb.dbo.##TEMPCOUNTERTABLE', 'U') IS NOT NULL
DROP TABLE dbo.##TEMPCOUNTERTABLE
CREATE TABLE dbo.##TEMPCOUNTERTABLE (
counter INT
, start_date DATETIME2
, end_date DATETIME2
)
--Perform this loop for each year desired
WHILE #YEAR <= #ENDYEAR
BEGIN
--Perform for each month of the year
WHILE (#MONTH <= 12)
BEGIN
-- SET NOCOUNT ON added to prevent extra result sets from
-- interfering with SELECT statements.
SET NOCOUNT ON;
SET #FIRSTDAYMONTH = DATEADD(MONTH, #MONTH - 1, DATEADD(YEAR, #YEAR-1900, 0))
SET #LASTDAYMONTH = DATEADD(MONTH, #MONTH, DATEADD(YEAR, #YEAR-1900, 0)-1)
INSERT INTO dbo.##TEMPCOUNTERTABLE(counter, start_date, end_date)
SELECT COUNT(*) AS counter
, #FIRSTDAYMONTH AS start_date
, #LASTDAYMONTH AS end_date
FROM dbo.member_table
WHERE start_date <= #LASTDAYMONTH
AND end_date >= #FIRSTDAYMONTH
--Increment through all the months of the year
SET #MONTH = #MONTH + 1
END -- End Monthly Loop
--Reset Month counter
SET #MONTH = 1
--Increment the desired years
SET #YEAR = #YEAR + 1
END -- End Yearly Loop
--Display the results
SELECT *
FROM dbo.##TEMPCOUNTERTABLE
-- Drop the temp table
IF OBJECT_ID('tempdb.dbo.##TEMPCOUNTERTABLE', 'U') IS NOT NULL
DROP TABLE dbo.##TEMPCOUNTERTABLE
GO
This should do the trick
with datesCte(monthStart,monthEnd) as
(
select cast('20120101' as date) as monthStart, cast('20120131' as date) as monthEnd
union all
select DATEADD(MONTH, 1, d.monthStart), dateadd(day, -1, dateadd(month, 1, d.monthStart))
from datesCte as d
where d.monthStart < '20140101'
)
select *
from datesCte as d
cross apply
(
select count(*) as cnt
from dbo.MemberDates as m
where m.startDate <= d.monthEnd and m.endDate > d.monthStart
) as x
order by d.monthStart
I need to show N records with each record being 1 weeks worth of summarized data. One of the inputs will be a date range.
Since each week should start with monday, I want to get a list of all the monday dates in the date range. Any suggestions to get that?
Note: This has to run on SQL 2005
declare #dt datetime
SET #dt = '2010-01-01'
declare #dtEnd datetime
SET #dtEnd = '2010-12-04'
DECLARE #day AS NVARCHAR(30)
WHILE (#dt < #dtEnd) BEGIN
-- insert into table(datefield)
-- values(#dt)
SET #day = UPPER(DATENAME(weekday,#dt))
IF #day = 'MONDAY'
--PRINT 'date is: ' + CAST(#dt AS VARCHAR(30))
BEGIN
PRINT 'date is: ' + CAST(#dt AS VARCHAR(30)) + ' ' + #day
END
SET #dt = DATEADD(day, 1, #dt)
END
The code you show seems to work, but it can be made more efficient. The code I show below will calculate the first Monday on or after the start date. Then it is used in the same loop but we simply add 7 days each time through the loop.
declare #dt datetime,
#dtEnd datetime
SET #dt = '2010-01-01'
SET #dtEnd = '2010-12-04'
Set #dt = DateAdd(Day, 7 - DateDiff(Day, 0, #dt) % 7, #dt)
WHILE (#dt < #dtEnd) BEGIN
PRINT 'date is: ' + CAST(#dt AS VARCHAR(30)) + ' ' + DateName(Weekday, #dt)
SET #dt = DATEADD(day, 7, #dt)
END
One thing I notice about your code is that you are using DateName, which works ok for displaying the weekday name, but not so good for making decisions (like you are doing). The problem with DateName is that it uses the language setting of the currently logged in user.
The following code shows this behavior.
set language 'spanish'
Select DateName(Weekday, GetDate())
set language 'us_english'
Select DateName(Weekday, GetDate())
Basically, if the language of a user was set to spanish, your code would fail but mine would not. Mine works because it uses the fact that January 1, 1900 was a Monday. It calculates the number of days and takes a mod of that and uses that mod to add the correct number of days. Since January 1, 1900 will always be a Monday, and we are not using the DateName function, this code will work without regard to the language setting.
