Data Warehousing :: DateDiff Function To Return Positive Value Irrespective Of Values Passed

Aug 7, 2015

I have a requirement to use DateDiff(Months,DateTime1,DateTime2) and this must return positive integer values.

Currently negative numbers are being returned because DateTime1 < DateTime2  or DateTime1 > DateTime2 .

The DateTime1 and  DateTime2  values are dynamic and either of them can be bigger than the other.

Any query solution so that always positive value is returned.

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Using Datediff Function To Return 1st Of Month. Different Results With T-SQL And SSIS

May 15, 2007

Hi



I regularly use the T-SQL date functions to return the 1st of a particualr month.



e.g.



SELECT DATEADD(m,DATEDIFF(m,0,getdate()),0)



returns 2007-05-01 00:00:00.000



i.e the first of the current month at midnight.

However, when I try to use a similar expression as a derived column in SSIS it returns a completely different date.



DATEADD("month",DATEDIFF("month",(DT_DATE)0,GETDATE()),(DT_DATE)0)



returns 30/05/2007 00:00:00





Any ideas why and how I can obtain the first of a particualr month using SSIS derived column?





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DATEDIFF Function Dosen't Return Anything If Used Inside ASPX Page!!

Feb 12, 2008

Dear all, I have the Following Query:SELECT DATEDIFF(day, GETDATE(), RECENT_RESERVATION)AS Expr1,RECENT_RESERVATION FROM EMP WHERE SUN=empName;when i run it inside the query analyzer, it returns two columns. but if run it inside The ASPX page it retuns only one column wich is  RECENT_RESERVATION date.Note: i am using one methode that takes care of reading from SQL and assigning the result into an array, it works fine everywhere, but with this query it dosen't work. Any suggestions??   

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Return Multiple Values From A Function

Jun 19, 2007

searched all over, couldn't find a solid answer...is it possible to return multiple values from a sql function with sql server 2005?

e.g., I want to do this:

select id, data, whatever, dbo.fnMyFunction(id, whatever) from table

and have the output have columns (id, data, whatever, col1, col2) where col1 and col2 are returned by fnMyFunction

possible? easier way? thanks in advance...

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Function With Expression To Return Values

May 15, 2007

I have created a function to return values, which works fine, but I can't do calculations in it.

CREATE FUNCTION [dbo].[tf_Asset_Portfolio](@deal_id int,
@as_of_date datetime)
RETURNS TABLE
AS
RETURN ( SELECT DISTINCT dbo.Assets.issue_id, SUM(DISTINCT dbo.Assets.par_amount) AS par_amount, SUM(DISTINCT dbo.Assets.par_amount) AS market_value
FROM dbo.Issue INNER JOIN
dbo.Assets ON dbo.Issue.issue_id = dbo.Assets.issue_id INNER JOIN
dbo.Issuer_Rating_History ON dbo.Issue.issuer_id = dbo.Issuer_Rating_History.issuer_id
WHERE (dbo.Issuer_Rating_History.as_of_date <= @as_of_date)
GROUP BY ALL dbo.Assets.issue_id, dbo.Assets.deal_id, dbo.Issue.default_date
HAVING (dbo.Assets.deal_id = @deal_id) )

I need to do calculations on market value based on the default date.
If default date isn't specified then it should be 100% of par amount.
If default date is less than one year ago - 65% of the par_amount.
If default date is one or more years ago - 0.
I have no idea about how to do this and everything I try wont work.
I created another function to do the calculations and this seems to work, but it only does one record instead of all of them.

