I have a report in which I have business days (so no weekends) calculated based on the calendar date in which the event occurred in the SQL query. IE,
Date_Completed: 22 Jan 2014 (<--- From system tables)
Day_completed = 16 <---- What I calculated in the query.
Now, I am trying to create a calendar-like report in SSRS for the purpose of tracking the event dates. I want to the make the business days and the corresponding calendar date as my column headers ie
1
1/1/2014
2
1/2/2014
So my problem is that the events do not occur on every business day, so I created a bunch of dummy values in my query (basically union all 1-40 business days...I'll work on making this dynamic in the future).
So now I have the 1-40 business day columns, and my values are aligning to them properly...however, the calendar dates are not matching because I do not know how to make SSRS skip weekend dates in my column header...
I tried something like:
=iif((weekday(dateadd("d",Fields!period_complete.Value,Fields!period_dt.Value))<>1
or weekday(dateadd("d",Fields!period_complete.Value,Fields!period_dt.Value))<>7),
dateadd("d",Fields!period_complete.Value,Fields!period_dt.Value),"")
but that did nothing. Anyone have any ideas on how I can get SSRS to spit out non-weekend, calendar dates for a quarter (Jan-Mar)?
I see from your question that you're trying to use SSRS expressions to achieve your desired result. However, when it comes to dynamic filtering, I've found myself turning to SQL to do the majority of the work. With that in mind, I think this will achieve your desired goal.
The query below will create a list of all dates for a year (equal to the current year based on the system time). Then, it will give you the day of the week as an integer and the day of the year. Additionally, it will filter out all Saturdays and Sundays, leaving you with only working days. Finally, it calculates which quarter we're in now (I'm using quarters by calendar year, but you could easily adapt it to rolling quarters or to your fiscal year quarters) as a way of filtering the list of days to the current quarter. Please let me know if you have questions.
Declare #Start date = Case
When Month(getdate()) <= 3
Then '01/01/' + CAST(YEAR(getdate()) as varchar(4))
When Month(getdate()) <= 6
Then '04/01/' + CAST(YEAR(getdate()) as varchar(4))
When Month(getdate()) <= 9
Then '07/01/' + CAST(YEAR(getdate()) as varchar(4))
Else '10/01/' + CAST(YEAR(getdate()) as varchar(4))
End;
Declare #End date = Case
When Month(getdate()) <= 3
Then '03/31/' + CAST(YEAR(getdate()) as varchar(4))
When Month(getdate()) <= 6
Then '6/30/' + CAST(YEAR(getdate()) as varchar(4))
When Month(getdate()) <= 9
Then '09/30/' + CAST(YEAR(getdate()) as varchar(4))
Else '12/31/' + CAST(YEAR(getdate()) as varchar(4))
End;
Set NoCount on;
With All_Dates as (
Select CAST('01/01/' + CAST(YEAR(getdate()) as varchar(4)) as Date) as Dates
Union All
Select DATEADD(day,1,Dates) as Dates
From All_Dates
Where Dates < '12/31/' + CAST(YEAR(getdate()) as varchar(4))
)
Select DATEPART(DAYOFYEAR,Dates) as Day_of_Year
, DATEPART(WEEKDAY,Dates) as Day_of_Week
, Dates
From All_Dates
Where DATEPART(WEEKDAY,Dates) between 2 and 6
and Dates >= #Start
and Dates <= #End
Option (MaxRecursion 400);
I need to convert character string to date format to fetch the first and last day of the month.
Input.
DECLARE #InpDat AS VARCHAR(20)
SET #InpDat = 201308
Expected output 2013-08-31
I need to get the first and last day of the given yearmonth. Can you please help me to get that.
I tried with the convert option but couldnt get it. ?
Use CAST(#InpDat + '01' AS DATE) to convert it to a date and DateAdd for the date arithmetic.
DECLARE #InpDat AS VARCHAR(20) SET #InpDat = '201308'
DECLARE #Month DATE SET #Month = CAST(#InpDat + '01' AS DATE)
SELECT #MONTH AS First,
DATEADD(day, -1, DATEADD(month, 1, #MONTH)) AS Last
DEMO
DECLARE #InpDat AS VARCHAR(20)
SET #InpDat = '201308'
SELECT CONVERT(DATETIME(#InpDat+'01') AS FirstDate,
DATEADD(dd,-1,DATEADD(mm,1,CONVERT(DATETIME(#InpDat+'01'))) AS LastDate
Why do you expect that a random number will convert to a date format without any cajoling? The database is a powerful tool but not magical.