SELECT * from table t
WHERE t.date between 'startDate' and 'endDate'
AND (SELECT DATEPART(dw,t.StudyDate))=2
I would like to contribute my codes related to the date conversion between local time and UTC time for both current and historical dates, including day light savings time consideration. I reviewed some recent postings but could not find the solution I am looking for so I summarize the solutions I found to develop my own solution. I am running my codes in SQL Server 2008R2. Test cases are provided as comments before each function. Feel free to check out the Credits section for related postings, try out my codes, and let me know if I miss anything. Thanks.
-- Test 1: select dbo.fn_isDayLightSavings(CAST('2014-01-01' AS DATE)), expect 0
-- Test 2: select dbo.fn_isDayLightSavings(CAST('2014-04-14' AS DATE)), expect 1
IF OBJECT_ID(N'fn_isDayLightSavings') IS NOT NULL AND OBJECTPROPERTY(OBJECT_ID(N'fn_isDayLightSavings'),'IsScalarFunction') = 1
DROP FUNCTION fn_isDayLightSavings;
GO
CREATE FUNCTION fn_isDayLightSavings (#dt AS DATETIME)
RETURNS TINYINT
AS
BEGIN
DECLARE #rtn TINYINT, #year INT, #dtsStartMonth DATETIME, #dtsEndMonth DATETIME, #dstStart DATETIME, #dstEnd DATETIME;
SET #year = DATEPART(YEAR, #dt);
-- In year 2007, US day light savings period changes from Apr-Oct to Mar-Nov.
if #year < 2007
BEGIN
-- Last Sunday of April at 2 AM
SET #dtsStartMonth = DATEADD(MONTH, 4, DATEADD(YEAR, #year - 1900, 0));
SET #dstStart = DATEADD(HOUR, 2, DATEADD(day, -(DATEPART(dw, #dtsStartMonth) - 1), #dtsStartMonth));
-- Last Sunday of October at 2 AM
SET #dtsEndMonth = DATEADD(MONTH, 10, DATEADD(YEAR, #year - 1900, 0));
SET #dstEnd = DATEADD(HOUR, 2, DATEADD(day, -(DATEPART(dw, #dtsEndMonth) - 1), #dtsEndMonth));
END
else
BEGIN
-- 2nd Sunday of March at 2 AM
SET #dtsStartMonth = DATEADD(MONTH, 2, DATEADD(YEAR, #year - 1900, 0));
SET #dstStart = DATEADD(HOUR, 2, DATEADD(day, ((15 - DATEPART(dw, #dtsStartMonth)) % 7) + 7, #dtsStartMonth));
-- 1st Sunday of November at 2 AM
SET #dtsEndMonth = DATEADD(MONTH, 10, DATEADD(YEAR, #year - 1900, 0));
SET #dstEnd = DATEADD(HOUR, 2, DATEADD(day, ((8 - DATEPART(dw, #dtsEndMonth)) % 7) + 7, #dtsEndMonth));
END
if #dt BETWEEN #dstStart AND #dstEnd SET #rtn=1 ELSE SET #rtn=0;
RETURN #rtn;
END
GO
-- Test 1: select dbo.fn_DateTime2UTC(CAST('2014-01-01 01:00:00' AS DATETIME), 0), expect '2014-01-01 07:00:00.000'
-- Test 2: select dbo.fn_DateTime2UTC(CAST('2014-01-01 01:00:00' AS DATETIME), 1), expect '2013-01-01 07:00:00.000'
-- Test 3: select dbo.fn_DateTime2UTC(CAST('2014-05-01 01:00:00' AS DATETIME), 0), expect '2014-05-01 06:00:00.000'
-- Test 4: select dbo.fn_DateTime2UTC(CAST('2014-05-01 01:00:00' AS DATETIME), 1), expect '2014-05-01 07:00:00.000'
IF OBJECT_ID(N'fn_DateTime2UTC') IS NOT NULL AND OBJECTPROPERTY(OBJECT_ID(N'fn_DateTime2UTC'),'IsScalarFunction') = 1
DROP FUNCTION fn_DateTime2UTC;
GO
CREATE FUNCTION fn_DateTime2UTC (#dt AS DATETIME, #ignoreDST AS TINYINT = 0)
-- do CAST(? AS DATETIMEOFFSET), if need datetimeoffset type
RETURNS DATETIME
AS
BEGIN
DECLARE #tzOffset INT, #utcDt DATETIME;
-- Get current time zone offset in minutes
SET #tzOffset = DATEPART(TZoffset, SYSDATETIMEOFFSET()) +
(CASE WHEN dbo.fn_isDayLightSavings(#dt)=1 THEN 60 ELSE 0 END);
if dbo.fn_isDayLightSavings(#dt)=0
set #utcDt = DATEADD(MINUTE, -#tzOffset, #dt);
else if #ignoreDST=0
set #utcDt = DATEADD(MINUTE, -#tzOffset, #dt);
else
