Viewing 15 posts - 31 through 45 (of 57 total)
SQL Server is as good as any to store this data but it depends on how you want to use it.
Don't store all columns for all stocks in one row....
March 16, 2015 at 5:15 am
ricardo_chicas (3/16/2015)
What is really annoying is that I even trimmed the query as to return the same column where I have the clustered index on..
I know that the solutions provided...
March 16, 2015 at 5:09 am
Is the range you are using recently added rows? Recently added values don't appear in the statistic for the table so optimiser can't use stats to estimate record count. However...
March 15, 2015 at 3:50 pm
You may see what looks like multiple entries in the plan cache for the same sproc if the sproc has multiple select statements. Each part of the query will have...
March 15, 2015 at 2:49 am
Create a separate table and add a row from within the trigger. Then create a job to scan that table, do whatever joins you need and send the email.
March 15, 2015 at 2:46 am
Check parameters are identical datatype to the column datatype. I encountered a situation where the parameter was an nvarchar and the column was varchar. It was table scanning instead of...
March 15, 2015 at 2:43 am
Select * from site
Left outer join sitedate on site.siteid = sitedate.siteid and sitedate.date between @startedate And @enddate
Where sitedate.date IS NULL
Left outer join putting the date filter not the join.
March 15, 2015 at 2:37 am
With no unique identifier you may struggle to delete some rows and keep others. You could run a query to all the unique rows into another table, then drop/rename or...
March 11, 2015 at 10:57 am
Checkout sys.dm_db_index_operational_stats to see what indexes are getting updated and used.
All indexes will get updated and sql will need to do the work during an insert to find where...
March 9, 2015 at 4:46 pm
After reading other posts spotted something I would consider a problem for the compiler.
select <many items>
from A
left join B on B.id_A = A.id
left join C...
March 9, 2015 at 9:12 am
The two statements are not the same obviously and the outputs are the same because you understand the data. I'm guessing the same ID's existing in all tables. Let outers...
March 9, 2015 at 5:58 am
Are you looking to replace the existing table with a cleaned up version or do you intend to leave the existing table as is and create a separate table or...
March 9, 2015 at 5:25 am
why do you want to remove the GROUP BY ? Is it stopping you doing something else or not working ?
Its not effecient but you can try sub queries.
SELECT
emp.empid,
...
March 6, 2015 at 9:03 am
Did you try assigning the encrypted value to a @variable and using the @variable in the where clause ?
March 6, 2015 at 8:53 am
Viewing 15 posts - 31 through 45 (of 57 total)