You have to parse the number into a date, then format the date the way you like. I might suggest this sequence of calls:
-- assume you can separate the year and month into separate parts.
-- use 01 as default day for now, we'll get last day shortly.
DECLARE #IntermediateDate AS DATE SET #IntermediateDate = DATEFROMPARTS(2013,08,01)
-- Advance the month by one, and subtract one day to find the last day of the month.
SET #IntermediateDate = DATEADD( dd, -1, DATEADD( mm, 1, #IntermediateDate))
Now format using your favorite formatting functions.
In SQL Statement in microsoft sql server, there is a built-in function to get week number but it is the week of the year.
Select DatePart(week, '2012/11/30') // **returns 48**
The returned value 48 is the week number of the year.
Instead of 48, I want to get 1, 2, 3 or 4 (week number of the month). I think the week number of the month can be achieved by modules with Month Number of this week. For e.g.
Select DATEPART(week, '2012/11/30')%MONTH('2012/11/30')
But I want to know is there other built-in functions to get WeekNumber of the month in MS SQL SERVER.
Here are 2 different ways, both are assuming the week starts on monday
If you want weeks to be whole, so they belong to the month in which they start:
So saturday 2012-09-01 and sunday 2012-09-02 is week 4 and monday 2012-09-03 is week 1 use this:
DECLARE #date date = '2012-09-01'
SELECT (day(datediff(d,0,#date)/7*7)-1)/7+1
If your weeks cut on monthchange so saturday 2012-09-01 and sunday 2012-09-02 is week 1 and monday 2012-09-03 is week 2 use this:
DECLARE #date date = '2012-09-01'
SELECT
datediff(ww,datediff(d,0,dateadd(m,datediff(m,7,#date),0)
)/7*7,dateadd(d,-1,#date))+1
I received an email from Gerald. He pointed out a flaw in the second method. This should be fixed now
I received an email from Ben Wilkins. He pointed out a flaw in the first method. This should be fixed now
DECLARE #DATE DATETIME
SET #DATE = '2013-08-04'
SELECT DATEPART(WEEK, #DATE) -
DATEPART(WEEK, DATEADD(MM, DATEDIFF(MM,0,#DATE), 0))+ 1 AS WEEK_OF_MONTH
No built-in function. It depends what you mean by week of month. You might mean whether it's in the first 7 days (week 1), the second 7 days (week 2), etc. In that case it would just be
(DATEPART(day,#Date)-1)/7 + 1
If you want to use the same week numbering as is used with DATEPART(week,), you could use the difference between the week numbers of the first of the month and the date in question (+1):
(DATEPART(week,#Date)- DATEPART(week,DATEADD(m, DATEDIFF(m, 0, #Date), 0))) + 1
Or, you might need something else, depending on what you mean by the week number.
Just look at the date and see what range it falls in.
Range 1-7 is the 1st week, Range 8-14 is the 2nd week, etc.
SELECT
CASE WHEN DATEPART(day,yourdate) < 8 THEN '1'
ELSE CASE WHEN DATEPART(day,yourdate) < 15 then '2'
ELSE CASE WHEN DATEPART(day,yourdate) < 22 then '3'
ELSE CASE WHEN DATEPART(day,yourdate) < 29 then '4'
ELSE '5'
END
END
END
END
Similar to the second solution, less code:
declare #date datetime = '2014-03-31'
SELECT DATEDIFF(week,0,#date) - (DATEDIFF(week,0,DATEADD(dd, -DAY(#date)+1, #date))-1)
Check this out... its working fine.
declare #date as datetime = '2014-03-10'
select DATEPART(week,#date) - DATEPART(week,cast(cast(year(#date) as varchar(4))+'-' + cast(month(#date) as varchar(2)) + '-01' as datetime))+1
WeekMonth = CASE WHEN (DATEPART(day,TestDate) - datepart(dw,TestDate))>= 22 THEN '5'
WHEN (DATEPART(day,TestDate) - datepart(dw,TestDate))>= 15 THEN '4'
WHEN (DATEPART(day,TestDate) - datepart(dw,TestDate))>= 8 THEN '3'
WHEN (DATEPART(day,TestDate) - datepart(dw,TestDate))>= 1 THEN '2'
ELSE '1'
END
There is no inbuilt function to get you the week number. I dont think dividing will help you anyway as the number of weeks in a month is not constant.
http://msdn.microsoft.com/en-us/library/bb675168.aspx
I guess you can divide the number(48) by 4 and take the modules of the same and project that as the week number of that month, by adding one to the result.