set #utcDt = DATEADD(MINUTE, -#tzOffset+60, #dt);
return #utcDt;
END
GO
-- Test 1: select dbo.fn_UTC2DateTime(CAST('2014-01-01 07:00:00.000' AS DATETIME), 0), expect '2014-01-01 01:00:00'
-- Test 2: select dbo.fn_UTC2DateTime(CAST('2013-01-01 07:00:00.000' AS DATETIME), 1), expect '2014-01-01 01:00:00'
-- Test 3: select dbo.fn_UTC2DateTime(CAST('2014-05-01 06:00:00.000' AS DATETIME), 0), expect '2014-05-01 01:00:00'
-- Test 4: select dbo.fn_UTC2DateTime(CAST('2014-05-01 07:00:00.000' AS DATETIME), 1), expect '2014-05-01 01:00:00'
IF OBJECT_ID(N'fn_UTC2DateTime') IS NOT NULL AND OBJECTPROPERTY(OBJECT_ID(N'fn_UTC2DateTime'),'IsScalarFunction') = 1
DROP FUNCTION fn_UTC2DateTime;
GO
CREATE FUNCTION fn_UTC2DateTime (#utcDt AS DATETIME, #ignoreDST AS TINYINT = 0)
RETURNS DATETIME
AS
BEGIN
DECLARE #tzOffset INT, #dt DATETIME;
-- Get current time zone offset in minutes
SET #tzOffset = DATEPART(TZoffset, SYSDATETIMEOFFSET()) +
(CASE WHEN dbo.fn_isDayLightSavings(#utcDt)=1 THEN 60 ELSE 0 END);
if dbo.fn_isDayLightSavings(#utcDt)=0
set #dt = DATEADD(MINUTE, #tzOffset, #utcDt);
else if #ignoreDST=0
set #dt = DATEADD(MINUTE, #tzOffset, #utcDt);
else
set #dt = DATEADD(MINUTE, #tzOffset-60, #utcDt);
return #dt;
END
GO
-- Test 1: select dbo.fn_UTC2DateTimeOffset(CAST('2014-01-01 07:00:00.000' AS DATETIME), 0), expect '2014-01-01 01:00:00.0000000 -06:00'
-- Test 2: select dbo.fn_UTC2DateTimeOffset(CAST('2013-01-01 07:00:00.000' AS DATETIME), 1), expect '2014-01-01 01:00:00.0000000 -06:00'
-- Test 3: select dbo.fn_UTC2DateTimeOffset(CAST('2014-05-01 06:00:00.000' AS DATETIME), 0), expect '2014-05-01 01:00:00.0000000 -05:00'
-- Test 4: select dbo.fn_UTC2DateTimeOffset(CAST('2014-05-01 07:00:00.000' AS DATETIME), 1), expect '2014-05-01 00:00:00.0000000 -06:00'
IF OBJECT_ID(N'fn_UTC2DateTimeOffset') IS NOT NULL AND OBJECTPROPERTY(OBJECT_ID(N'fn_UTC2DateTimeOffset'),'IsScalarFunction') = 1
DROP FUNCTION fn_UTC2DateTimeOffset;
GO
CREATE FUNCTION fn_UTC2DateTimeOffset (#utcDt AS DATETIME, #ignoreDST TINYINT = 0)
RETURNS DATETIMEOFFSET
AS
BEGIN
DECLARE #tzOffset INT, #dt DATETIMEOFFSET;
-- Get current time zone offset in minutes
SET #tzOffset = DATEPART(TZoffset, SYSDATETIMEOFFSET()) +
(CASE WHEN dbo.fn_isDayLightSavings(#utcDt)=1 THEN 60 ELSE 0 END);
if dbo.fn_isDayLightSavings(#utcDt)=0
set #dt = SWITCHOFFSET(CAST(#utcDt AS DATETIMEOFFSET), #tzOffset);
else if #ignoreDST=0
set #dt = SWITCHOFFSET(CAST(#utcDt AS DATETIMEOFFSET), #tzOffset);
else
set #dt = SWITCHOFFSET(CAST(#utcDt AS DATETIMEOFFSET), #tzOffset-60);
return #dt;
END
GO
-- Credits:
-- Determination of day light savings start and end time, Jamie F., 11/1/2013
-- (http://stackoverflow.com/questions/19732896/how-to-create-daylight-savings-time-start-and-end-function-in-sql-server)
-- Day light savings time calendar, USNO
-- (http://aa.usno.navy.mil/faq/docs/daylight_time.php)
-- Time zone offset, Microsoft
-- (http://msdn.microsoft.com/en-us/library/ms174420.aspx)
-- Local system date time with time zone, Robert Cantor, 12/7/2013
-- (http://stackoverflow.com/questions/1205142/tsql-how-to-convert-local-time-to-utc-sqlserver2008)
-- Convert date with time zone and day light savings, Eric Z Beard, 8/24/2008
-- (http://stackoverflow.com/questions/24797/effectively-converting-dates-between-utc-and-local-ie-pst-time-in-sql-2005#25073)
Kudos to you for assembling that knowledge into one place. Assuming it is accurate, it's easy to pick up and use immediately.