CREATE FUNCTION dbo.tf_Asset_Portfolio2
(@deal_id int,
@as_of_date datetime)
RETURNS @Market TABLE
(issue_id int, par_amount money, market_value money)
AS
BEGIN
DECLARE @ReturnDate datetime
DECLARE @DD datetime
DECLARE @PA money
DECLARE @MV money
DECLARE @ID int
DECLARE @DateD int

SELECT TOP 1
@ReturnDate = LAST_BATCH
FROM master..sysprocesses
WHERE SPId = @@SPID

SELECT @ID = issue_id FROM Assets WHERE Assets.deal_id = @deal_id
SELECT @PA = SUM(DISTINCT par_amount) FROM Assets WHERE Assets.issue_id = @ID AND Assets.deal_id = @deal_id
SELECT @DD = default_date FROM Issue WHERE Issue.issue_id = @ID

SET @DateD = DateDiff("yyyy", @DD, @ReturnDate)

If @DD = Null
BEGIN
SET @MV = @PA
END
Else If @DD > @ReturnDate
BEGIN
SET @MV = @PA
END
Else If @DateD < 1
BEGIN
SET @MV = @PA * .65
END
Else If @DateD >= 1
BEGIN
SET @MV = 0
END

insert into @Market
(issue_id, par_amount, market_value)
values
(@ID,@PA,@MV)

RETURN
END

I need to combine the functionality of being able to return mutliple records that isn't in the 2nd function and being able to calculate the market value which isn't in the first one. Please help. Thank you in advance.

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Fetch Data Of User Who Have Created Profile Within 7 Days - DateDiff Function

May 2, 2015

I have added one webpage designed in ASP.Net with C# and sql server 2005 as database. There is table for user registration in which there is a column for ProfileCreationDate the data type of that column is date time .

I would like to fetch data of those user who have created profile within 7 days. For getting desired result I am trying this query.

select Name ,Profession,ProfileCreationDate from tblRegistration where DATEDIFF ( Day , '" + System.DateTime.Now + "',ProfileCreationDate)<7 order by ProfileCreationDate DESC

System.DateTime.Now is a function for getting current date time in C#

The query is neither giving error nor giving desired result.

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How To Return Multiple Table Values From A Function

May 5, 2004

Hi,
We Have Been Trying To Convert Some Pf The Procs Into Functions Of Late,but There Is A Problem :-we Have Been Unable To Return More Than 1 Table Value From A Function.

Create Function F_clusters()
Returns @ki Table(names Nvarchar(200),total Int),
As
Begin
Insert @ki
Select Names,count(distinct Chremail) As From Customer
Where Chremail Is Not Null
Return
End

This Works Fine :-
And Gives The Reqd. Results.

But,

If I Am Using The Same Function To Return Two Tables Then It Doesn't Work,could You Pls Chk.



Create Function F_clusters()
Returns @ki Table(names Nvarchar(200),total Int),@k2 Table(names Nvarchar(200),total Int)
As
Begin
Declare @cnt Int
Set @cnt = 1
While @cnt <= 2
If @cnt =1
Begin
Insert @ki
Select Names,count(distinct Chremail) As From Customer
Where Chremail Is Not Null
Set @cnt = @cnt + 1
End
If @cnt =2
Begin
Insert @k2
Select @naamre,count(distinct(a.intcustomerid)) As Pura_ginti From Trcustomerpreference03july A Inner Join Cleancustomer B
On A.intcustomerid = B.intcustomerid
Where Chremail <> ' ' And Chremail Is Not Null
And Intpreferenceid In (6,7,2,3,12,10)
Set @cnt2 = @cnt2 + 1
End
End
Return
End


Can We Return Two Tables Or Is It Not Possible ?
Pls Chk Into This And Tell Me.

Thanks.

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Positive Negative Values In SUM

Apr 25, 2008





Hi,


I have table with a column which has both negative and positive values.

I want to calculate the sum of positve values and the sum of negative values from that column separately. Is it possible.

Plase help me.


Ex: My Input Values are

column1
-186.5499257
-141.8985222
-95.3846883
-71.1359768
-69.39071172
30.44750844
40.33666032
53.81527872
98.2967252
112.948667

OutPut goes like This

Column1 Column2
-186.5499257 null
-141.8985222 null
-95.3846883 null
-71.1359768 null
-69.39071172 -564.36
30.44750844 null
40.33666032 null
53.81527872 null
98.2967252 null
112.948667 335.84

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May 12, 2015

I've to write a function to return a comma delimited values from a table columns

If a table has Tab1 ( Col1,Col2,Col3).

