SSRS 2008 - returning a date 3 months from today from a list of dates - reporting-services

I have a Report Item that contains dates, that I want to compare against today's date, if it's within 3 months or today then either return that date or return a 'Yes'.
Is there a simple way of carrying this out?


Use DateDiff to Get the difference Between two dates
IF DATEDIFF( MONTH,#ReportDate, Getdate())<=3
BEGIN
SELECT #ReportDate as ReportDate
END
ELSE
BEGIN
SELECT "Yes"
END


Here's a method you can use in your query:
DECLARE #Date DATE = '2018-08-01'
SELECT CASE WHEN DATEADD(MONTH,3,GETDATE()) >= #Date
THEN CAST(GETDATE() AS VARCHAR) ELSE 'Yes'
END AS ColumnName
SSRS expression might look something like this:
=IIF("2018-05-01" > DATEADD(DateInterval.Month, 3, Today), "Yes", Today)
Where you would obviously replace 2018-05-01 with your date field.

Related

In Case statement update date not greater than today date

Hi as per current statement i am updating the date in table as per below statement. I want to update the date that will be not more than today date. If the date will be more than today date then it will update with today date.
Update visiting
set OriginArrvDate
WHEN DATEPART(weekday,ORDDATE)=6 THEN DATEADD(dd,3,ORDDATE)
WHEN DATEPART(weekday,ORDDATE)=7 THEN DATEADD(dd,3,ORDDATE)
WHEN DATEPART(weekday,ORDDATE)=1 THEN DATEADD(dd,2,ORDDATE)
WHEN DATEPART(weekday,ORDDATE)=2 THEN DATEADD(dd,1,ORDDATE)
ELSE ORDDATE END
where
orddate>=CONVERT(DATE,DATEADD(dd,-100,getdate()))
In above script i am trying that when after calculating the date in ( DATEADD(dd,3,ORDDATE) ) if output will be greater than today date then it will be update with today date.
Example : ORDATE is '2021-09-26' so as per current condition ORDDATE will be 2021-09-29 but i want the result after calculating the date not more than today date. So the date will be ' 2021-09-27'
If the ORDATE calculation below today date then it will be fine and not change in date required.
Thanks for your reply
UPDATE : -
As per suggestion i have add data with Script on sqlfiddle . If you review the data for visiting table for Oid 1 and 5 . Both updated data for OriginArrvDate is more than today date. I want to update the OriginArrvDate with today date when new ordDate is more than current date.

I would use a CTE which makes it very easy to test your logic by applying a CASE expression (not a CASE statement) to the output of your existing CASE expression.
;WITH x AS
(
SELECT Oid, ORDDATE, OriginArrvDate,
CalcDate = CASE DATEPART(weekday, ORDDATE)
WHEN 6 THEN DATEADD(DAY, 3, ORDDATE)
WHEN 7 THEN DATEADD(DAY, 3, ORDDATE)
WHEN 1 THEN DATEADD(DAY, 2, ORDDATE)
WHEN 2 THEN DATEADD(DAY, 1, ORDDATE)
ELSE ORDDATE END
FROM dbo.visiting
WHERE orddate >= CONVERT(DATE,DATEADD(DAY, -100, getdate()))
AND OriginArrvDate IS NULL
)
--SELECT *,
/* -- */ UPDATE x SET
OriginArrvDate = CASE
WHEN CalcDate > GETDATE() THEN GETDATE()
ELSE CalcDate END
--FROM x;
Thanks for the fiddle, but I wrote a db<>fiddle here because I couldn't get SQLFiddle to properly show the update.
In addition to using DAY instead of dd I would also make sure to always specify a table's schema.

How to get 'Monday Date' in MySQL?

I have an Oracle SQL Query, which I am trying to re-write in MySQL.
PS : The dates are just arbitrary here, in the actual scenario I use this custom query inside Tableau where it accepts user defined Dates.The "Reference Start Date" is being compared to "Start Date".
Part of the Oracle SQL query:
CASE WHEN psr.dt_orgn IS NULL THEN NULL
ELSE
CASE WHEN CAST('2017-05-22' AS DATE) >= CAST('2017-06-22' AS DATE) THEN TRUNC(psr.dt_orgn,'IW')
ELSE TRUNC(psr.dt_orgn + ((CAST('2017-06-22' AS DATE)-CAST('2017-05-22' AS DATE))*(INTERVAL '1' DAY)),'IW')
END
END) week
The above uses an existing Calender table in a Database to convert dates into 'Week Start Dates' i.e Monday Date of that Week
MySQL version:
CASE WHEN psr.originated IS NULL THEN NULL
ELSE
CASE WHEN CAST('2017-05-22' AS DATE) >= CAST('2017-06-22' AS DATE) THEN DATE_ADD(psr.originated,
INTERVAL - WEEKDAY(psr.originated) DAY)
ELSE DATE_ADD(psr.originated + ((CAST('2017-06-22'AS DATE) - CAST('2017-05-22' AS DATE)) * (INTERVAL '1' DAY))
END
END) week
I am using a solution I found to get 'Week Start Date' in MySQL as follows:
DATE_ADD(mydate, INTERVAL(1-DAYOFWEEK(mydate)) DAY)
And I am getting error while doing this.
My question is how do I apply the same to the 2nd else condition with the '1 day Interval' in the MySQL query?
Or is there an easier solution to get week start date in MySQL?

I guess you're asking how to take any arbitrary DATE value and return the DATE value of the most recent Monday. In Oracle you can use TRUNC(datestamp, 'W') to do that.
Here's one way to do it in MySQL:
FROM_DAYS(TO_DAYS(datestamp) -MOD(TO_DAYS(datestamp) -2, 7))
If you wanted the Sunday, you would use
FROM_DAYS(TO_DAYS(datestamp) -MOD(TO_DAYS(datestamp) -1, 7))
I wrote it up here. http://www.plumislandmedia.net/mysql/sql-reporting-time-intervals/

You can get the day of a date using the function on MySQL called dayname(), for example dayname("2006-04-24") will return you Monday. In this case you only need to put the row that contains your date into the function. After that you can easily compare two dates.

Count first day of month between two dates

Could someone advise how I can go about counting the number of first day of month between two dates?
For example: 02/01/2015 to 05/05/2015 will count as 4.

It can be accomplished easily like this:
DECLARE #sd DATE = '02/02/2015', #ed DATE = '05/01/2015'
SELECT DATEDIFF(mm, #sd, #ed) + CASE WHEN DAY(#sd) = 1 THEN 1 ELSE 0 END

If you have a sequence of dates stored somewhere (like in a calendar table) you can easily count the dates matching the day(date) predicate. If you don't have any suitable table with a date sequence you can use a recursive common table expression to generate one on the fly like this:
declare #start date
set #start = '02/01/2015'
declare #end date
set #end = '05/05/2015'
;with dates (date) as (
select #start date
union all
select dateadd(day, 1, date) as date from dates where date < #end
)
select count(*) no_of_firsts from dates where day(date) = 1
Sample SQL Fiddle

How to get week number of the month from the date in sql server 2008

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())

How to use DATENAME in WHERE clause

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