Here's a suggestion for getting the first and last days of the week for a month:
-- Build a temp table with all the dates of the month
drop table #tmp_datesforMonth
go
declare #begDate datetime
declare #endDate datetime
set #begDate = '6/1/13'
set #endDate = '6/30/13';
WITH N(n) AS
( SELECT 0
UNION ALL
SELECT n+1
FROM N
WHERE n <= datepart(dd,#enddate)
)
SELECT DATEADD(dd,n,#BegDate) as dDate
into #tmp_datesforMonth
FROM N
WHERE MONTH(DATEADD(dd,n,#BegDate)) = MONTH(#BegDate)
--- pull results showing the weeks' dates and the week # for the month (not the week # for the current month)
select MIN(dDate) as BegOfWeek
, MAX(dDate) as EndOfWeek
, datediff(week, dateadd(week, datediff(week, 0, dateadd(month, datediff(month, 0, dDate), 0)), 0), dDate) as WeekNumForMonth
from #tmp_datesforMonth
group by datediff(week, dateadd(week, datediff(week, 0, dateadd(month, datediff(month, 0, dDate), 0)), 0), dDate)
order by 3, 1
A dirty but easy one liner using Dense_Rank function. Performance WILL suffer, but effective none the less.
DENSE_RANK()over(Partition by Month(yourdate),Year(yourdate) Order by Datepart(week,yourdate) asc) as Week
Here is the query that brings the week number on whatever the startday and endday of the week it may be.
SET DATEFIRST 2
DECLARE #FROMDATE DATE='12-JAN-2015'
-- Get the first day of month
DECLARE #ALLDATE DATE=DATEADD(month, DATEDIFF(month, 0, #FROMDATE), 0)
DECLARE #FIRSTDATE DATE
;WITH CTE as
(
-- Get all dates in that month
SELECT 1 RNO,CAST(#ALLDATE AS DATE) as DATES
UNION ALL
SELECT RNO+1, DATEADD(DAY,1,DATES )
FROM CTE
WHERE DATES < DATEADD(MONTH,1,#ALLDATE)
)
-- Retrieves the first day of week, ie, if first day of week is Tuesday, it selects first Tuesday
SELECT TOP 1 #FIRSTDATE = DATES
FROM CTE
WHERE DATEPART(W,DATES)=1
SELECT (DATEDIFF(DAY,#FIRSTDATE,#FROMDATE)/7)+1 WEEKNO
For more information I have answered for the below question. Can check that.
How do I find week number of a date according to DATEFIRST
floor((day(#DateValue)-1)/7)+1
Here you go....
Im using the code below..
DATEPART(WK,#DATE_INSERT) - DATEPART(WK,DATEADD(DAY,1,DATEADD(s,-1,DATEADD(mm, DATEDIFF(m,0,#DATE_INSERT),0)))) + 1
Try Below Code:
declare #dt datetime='2018-03-15 05:16:00.000'
IF (Select (DatePart(DAY,#dt)%7))>0
Select (DatePart(DAY,#dt)/7) +1
ELSE
Select (DatePart(DAY,#dt)/7)
There is an inbuilt option to get the week number of the year
**select datepart(week,getdate())**
You can simply get week number by getting minimum week number of month and deduct it from week number. Suppose you have a table with dates
select
emp_id, dt , datepart(wk,dt) - (select min(datepart(wk,dt))
from
workdates ) + 1 from workdates
Solution:
declare #dt datetime='2018-03-31 05:16:00.000'
IF (Select (DatePart(DAY,#dt)%7))>0
Select (DatePart(DAY,#dt)/7) +1
ELSE
Select (DatePart(DAY,#dt)/7)
declare #end_date datetime = '2019-02-28';
select datepart(week, #end_date) - datepart(week, convert(datetime, substring(convert(nvarchar, convert(datetime, #end_date), 127), 1, 8) + '01')) + 1 [Week of Month];
Here is the tried and tested solution for this query in any situation - like if 1st of the month is on Friday , then also this will work -
select (DATEPART(wk,#date_given)-DATEPART(wk,dateadd(d,1-day(#date_given),#date_given)))+1
above are some solutions which will fail if the month's first date is on Friday , then 4th will be 2nd week of the month
Logic here works as well 4.3 weeks in every month. Take that from the DATEPART(WEEK) on every month but January. Just another way of looking at things. This would also account for months where there is a 5th week
DECLARE #date VARCHAR(10)
SET #date = '7/27/2019'
SELECT CEILING(DATEPART(WEEK,#date)-((DATEPART(MONTH,#date)-1)*4.3333)) 'Week of Month'
Below will only work if you have every week of the month represented in the select list. Else the rank function will not work, but it is a good solution.