E.g. as below ( the columnName content I want to use as columns for my pivot table

CREATE FUNCTION [RPT].[GetListOfCol]
(
@vCat NVARCHAR(4000)
)
RETURNS @PList
AS
BEGIN
SELECT @PList += N', [' + [ColumnName] +']'
FROM [ETL].[TableDef]
WHERE [IsActive] = 1
AND [Category] = @vCat
RETURN;
END;

I want out put to be as below, I am getting this output from the select query independently by declaring @Plist variable and passing @vcat value, now I want it to be returned from a function when called from a select query output ,Colum1,column2,....

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Jun 29, 2015

I have tables and a function as representated by the code below. The names  for objects here are just for representation and not the actual names of objects. Table RDTEST may have one or multiple values for RD for each PID. So the function GIVERD will return one or multiple values of RD for each value of PID passed to it.

When I run the following query, I get the required result except the rows for CID 500 for which PID is NULL in table T1. I want the rows for CID 500 as well with PID values as NULL.

SELECT  A.CID, 
A.ANI,
A.PID,
B.RD
FROM T1 AS A CROSS APPLY GIVERD(A.PID) B

CREATE TABLE [DBO].[RDTEST](
[PID] [INT] NULL,
[RD] [INT] NULL
)

[Code] ....

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Feb 3, 2012

How do I count the number of positive dollar values in a query? I have various negative and positive dollar values and want to count how many positive I have then I will know how many are negative. I cannot put >0 or >0.00 in the where because it still returns everything cuz the programmer set the datatype to be money

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Oct 29, 2007

I have a table that displays a bunch of data based on their percentage changes.

There is a parameter (let's say 5%) that the user can set to display only the data which either increased more than 5% or decreased more than 5%.

I want to have a space between the positive changes and the negative changes. ie, to create a blank row between the +'s and the -'s. Does anyone know of a way to do this?

Thanks,
Steven

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Aug 19, 2015

I have a stored procedure that selects the unique Name of an item from one table. 

SELECT DISTINCT ChainName from Chains

For each ChainName, there exists 0 or more StoreNames in the Stores. I want to return the result of this select as the second field in each row of the result set.

SELECT DISTINCT StoreName FROM Stores WHERE Stores.ChainName = ChainName

Each row of the result set returned by the stored procedure would contain:

ChainName, Array of StoreNames (or comma separated strings or whatever)

How can I code a stored procedure to do this?

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Sep 18, 2015

MS SQL Server 2008 R2

I have both positive and negative values in a single column, where I want sum total of positive values & negative values. Is there any Expression for this to sort out.

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Feb 14, 2006

I just discover the result for my query that had use the function dateDiff seems to be in-correct, no matter what's the date, the dateDiff always return a zero. Am I using it in-correctly?
select DateDiff(mm,11/1/2004, 12/31/2005)
SELECT component_id, component_description,SUM(CASE DateDiff(mm,date_complete,'12/31/2004')WHEN 2 THEN component_qty ELSE 0 END) AS mm1,SUM(CASE DateDiff(mm,date_complete,'12/31/2004')WHEN 1 THEN component_qty ELSE 0 END) AS mm2,SUM(CASE DateDiff(mm,date_complete,'12/31/2004')WHEN 0 THEN component_qty ELSE 0 END) AS mm3,sum(component_qty) as totalFROM view_jobcomponent WHERE date_complete between '10/1/2004' and '12/31/2004' GROUP BY component_id, component_descriptionorder by component_id, component_description

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Jun 12, 2008

Hi GUYS

I am trying to use the datediff function to return a Y if difference is greater than 0 and a N if difference is less than or equal to 0

select if datediff(dd,maturity_date,log_changed) > 0 'Y','N')

and i am getting a syntax error.Can someone show me where i am getting it wrong.

In god we trust,everything else we test.

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Jun 24, 2004

I've been trying to create a function that returns the difference, in months, between two dates. The DateDiff function (m) returns an integer, but I really need a decimal. I could return the days instead, but it would be difficult to figure out how the number of months from this, especially when the dates span multiple years and I need to adjust for leap year. Does anyone know of a resource that might have a solution for this?

