Help Needed For Datediff Function For SQL Query

Apr 9, 2007

Hi Experts,
I am working on SSRS 2005, and I am facing a problem in counting the
no of days.
My database has many fields but here I am using only two fields
They are Placement_Date and Discharge_Date
If child is not descharged then Discharge_Date field is empty.

I am writing below query to count the number of days but is is not
working it is showing the error
"The conversion of a char data type to a datetime data type resulted
in an out-of-range datetime value."

select case
when convert(datetime,Discharge_Date,103) = '' then
datediff(day,CONVERT(datetime,Placement_Date,103), GETDATE())
else
datediff(day,CONVERT(datetime,Placement_Date,
103),CONVERT(datetime,Discharge_Date,103))
end NoOfDays
from Placement_Details
So please tell me where I am wrong?


Any help will be appriciated.
Regards
Dinesh

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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 5, 2007

I am trying to figure out how much coverage (supervision time) an employee has based on their managers schedule.

Example
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Name: Manager, SundayStart: 8 AM, SundayEnd: 4 PM


############# 8 AM |Covered Time| 5 PM
Employee ##########|------------|-----
Manger ######------|------------|


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Anyway to figure this out? This is what I have so far. Sorry, but im in access 03.


SELECT tblCAESchedule.Name, tblCAESchedule.Title, tblCAESchedule.SunStart, tblCAESchedule.SunEnd,
IIf(DateDiff("n",[tblCAESchedule].[sunstart],[tblCAESchedule].[sunend])>=1,DateDiff("n",[tblCAESchedule].[sunstart],[tblCAESchedule].[sunend]),DateDiff("n",[tblCAESchedule].[sunstart],[tblCAESchedule].[sunend])+1440)/60 AS SunHours,
tblSupSchedule.Name, tblSupSchedule.SunStart, tblSupSchedule.SunEnd,
Abs(DateDiff("n",[tblCAESchedule].[sunstart],[tblSupSchedule].[sunstart])/60) AS StartDif,
Abs(DateDiff("n",[tblCAESchedule].[sunend],[tblSupSchedule].[sunend])/60) AS EndDif,
Abs(DateDiff("n",[tblCAESchedule].[sunstart],[tblSupSchedule].[sunstart])/60)+Abs(DateDiff("n",[tblCAESchedule].[sunend],[tblSupSchedule].[sunend])/60) AS ShiftDif
FROM tblCAESchedule INNER JOIN tblSupSchedule ON tblCAESchedule.Skill = tblSupSchedule.Skill
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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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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
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4

but 49/12 is 4.08333=4.1

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

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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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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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sample date
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1/27/06 17:00 1/27/06 17:54
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Code Snippet
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563136    199535840A    D2119562408    199615461C    D2119562806    199748610A    D2119547463    199748610A    D2368562278    200255598A    D2368562286    200255598A    D2468
Field2 can have the same value.
What I need is to return all records where field3 is not 'D2468' ever for field2.  So with the above data, I'd want the first 4 records returned, but not 5 and 6 because for those field2 is the same value and in 6, field3 is 'D2468'.
So, I can't simply say:
SELECT a.field2FROM table1 a inner join table2 b on a.field2 = b.field2    and a.field5 = b.field5where a.field3 not like 'D2468'
because it will still return record 5. 
Can anyone help me with this?  Thank you!
 

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