This algorithm requires an existing Prime numbers table. You can easily create one of your own or importing the primes ranging from 2 to 3,037,000,493 from the Internet. If you only is interested in primes with INT range {2.
Read more →
Late this evening I stumbled across a post where OP wanted to get all IDENTITY values for the latest INSERT statement, and correctly OP stated that SCOPE_IDENTITY() only returns last IDENTITY value of the batch.
Read more →
Since this blog post http://weblogs.sqlteam.com/peterl/archive/2007/09/24/Finding-group-of-records-with-a-certain-status.aspx I have kept the technique in similar situations and perhaps never bothered to rethink my prerogative. Well, today I have. Recently I blogged here http://weblogs.
Read more →
Here http://weblogs.sqlteam.com/peterl/archive/2009/07/12/How-to-calculate-the-number-of-weekdays-in-a-month.aspx is a specialized version for a month, and here http://weblogs.sqlteam.com/peterl/archive/2009/07/12/How-to-calculate-number-of-weekdays-in-a-year.aspx is a specialized version for a year. This code below calculates the number of weekdays for any given period.
Read more →
Here http://weblogs.sqlteam.com/peterl/archive/2009/07/12/How-to-calculate-the-number-of-weekdays-for-any-given.aspx you can find a specialized version that calculates number of weekdays for any given period. CREATEFUNCTION dbo.fnMonthWeekDays ( @Year SMALLINT, @Month TINYINT ) RETURNSTINYINT AS BEGIN RETURN ( SELECT 20 + SUM(1 - DATEDIFF(DAY, '17530101', dt) % 7 / 5) FROM ( SELECT DATEADD(MONTH, 12 * @Year - 22800 + @Month, -1) AS dt WHERE ISDATE(10000 * @Year + 100 * @Month + 31) = 1 UNION ALL SELECT DATEADD(MONTH, 12 * @Year - 22800 + @Month, -2) WHERE ISDATE(10000 * @Year + 100 * @Month + 30) = 1 UNION ALL SELECT CASE ISDATE(10000 * @Year + 100 * @Month + 29) WHEN 1 THEN DATEADD(MONTH, 12 * @Year - 22800 + @Month, -3) ELSE '18991231' END ) AS d WHERE @Year BETWEEN 1753 AND 9999 AND@Month BETWEEN 1 AND 12 ) END
Legacy Comments
Sowmya
2009-10-22
re: How to calculate the number of weekdays in a month Can u please explain the logic ?
Read more →
This function calculates the number of weeksdays in a year, and has error-checking for invalid years. It is also language independant. Here http://weblogs.sqlteam.com/peterl/archive/2009/07/12/How-to-calculate-the-number-of-weekdays-for-any-given.aspx you can find a specialized version that calculates number of weekdays for any given period.
Read more →
It seems JDEdwards dates are stored NUMERIC(6, 0) in this format 107299, where 107 is the number of years after 1900, and 299 is the 299th day of the year. So how to convert JDEdwards to DATETIME?
Read more →
Recently I posted a function which returned the Nth weekday of a month, either from the beginning of month or from the end of month. Function is found here http://weblogs.sqlteam.com/peterl/archive/2009/06/17/How-to-get-the-Nth-weekday-of-a-month.aspx I have fiddled around with it and have now extended the function to find the Nth weekday not only for a month, but also for a quarter or a year.
Read more →
You call this function with three parameters: 1. Any date of the month in question 2. The weekday to calculate; Monday 1, Tuesday 2, Wednesday 3, Thursday 4, Friday 5, Saturday 6 and Sunday 7 3.
Read more →
I did some tests today to measure the different approaches for finding records present in one table but not in another. The results of CPU and Duration are presented below with some help from SQL Profiler.
Read more →
CREATEFUNCTION dbo.fnCheckDanSSN ( @SSN CHAR(10) ) RETURNSBIT AS BEGIN IF @SSN NOT LIKE '[0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9]' RETURN 0 DECLARE @x CHAR(6) SET @x = SUBSTRING(@SSN, 5, 2) + SUBSTRING(@SSN, 3, 2) + SUBSTRING(@SSN, 1, 2) IF SUBSTRING(@SSN, 7, 1) IN('0', '1', '2', '3') AND ISDATE('19' + @x) = 0 RETURN 0 IF SUBSTRING(@SSN, 7, 1) IN('4', '9') SUBSTRING(@SSN, 5, 2) BETWEEN '00' AND '36' AND ISDATE('20' + @x) = 0 RETURN 0 IF SUBSTRING(@SSN, 7, 1) IN('4', '9') SUBSTRING(@SSN, 5, 2) BETWEEN '37' AND '99' AND ISDATE('19' + @x) = 0 RETURN 0 IF SUBSTRING(@SSN, 7, 1) IN('5', '6', '7', '8') SUBSTRING(@SSN, 5, 2) BETWEEN '00' AND '57' AND ISDATE('20' + @x) = 0 RETURN 0 IF SUBSTRING(@SSN, 7, 1) IN('5', '6', '7', '8') SUBSTRING(@SSN, 5, 2) BETWEEN '58' AND '99' AND ISDATE('18' + @x) = 0 RETURN 0 DECLARE @Digits INT SET @Digits = 4 * SUBSTRING(@SSN, 1, 1) + 3 * SUBSTRING(@SSN, 2, 2) + 2 * SUBSTRING(@SSN, 3, 2) + 7 * SUBSTRING(@SSN, 4, 2) + 6 * SUBSTRING(@SSN, 5, 2) + 5 * SUBSTRING(@SSN, 6, 2) + 4 * SUBSTRING(@SSN, 7, 2) + 3 * SUBSTRING(@SSN, 8, 2) + 2 * SUBSTRING(@SSN, 9, 2) + 1 * SUBSTRING(@SSN, 10, 2) RETURN1 - SIGN(@Digits % 11) END
Legacy Comments
Saggi Neumann
2009-05-30
re: How to check Danish personal number This is how we validate ID numbers in Israel.