Thanks,
Rob

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Aug 30, 2006

I have a query and am trying to just return the difference between two dates but not include weekends.For instance, if I have 08/21/2006 - 08/28/2006, there are 6 weekdays.  I tried this, but I am getting 7 as a result.SELECT DATEDIFF(weekday, request_start_date, request_end_date) AS days_off, request_id
FROM request Any help would be greatly appreciated.

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Nov 23, 2006

Hi, I am facing problem rite now.. I want to calculate the date different minutes between 23:00:00 and 01:00:00.
My code :
datediff(Minute,'01:00:00','23:00:00')
The result is 1320 minutes. (22 hours)... But, the result that I want is 120 minutes (2 hours)....
Can anybody help ???
Thanks in advance...

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Nov 12, 2007

Hi,I'm using the datediff function to display the ages of the users in my database. However the age rounds up once they are 35.5 etc...I could create another function which works similar to the DateDiff function, but use math.floor to always round down, but I need to use this function in a SQL statement WHERE clause. Is there any way around this?Thanks,Curt. 

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Datediff Function

Apr 25, 2007

hi
select datediff(m,'3/5/2003',getdate as experience

output is
Experience
----------
49

select (datediff(m,'3/5/2003',getdate()))/12 as experience

output is
Experience
----------
4

but 49/12 is 4.08333=4.1

how i can get this
Actually my task is to output as years.months

Thanks in advance


Malathi Rao

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Please Help With Datediff Function

Oct 5, 2006

I am trying to break up the age into column from a dob field for a cross tab report. I can query the datediff with an alias but can not individually change the columns I need like a virtual or temp column, but can't figure out how to accomplish it.

use 'Database'

Select datediff(Year, dob, getdate()) As "Under 22", datediff(Year, dob, getdate()) As "22-45",

datediff(Year, dob, getdate()) As "46-65", datediff(Year, dob, getdate()) As "Over 65"

from 'Table'

WHERE (DATEDIFF(yy, dob, GETDATE()) < 22 OR

DATEDIFF(yy, dob, GETDATE()) BETWEEN 22 AND 45 OR

DATEDIFF(yy, dob, GETDATE()) BETWEEN 46 AND 65 OR

DATEDIFF(yy, dob, GETDATE()) > 66) AND (status = 'Current' OR

status = 'Previous' OR

status = 'non-billable')

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Nov 3, 2004

I have a generic page with some fields on it, I would like to be able to "groom" the values so that the data is nice and consistent.

Example: input 1 -- user types in "bob" or "BOB" or "Bob" or "JOE" etc....

the resulting record in Sql would be "Bob", "Fred", "Joe"
nice and formted out.

Any suggestions?

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A More Precise DateDiff Function

Feb 7, 2007

This will give you the time information about how apart two dates are.CREATE FUNCTION dbo.fnTimeApart
(
@FromTime DATETIME,
@ToTime DATETIME
)
RETURNS @Time TABLE ([year] SMALLINT, [month] TINYINT, [day] TINYINT, [hour] TINYINT, [minute] TINYINT, [second] TINYINT, [millisecond] SMALLINT)
AS
BEGIN
DECLARE@Temp DATETIME,
@Mts INT,
@year SMALLINT,
@month TINYINT,
@day TINYINT,
@hour TINYINT,
@minute TINYINT,
@second TINYINT,
@millisecond SMALLINT

IF @FromTime > @ToTime
SELECT@Temp = @FromTime,
@FromTime = @ToTime,
@ToTime = @Temp

SET@Mts =CASE
WHEN DATEPART(day, @FromTime) <= DATEPART(day, @ToTime) THEN 0
ELSE -1
END + DATEDIFF(month, @FromTime, @ToTime)

SELECT@year = @Mts / 12,
@month = @Mts % 12,
@Temp = DATEADD(month, @Mts, @FromTime)

SELECT@day = datediff(hour, @Temp, @ToTime) / 24,
@Temp = DATEADD(day, @day, @Temp)

SELECT@hour = DATEDIFF(minute, @Temp, @ToTime) / 60,
@Temp = DATEADD(hour, @hour, @Temp)

