Viewing 15 posts - 9,016 through 9,030 (of 13,876 total)
Instead of SELECT @RetVal in my above example,
select left(@RetVal,len(@Retval)-1)
Returns everything apart from the rightmost character. This may be what you need ...
--Edit: fixed inconsistent formatting
October 13, 2014 at 10:36 am
Comment deleted. I answered the wrong question! +1 to Alvin's answer.
October 13, 2014 at 10:24 am
There's a post which may help you here.
October 13, 2014 at 5:55 am
Here is a solution which relies on Jeff Moden's infamous '8k Splitter'[/url].
It assumes that the DelimitedSplit8K function has been created in the database you are going to call it from.
if...
October 13, 2014 at 5:02 am
When you call this function, will you always pass exactly 2 arguments (eg 1 2), or do you want to pass a parameter list with a variable list of params?
October 13, 2014 at 4:06 am
Have you recently renamed your SQL Server instance?
October 13, 2014 at 12:17 am
cbrammer1219 (10/10/2014)
October 10, 2014 at 8:51 am
cbrammer1219 (10/10/2014)
I have used this for files, but with table names I wouldn't know what or where to start, from what I have read says table names can't be dynamic.
In...
October 10, 2014 at 8:30 am
Deleted comment
October 10, 2014 at 2:56 am
With the numbers' you're talking about - you're going to have to address a way to compare like attribute to like attribute. Since you said that the attribue ID's might...
October 9, 2014 at 12:28 pm
I am wondering whether it would be worth producing properly normalised tables in each of the databases. I'm thinking along the lines of
(CustId, AttId, AttValue)
These three values could form...
October 9, 2014 at 12:14 pm
Paul Stasny (10/9/2014)
Thanks for your reply. I'm not clear on what you are suggesting. The only thing that links the two databases is the custid and custcode values. Based on...
October 9, 2014 at 12:05 pm
Perhaps you can change your approach slightly – how about modifying your queries to run independently and produce lists of row counts, sums, checksums, whatever and then comparing those results?
October 9, 2014 at 11:22 am
Rick Harper (10/9/2014)
An old friend has come up with the answer so the riddle is solvedI just replace:
DECLARE @strdate VARCHAR
with
DECLARE @strdate VARCHAR(15)
and everything works fine.
Never mind your old friend. Your...
October 9, 2014 at 5:42 am
Try running this and the answer will become clear:
DECLARE @strdate varchar
SET @strdate = '2004-01-01'
select @strdate
By declaring your variable as Varchar but without a length, SQL Server assumes a length of...
October 9, 2014 at 5:24 am
Viewing 15 posts - 9,016 through 9,030 (of 13,876 total)