Slet records der er mere en 5 dage gamle

Du skal kigge på funktionen datediff.

Smid GetDate() og din dato (RegTime???) ind, og hvis resulatet er "> 5", så slet.

Hvis du ikke lige har booksonline i nærheden, kommer her et udsnit:

DATEDIFF
Returns the number of date and time boundaries crossed between two specified dates.

Syntax
DATEDIFF ( datepart , startdate , enddate )

Arguments
datepart

Is the parameter that specifies on which part of the date to calculate the difference. The table lists dateparts and abbreviations recognized by Microsoft® SQL Server™.

Datepart Abbreviations
Year yy, yyyy
quarter qq, q
Month mm, m
dayofyear dy, y
Day dd, d
Week wk, ww
Hour hh
minute mi, n
second ss, s
millisecond ms


startdate

Is the beginning date for the calculation. startdate is an expression that returns a datetime or smalldatetime value, or a character string in a date format.

Because smalldatetime is accurate only to the minute, when a smalldatetime value is used, seconds and milliseconds are always 0.

If you specify only the last two digits of the year, values less than or equal to the last two digits of the value of the two digit year cutoff configuration option are in the same century as the cutoff year. Values greater than the last two digits of the value of this option are in the century that precedes the cutoff year. For example, if the two digit year cutoff is 2049 (default), 49 is interpreted as 2049 and 2050 is interpreted as 1950. To avoid ambiguity, use four-digit years.

For more information about specifying time values, see Time Formats. For more information about specifying dates, see datetime and smalldatetime.

enddate

Is the ending date for the calculation. enddate is an expression that returns a datetime or smalldatetime value, or a character string in a date format.

Return Types
integer

Remarks
startdate is subtracted from enddate. If startdate is later than enddate, a negative value is returned.

DATEDIFF produces an error if the result is out of range for integer values. For milliseconds, the maximum number is 24 days, 20 hours, 31 minutes and 23.647 seconds. For seconds, the maximum number is 68 years.

The method of counting crossed boundaries such as minutes, seconds, and milliseconds makes the result given by DATEDIFF consistent across all data types. The result is a signed integer value equal to the number of datepart boundaries crossed between the first and second date. For example, the number of weeks between Sunday, January 4, and Sunday, January 11, is 1.

Examples
This example determines the difference in days between the current date and the publication date for titles in the pubs database.

USE pubs
GO
SELECT DATEDIFF(day, pubdate, getdate()) AS no_of_days
FROM titles
GO

dfens løsning er ikke den optimale rent performance mæssigt.
Løsningen beregner forskellen på hver enkelt række i databasen.

Dit eget forslag er bedre. Der beregner du tidspunktet for 5 dage siden og sammenligner det med indholdet i databasen. Det vil sige at det kan udnytte evt. indexes du har på kolonnen. Måske er det lidt flueknepperi, da det alligevel sikkert er sletningen der tager tid.

hej jeg er selv kommet frem til følgende

DELETE FROM Tabel
WHERE RegTime <= (SELECT DATEADD(day,-5,getdate())

Selvfølgelig....

runesoft har ret.

"DELETE FROM Tabel WHERE RegTime <= getdate()-5 "

ville gøre det samme...Jeg ved ikke rent performance-mæssigt om der er forskel, men det ser pænere ud...

knowit-mmp->  Udseende er en smagssag :)