SELECT DENSE_RANK() OVER (PARTITION BY MONTH(DATEFIELD)
ORDER BY DATEPART(WEEK,DATEFIELD) ASC) AS WeekofMont
try this one
declare #date datetime = '20210928'
select convert(int,(((cast(datepart(day,#date) as decimal(4,2))/7)-(1.00/7.00))+1.00))
select datepart(week,#date)-datepart(week,dateadd(day,1,eomonth(dateadd(m,-1,#date))))+1
or
select datepart(week,#date)-datepart(week,dateadd(d,-datepart(d,#date)+1,#date))+1
steps:
1,get the first day of month
2,week of year of the date - week of year of the first day of the month
3,+1
2023-1-1 is sunday
SET DATEFIRST 1;
DECLARE #date date = '2023-1-02';
select datepart(week,#date)-datepart(week,dateadd(day,1,eomonth(dateadd(m,-1,#date))))+1
return 2
SET DATEFIRST 7;
DECLARE #date date = '2023-1-02';
select datepart(week,#date)-datepart(week,dateadd(day,1,eomonth(dateadd(m,-1,#date))))+1
return 1
Code is below:
set datefirst 7
declare #dt datetime='29/04/2016 00:00:00'
select (day(#dt)+datepart(WEEKDAY,dateadd(d,-day(#dt),#dt+1)))/7
select #DateCreated, DATEDIFF(WEEK, #DateCreated, GETDATE())
Am using Sql Server 2008, I have a column named Date in my table, and I want to get the datas for the particular date.... I need to give this Date in my WHERE condition.
for example, if I want to get the records for the particular month in the given date, how can I use this Date in WHERE condition.
DATANAME(MONTH,'#Date')
if I give like this in my query I can get the month from the given DATE, the same way I tried by putting in WHERE condition like,
WHERE DATE= DATANAME(MONTH,'#Date')
here it reports conversion error...how can I display the datas for a particular month, can anyone help me
If you want a month of data for a table you should check against an interval. The query is not able to use indexes on the date column if you are applying functions on the column.
Use something like this to get data for April 2012.
-- The date parameter
declare #Date datetime
set #Date = '2012-04-11'
declare #FromDate datetime
declare #ToDate datetime
-- set FromFate to first of april
set #FromDate = dateadd(month, datediff(month, 0, #Date), 0)
-- set ToDate to first of may
set #ToDate = dateadd(month, 1+datediff(month, 0, #Date), 0)
select *
from YourTable
where [Date] >= #FromDate and [Date] < #ToDate
If you want to show data for a particular year and month you can use the YEAR and MONTH functions:
SELECT ...
FROM ...
WHERE YEAR(mydate) = 2012 AND MONTH(mydate) = 3 -- March, 2012
To me it seems that your field Date is not of type varchar or nvarchar, so using a condition where a Datetime = string is obviously wrong.
Have you tried
WHERE DATE= #Date
Shouldn't it be:
DATENAME(MONTH, #Date)
Instead of:
DATANAME(MONTH,'#Date')
(Notice "DATA" vs "DATE" and #Date isn't in quotations)
Then to use this against a date/datetime column you would have to cast both sides like below:
WHERE datename(Month, [Date]) = datename(Month, [Date])
Warning: The above does not use any indexes so isn't as efficient as "WHERE Date = Date"
First: Remove '' from variable. #Date, not '#Date'
If you want to find dates from specific month. (You have to remember about year condition also)
WHERE DATANAME(MONTH, #Date) = 'April'
if you want to find exact date:
WHERE DATE = #date