SELECT@minute = DATEDIFF(second, @Temp, @ToTime) / 60,
@Temp = DATEADD(minute, @minute, @Temp)

SELECT@second = DATEDIFF(millisecond, @Temp, @ToTime) / 1000,
@Temp = DATEADD(second, @second, @Temp),
@millisecond = DATEDIFF(millisecond, @Temp, @ToTime)

INSERT@Time (year, month, day, hour, minute, second, millisecond)
SELECT@year,
@month,
@day,
@hour,
@minute,
@second,
@millisecond

RETURN
ENDAnd to test the functionSELECTd.FromDate,
d.ToDate,
x.*
FROM(
SELECT'19690906' AS FromDate, '19760608' AS ToDate UNION ALL
SELECT'19991231', '20000101' UNION ALL
SELECT'20070207', '20070208' UNION ALL
SELECT'20000131', '20000228' UNION ALL
SELECT'20070202', '20070201' UNION ALL
SELECT'20070207', '20070307' UNION ALL
SELECT'20000131', '20000301' UNION ALL
SELECT'20011231 15:24:13.080', '20020101 17:15:56.343' UNION ALL
SELECT'20011231 17:15:56.343', '20020101 15:24:13.080' UNION ALL
SELECT'20020101 15:24:13.080', '20011231 17:15:56.343' UNION ALL
SELECT'20000131', '20000229'
) AS d
CROSS APPLYdbo.fnTimeApart(d.FromDate, d.ToDate) AS x
ORDER BYd.FromDate,
d.ToDateAnd the output isFromDateToDateyearmonthdayhourminutesecondmillisecond
19690906197606086920000
19991231200001010010000
200001312000022800280000
200001312000022900290000
20000131200003010110000
20011231 15:24:13.08020020101 17:15:56.34300115143263
20011231 17:15:56.34320020101 15:24:13.08000022816736
20020101 15:24:13.08020011231 17:15:56.34300022816736
20070202200702010010000
20070207200702080010000
20070207200703070100000
Peter Larsson
Helsingborg, Sweden

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Feb 9, 2007

I have two date columns one is sent_date and other is approved_date
my requirment is to find the difference between the two dates
which can be minutes/hrs/days.
using datediff function iam able to get it in minusts or hrs but my
output should be of the format hh:mm
23:10 (ie 23 hrs and 10 min) or say
48:00 (for 2 days)

sample date
sent_date approved_date
2/28/06 11:06 2/28/06 11:39
2/2/06 17:42 2/2/06 18:03
2/8/06 16:55 2/8/06 17:38
1/27/06 17:00 1/27/06 17:54
1/26/06 12:08 1/26/06 12:09
2/28/06 15:46 2/28/06 16:26
1/23/06 10:01 1/23/06 10:43
1/26/06 13:46 1/26/06 13:59
1/13/06 13:51 1/13/06 14:47

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Nov 2, 2007

recently i created a system using php and mysql to record faults that the ict technicians could get reported faults out of. so far it does everything i want but some senior members of staff now want the front end to give a report of status, one of the things they wish to see is how many jobs have been completed this week month and year

within my tables is a field date. naturally it has the date submited contained within it. the code i have been using to try and get an output is

mysql_query("SELECT date, datediff('day', date, CURRENTDATE) FROM softwarerepairtable WHERE datediff<= '7' and complete='yes' ");

the complete feild is another contained within the datebase. i have also tried SELECT * datediff('day', date, CURRENTDATE) FROM softwarerepairtable WHERE datediff<= '7' and complete='yes' ");

unfortunatly my limited knowledge of php and sql is holding me back on this. any help guys would be appriacted

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Query On Datediff Function

May 18, 2007

Hi,



I am trying to use DataDiff function and I have used the following queries:



1.

select datediff(ms, '2007-05-18 19:35:07.320','2007-05-18 19:35:07.320') as test

Expected Result: 0 milliseconds

Actual Result: 0 milliseconds



2.

select datediff(ms, '2007-05-18 19:35:07.320','2007-05-18 19:35:07.321') as test

Expected Result: 1 milliseconds

Actual Result: 0 milliseconds



3.