Read more →
CREATEFUNCTION dbo.fnCalculateFinSSN ( @SSN CHAR(10) ) RETURNSCHAR(11) AS BEGIN IF LOWER(@SSN) NOT LIKE '[0-9][0-9][0-9][0-9][0-9][0-9][-+a][0-9][0-9][0-9]' RETURN NULL IF SUBSTRING(@SSN, 7, 1) IN('-', '+') AND ISDATE('19' + SUBSTRING(@SSN, 1, 6)) = 0 RETURN NULL IF LOWER(SUBSTRING(@SSN, 7, 1)) = 'a' AND ISDATE('20' + SUBSTRING(@SSN, 1, 6)) = 0 RETURN 0 DECLARE @Digits INT SET @Digits = CAST(SUBSTRING(@SSN, 1, 6) + SUBSTRING(@SSN, 8, 3) AS INT) % 31 RETURN @@SSN + SUBSTRING('0123456789abcdefghijklmnopqrstuvwxy', @Digits + 1, 1) END
Read more →
CREATEFUNCTION dbo.fnCheckFinSSN ( @SSN CHAR(11) ) RETURNSBIT AS BEGIN IF LOWER(@SSN) NOT LIKE '[0-9][0-9][0-9][0-9][0-9][0-9][-+a][0-9][0-9][0-9][0-9a-y]' RETURN 0 IF SUBSTRING(@SSN, 7, 1) IN('-', '+') AND ISDATE('19' + SUBSTRING(@SSN, 1, 6)) = 0 RETURN 0 IF LOWER(SUBSTRING(@SSN, 7, 1)) = 'a' AND ISDATE('20' + SUBSTRING(@SSN, 1, 6)) = 0 RETURN 0 DECLARE @Digits INT SET @Digits = CAST(SUBSTRING(@SSN, 1, 6) + SUBSTRING(@SSN, 8, 3) AS INT) % 31 IF SUBSTRING('0123456789abcdefghijklmnopqrstuvwxy', @Digits + 1, 1) <> LOWER(RIGHT(@SSN, 1)) SET@Digits = -1 RETURN @Digits + 1 END
Read more →
Wow! This seems to be blog post number 100. Today I stumbled across an interesting problem where OP had a set of records. Every record were "typed" and "grouped". Now OP wanted to get distinct group id's for those groups having at least one record typed 33, but not having a record typed 11.
Read more →
CREATEFUNCTION dbo.fnISOWEEK ( @theDate DATETIME ) RETURNSTINYINT AS BEGIN RETURN ( SELECT CASE WHEN @theDate >= '99990104' THEN (DATEPART(DAYOFYEAR, @theDate) - 4) / 7 WHEN @theDate >= '99990101' THEN 52 WHEN NextYear <= @theDate THEN 0 WHEN CurrentYear <= @theDate THEN DATEDIFF(DAY, CurrentYear, @theDate) / 7 ELSE DATEDIFF(DAY, PreviousYear, @theDate) / 7 END + 1 FROM ( SELECT DATEADD(DAY,(DATEDIFF(DAY, '17530101', DATEADD(YEAR, -1, Jan4)) / 7) * 7, '17530101') AS PreviousYear, DATEADD(DAY,(DATEDIFF(DAY, '17530101', Jan4) / 7) * 7, '17530101') AS CurrentYear, DATEADD(DAY,(DATEDIFF(DAY, '17530101', DATEADD(YEAR, 1, Jan4)) / 7) * 7, '17530101') AS NextYear FROM ( SELECT DATEADD(YEAR, DATEDIFF(YEAR, 0, @theDate), 3) AS Jan4 )AS x ) AS d ) END
Read more →
I was browsing some of my older solutions, and I stumbled across this interesting piece. The object was to mark any employees with Duty 'B' for those employees having both Duty 'O' and Duty 'D'.
Read more →
This approach is not dependent on the 32-level recursion as most other algorithms are for this problem.
CREATEFUNCTION dbo.fnGCD ( @a INT, @b INT ) RETURNSINT AS BEGIN DECLARE @c INT IF @a IS NULL OR @b IS NULL OR (@a = 0 AND @b = 0) RETURN NULL IF @a = 0 OR @b = 0 RETURN ABS(@a) + ABS(@b) IF ABS(@a) < ABS(@b) SELECT @c = ABS(@a), @a = ABS(@b), @b = @c ELSE SELECT @a = ABS(@a), @b = ABS(@b) SET @c = @a % @b WHILE @c > 0 SELECT @a = @b, @b = @c, @c = @a % @b RETURN @b END
Legacy Comments
RBarryYoung
2009-04-15
re: Greatest Common Divisor function That's the iterative version of the chinese remainder algorithim, isn't it Peso?
Read more →
Today I am showing you the difference between @@ERROR, BEGINTRY/CATCH and XACT_ABORT. The three alternatives to error handling works little different and let us intercept the error handling in different stages in the process.
Read more →
Today I am going to show you one way to calculate a decimal year. This is not a easy as it sounds because some years (leap years) include a leap day so the number of days in a year is not consistent.
Read more →
The scenario is very simple. You want to get all ID’s from one table that does not exists in another table. This is how the two simple tables are set up.
Read more →