Viewing 15 posts - 46 through 60 (of 119 total)
If you closely watch the link the store procedure is in msdb did you update it in msdb Database, please don't do that , create a wrapper proc for sp_send_dbmail...
March 24, 2014 at 5:40 am
i think this link will help you
March 24, 2014 at 4:40 am
how about a recursive idea
WITH cte AS (
SELECT a.id, a.column1, CAST(0 AS BIGINT) Base, CHARINDEX(' ', a.column1, 0) AS MID
FROM...
March 24, 2014 at 4:23 am
might be this will be handy and more readable
with cte as (
SELECT DISTINCT t1.DealerID, t2.tire,
TotalRepairs = SUM(1) OVER (PARTITION BY t1.DealerID),
RepairsOfThisTire = SUM(1) OVER(PARTITION...
March 24, 2014 at 12:50 am
might be a little tweak of Lynm
WITH Groups AS
(
SELECT 1 AS GroupId, 'Oracle' AS GroupName, 0 AS IdParentGroup UNION...
March 24, 2014 at 12:11 am
Hi there,
Thanks, I think you try to pick the missing crewid well need some more data, minimum crewid maximum crewid and some sample crewid to know how the crewid...
March 23, 2014 at 7:35 pm
Hi lowel,
thanks for you, I added a identity column, in the results is it possible to produce that kind...
March 23, 2014 at 7:25 pm
can you show Query, how did you try to pick the lowest value
March 21, 2014 at 5:51 am
Ok, I will explain it clearly please do not consider the Data3 column
idDAta1Data2Data3
1Name81
1possible Shifts307
1First shift228
1Second Shift249
1Third Shift2210
This is my First ID Based set, it might be the first version...
March 21, 2014 at 12:30 am
i want the result as follows
SELECT * FROM
(
VALUES
(1,'Name','8',1),
(1,'possible Shifts','30',1),
(1,'First shift','22',1),
(1,'Second Shift','24',1),
(1,'Third Shift','22',1),
(1,'Name','8',2),
(1,'possible Shifts','30',2),
(1,'First shift','22',2),
(1,'Second Shift','24',2),
(1,'Name','8',3),
(1,'possible Shifts','30',3),
(1,'1st shift','18',3),
(1,'2nd Shift','18',3)
) AS vtable (id,[DAta1],[Data2],[DVersion])
March 20, 2014 at 8:21 pm
Thanks for the reply, but that doesn't work for me any other point
March 20, 2014 at 10:57 am
Excellent question New thing learned today, the 6th option is a tricky one
January 17, 2014 at 2:49 am
it seems the following values are give only 4 is the correct answers
DECLARE @AA VARCHAR(12) = 'AAA'
,@XXX ...
January 8, 2014 at 11:06 pm
intresting one i have one more option
how about this one
SELECT @Result = NULL;
-- Option 6:
SELECT @Result = CASE
WHEN @aa =@XXX AND @XXXCount >...
January 8, 2014 at 10:23 pm
Viewing 15 posts - 46 through 60 (of 119 total)