Viewing 15 posts - 14,041 through 14,055 (of 15,381 total)
There may be somebody else with knowledge of some tool that can do this but it seems pretty tough. It is kind of like asking "what data is missing from...
September 14, 2011 at 9:39 am
Wow that is nasty!!! I can give you some general direction but beyond that the details are going to be blood sweat and tears. I would approach this as multi...
September 14, 2011 at 9:38 am
That seems like a totally impossible (or at least inaccurate) task. How would you be able to locate processes or services that might be logging in using a given account?...
September 14, 2011 at 9:24 am
Pham Tuan Phat (9/14/2011)
Suppose I have a trigger:
CREATE TRIGGER myTrigger ON myTable
AFTER UPDATE
AS
BEGIN
If UPDATE(column1)
begin
-- Here I want...
September 14, 2011 at 9:18 am
Unless your data actually contains the single quotes. Here is sample data to see if this can help fit your needs.
create table #Temp
(
grp_key int,
grp_name varchar(25),
Client_List varchar (80)
)
insert #Temp (grp_key, grp_name,...
September 14, 2011 at 9:15 am
select column from table where Client_List like '9%'
September 14, 2011 at 9:09 am
wburnett_1 (9/13/2011)
Understood, that actually works better for this case, in that it should be more like drawing names from a hat...
Not really, most of the time the order will...
September 13, 2011 at 10:56 am
You can't do that quite like that. You have a couple of choices. You can either live with the location of the column, which as far as sql is concerned...
September 13, 2011 at 10:53 am
todinhkhoi (9/9/2011)
September 13, 2011 at 7:57 am
Nice job setting up the ddl and sample data. You have to use a recursive cte for this.
declare @IntroducerID int = 2
;with cte as (
select ID, IntroducerID from #temp...
September 13, 2011 at 7:54 am
gmrose (9/12/2011)
September 12, 2011 at 6:59 pm
Here is a quick and dirty proof.
CREATE TABLE PS_TestForOnline
(
number int identity,
);
INSERT INTO PS_TestForOnline default values
go 100
select * from PS_TestForOnline
select MAX(number) from
(
select top 95 percent number from PS_TestForOnline order by number
)...
September 12, 2011 at 3:48 pm
Well then....
select MAX(number) from
(
select top 95 percent number from PS_TestForOnline order by number
) MyAlias
You probably want a bit more sample data to prove this but it should be what you...
September 12, 2011 at 3:44 pm
OK so how did you come up with 754.25 from your sample data? It would seem that with your data 95% of them are no greater than 20. If you...
September 12, 2011 at 3:11 pm
Evil Kraig F (9/12/2011)
September 12, 2011 at 2:48 pm
Viewing 15 posts - 14,041 through 14,055 (of 15,381 total)