select datediff(ms, '2007-05-18 19:35:07.320','2007-05-18 19:35:07.322') as test

Expected Result: 2 milliseconds

Actual Result: 3 milliseconds



4.

select datediff(ms, '2007-05-18 19:35:07.320','2007-05-18 19:35:07.323') as test

Expected Result: 3 milliseconds

Actual Result: 3 milliseconds



5.

select datediff(ms, '2007-05-18 19:35:07.320','2007-05-18 19:35:07.324') as test

Expected Result: 4 milliseconds

Actual Result: 3 milliseconds



6.

select datediff(ms, '2007-05-18 19:35:07.320','2007-05-18 19:35:07.325') as test

Expected Result: 5 milliseconds

Actual Result: 6 milliseconds



7.

select datediff(ms, '2007-05-18 19:35:07.320','2007-05-18 19:35:07.326') as test

Expected Result: 6 milliseconds

Actual Result: 6 milliseconds



8.

select datediff(ms, '2007-05-18 19:35:07.320','2007-05-18 19:35:07.327') as test

Expected Result: 7 milliseconds

Actual Result: 6 milliseconds



9.

select datediff(ms, '2007-05-18 19:35:07.320','2007-05-18 19:35:07.328') as test

Expected Result: 8 milliseconds

Actual Result: 6 milliseconds



10.

select datediff(ms, '2007-05-18 19:35:07.320','2007-05-18 19:35:07.329') as test

Expected Result: 9 milliseconds

Actual Result: 10 milliseconds



Does any one know, why datediff does not return the Expected Result? There does not seem to be any consistency.



Thanks,

Tim

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Jan 25, 2008

I want to create and expression that basically says if a datetime value is today minus 2 days then do another formula.

Right now it looks like this...




Code Snippet
=iif(Fields!Channel_Id.Value="SFE1" and Fields!Report_Date.Value=Today-2,0,(Fields!Total_Apps.Value-Fields!total_apps_null_status.Value))





Basically, it spits an error saying [BC30452] Operator '-' is not defined for types 'Date' and 'Integer'.

Should this be done with DateDiff? if so, how can you specify two days ago?

Thanks much

C-

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However, I don't know how to tell if the date value in the function came from an input or was the default.

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Oct 17, 2007


Greetings, all!

I had a question about how SRS passes multiple values to a query.

For example:

Say I had a "select box" (Sorry, can't think of the proper term right now) with the following values:

F1
F2
F3
F4

The user then selects F1 and F4.

When SRS passes the selected values to the query or procedure, how are they passed? As a delimited value?

Examples of what I mean:

@Parameter = 'F1', 'F4'
@Parameter = F1 [Some unknown delimiting symbol] F4
@Parameter = 'F1,F4'

Does anyone know how this is done? Or is it done in some other odd way?

Thanks!

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Feb 3, 2015

I'm trying to use the DateDiff function within my select statement, but I'd like to add the parameter of greater than 30 days. This will have the query only return records where my bill stop date is greater than 30 days from the completion date. I currently have the datediff function within my select statement as

DATEDIFF (d,A.StopBillDate, a.CompletionDate) as [DIFFERENCE]

I would prefer to keep the datediff function within the select statement so as to have difference in days appear as a column within my output.I have been unable to add the parameter of > 30 days to the query without getting an error.

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Apr 9, 2007

Hi Experts,I am working on SSRS 2005, and I am facing a problem in counting theno of days.My database has many fields but here I am using only two fieldsThey are Placement_Date and Discharge_DateIf child is not descharged then Discharge_Date field is empty.I am writing below query to count the number of days but is is notworking it is showing the error"The conversion of a char data type to a datetime data type resultedin an out-of-range datetime value."select casewhen convert(datetime,Discharge_Date,103) = '' thendatediff(day,CONVERT(datetime,Placement_Date,103), GETDATE())elsedatediff(day,CONVERT(datetime,Placement_Date,103),CONVERT(datetime,Discharge_Date,103))end NoOfDaysfrom Placement_DetailsSo please tell me where I am wrong?Any help will be appriciated.RegardsDinesh

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