However, the trouble is that without an extensive set of unit tests you can't be sure of accuracy.
Might I suggest the best option for date/time conversions would be to use an existing highly tested open source library such as http://nodatime.org/. Date/time conversions have so many intracacies and details that could not possibly be contained in a few dozen lines of SQL code.
nodatime is a .NET library that should easily be accessible via the SQL Server CLR user-defined functions feature.
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 am writing a function in T-SQL returning the last day of the month regardless of the date input.
Here is my code:
Alter Function dbo.FN_Get_Last_Day_in_Month2
(#FN_InputDt Datetime)
Returns smalldatetime
as
Begin
Declare #Result smalldatetime
Set #Result =
case when #FN_InputDt <> 01-01-1900 then
DATEADD(m, DATEDIFF(M, 0,#FN_InputDt)+1, -1)
Else 0 End
Return #Result
End
The code is not working correctly, here is a test that shows the bad behavior:
SELECT dbo.fn_get_last_day_in_month (07-05-2010)
Here is the (incorrect) result:
2010-07-31 00:00:00
What is 07-05-2010...May 7th or July 5th? You need to use a safe date format, take a look at Setting a standard DateFormat for SQL Server
example from How to find the first and last days in years, months etc
DECLARE #d DATETIME
SET #d = '20100705' -- notice ISO format
SELECT
DATEADD(yy, DATEDIFF(yy, 0, #d), 0) AS FirstDayOfYear,
DATEADD(yy, DATEDIFF(yy, 0, #d)+1, -1) AS LastDayOfYear,
DATEADD(qq, DATEDIFF(qq, 0, #d), 0) AS FirstDayOfQuarter,
DATEADD(qq, DATEDIFF(qq, 0, #d)+1, -1) AS LastDayOfQuarter,
DATEADD(mm, DATEDIFF(mm, 0, #d), 0) AS FirstDayOfMonth,
DATEADD(mm, DATEDIFF(mm, 0, #d)+1, -1) AS LastDayOfMonth,
#d - DATEDIFF(dd, ##DATEFIRST - 1, #d) % 7 AS FirstDayOfWeek,
#d - DATEDIFF(dd, ##DATEFIRST - 1, #d) % 7 + 6 AS LastDayOfWeek
for just the day use day or datepart
select DAY(getdate()),
DATEPART(dd,GETDATE())
Cast the return value to a SQL datetime type, and then call the "DAY" function to get the day in as an integer. See the function reference here:
http://msdn.microsoft.com/en-us/library/ms176052.aspx
Not sure which database you're using, but this should be a standard function across all databases.
I'd return a DATETIME, I've had trouble with SMALLDATETIME in the past.
DECLARE #Result DATETIME
SET #Result = DATEADD(m , 1, #FN_Input);
RETURN CAST(FLOOR(CAST(DATEADD(d, DATEPART(d, #Result) * -1, #Result) AS FLOAT)) AS DATETIME)
Also, I think you may be a victim of SQL's complete disregard of date formatting. Always, always, always, when typing a string into test a SQL function use the following format;
'05 Jul 2010'
Your function probably works but it interpreted your date as 5th July - not 7th May.
DECLARE #date DATETIME = '20130624';
SELECT Day(EOMONTH ( #date )) AS LastDay;
GO