Viewing 15 posts - 286 through 300 (of 2,171 total)
SELECT 1 + DATEDIFF(DAY, 0, '20091217') % 7
February 19, 2010 at 1:28 am
piet_dj (2/10/2010)
Do you have any useful info on calculating multiple linear regression with T-SQL?
Here http://developerworkshop.net/software.html
It also handles logarithmic, exponential and power regression.
February 10, 2010 at 12:48 pm
UPDATEbt
SETbt.CustCode =CASE
WHEN x.LowerNum IS NULL THEN 'N'
ELSE 'Y'
END
FROMdbo.BatchTrans AS bt
LEFT JOIN(
SELECTLEFT(BottomRange + REPLICATE('0', 50), 50) AS LowerNum,
LEFT(TopRange + REPLICATE('9', 50), 50) AS HigherNum
FROMdbo.ForeignBins
) AS x ON x.LowerNum <= bt.CardNum
AND x.HigherNum...
January 26, 2010 at 6:10 am
How large are the BottomRange and TopRange intervals?
January 26, 2010 at 5:59 am
Please beware that a correlated subquery is an outer join.
If you have a value in target table before, and the correlated subquery doesn't return any value, the target table value...
January 25, 2010 at 6:00 am
I think OP wants several methods to concatenate result, which should work in all three RDMS's.
What we provided is a way for SQL Server to deal with this.
It's a monday...
January 25, 2010 at 4:33 am
OP, the senior application engineer, told us that ISADMIN columns had zeros in all the records. It is not true. It is column ISCONT that has zeros in all columns.
January 25, 2010 at 4:20 am
Here is a solution.
DECLARE@Emp TABLE
(
eno int,
ename varchar(5),
isemp bit,
iscont bit,
isadmin bit,
ishr bit
)
insert@emp
select1, 'aa', 1, 0, 0, 1 union all
select1, 'aa', 0, 0, 1, 1 union all
select1, 'aa', 1, 0, 0, 1...
January 25, 2010 at 4:17 am
No, ISCONT has zeros in all the records.
January 25, 2010 at 4:12 am
You can do this with SQL Server 2008 and the new table value parameters.
If you are working on a set of data, you can pass that set of data to...
January 15, 2010 at 1:13 am
shannonjk (1/7/2010)
How would I use this in my code?
It's quite tricky to answer because we haven't seen your code yet.
January 9, 2010 at 4:40 pm
SELECT ROW_NUMBER() OVER PARTITON BY SSN ORDER BY Amount) AS row_num,
SSN, Amount
FROM Table1
January 9, 2010 at 4:35 pm
See the GROUP BY part above. I added it now.
January 4, 2010 at 8:39 am
Well, I have some time left...
SELECTc.ConstructNo,
b.ImportanceBands,
COUNT(cib.ConstructNo) AS [Count]
FROM(
SELECTConstructNo
FROMCompetencyImportanceData
GROUP BYConstructNo
) AS c
CROSS JOIN(
SELECTImportanceBands
FROMImportanceLevelBands
GROUP BYImportanceBands
) AS b
LEFT JOINCompetencyImportanceData AS cib ON cib.ConstructNo = c.ConstructNo
AND cib.ImportanceScoreBand = b.ImportanceBands
GROUP BYc.ConstructNo,
b.ImportanceBands
January 4, 2010 at 6:30 am
Thank you. I thought I should post the link so that no more double unneccessary work is done.
I suggested your approach first, but after som clarifications, I saw that OP...
January 4, 2010 at 6:29 am
Viewing 15 posts - 286 through 300 (of 2,171 total)