Viewing 15 posts - 46 through 60 (of 79 total)
Eirikur Eiriksson (8/13/2014)
Quick thought, if the indexes cannot be partition-aligned, how about dropping them before the switch and then re-create them?😎
That's definitely a possibility. Thanks for the...
August 13, 2014 at 8:08 am
Jeff Moden (8/12/2014)Why is it non-partitioned? Is it because it needs to be truly unique?
Events are received and stored in the partitioned table by date. Most of the...
August 13, 2014 at 8:08 am
Thanks - That's what I thought but hoped wasn't the case. Without the non-partitioned index it is very expensive to query for an individual item when we don't know the...
August 12, 2014 at 4:20 pm
Ed Pollack (8/6/2014)
August 6, 2014 at 9:36 am
The article is well written and the author must be commended for it.
I tried implementing in-memory tables and compiles stored procedures a couple of months ago, when SQL Server 2014...
August 5, 2014 at 3:45 pm
Transaction ID was not the best term - I guess I should have used Event ID :-). Event ID is naturally increasing, but there are many changes to an...
July 23, 2014 at 12:28 pm
Identity as a clustered index was not my first choice. The problem is that this is a high volume, high concurrency OLTP with thousands of data inserts per minute. ...
July 23, 2014 at 12:05 pm
Thanks everyone for the replies.
In regards to the XML storage as VARCHAR(MAX), we can receive XML that does not go through XSD validation and may not even be a valid...
July 23, 2014 at 11:02 am
Thanks, Lynn. That's a nice solution.
July 15, 2014 at 1:13 pm
I ended up finding the result. Here's the query if others are interested:
;WITH
S AS
(
SELECTDISTINCT value, startpoint
FROM@table S1
WHERENOT EXISTS(
SELECT*
FROM@table S2
WHERES2.value = S1.value
ANDS2.startpoint < S1.startpoint
ANDS2.stoppoint >= S1.startpoint
)
),
E AS
(
SELECTDISTINCT value, stoppoint
FROM@table...
July 10, 2014 at 3:10 pm
This is inside a stored procedure. Results are not returned to the screen. I'm currently running the procedure in query analyzer to test it, but it's going to be...
June 20, 2014 at 6:33 pm
There aren't really border values in the data. The partition function would be integer. In the example I gave where there are 10 tenants and one extra partition, I...
April 23, 2014 at 8:31 am
I did not mean to do deletes, but to do a switch out, followed by loading directly into the table.
Jeff suggested two staging tables, one for switching out...
April 23, 2014 at 6:55 am
What is the advantage of loading data in the staging, non-partitioned table? Wouldn't you obtain the same result by loading the fact data directly into the partitioned table?
April 22, 2014 at 4:22 pm
Jeff,
I don't disagree with you that I should and will do my own testing should I need to. Unfortunately I don't have the time to run tests on every...
April 21, 2014 at 8:08 am
Viewing 15 posts - 46 through 60 (of 79 total)