Viewing 15 posts - 7,126 through 7,140 (of 7,608 total)
Sean Lange (10/9/2012)
maryjane9110024 (10/9/2012)
October 10, 2012 at 11:16 am
It's also possible you need the other index instead, rather than in addition to, the one you have know.
You always need to look at the usage stats too when looking...
October 10, 2012 at 11:07 am
Or just add the required permission(s) to the model db. Every db created after that will automatically have the same permission(s).
October 9, 2012 at 11:47 am
jarid.lawson (10/8/2012)
ScottPletcher (10/8/2012)
SELECT lastname
FROM HR.Employees
WHERE
lastname LIKE '%[/]%' AND
lastname NOT LIKE '%[/]%[/]%'
Still have to add the line that removes / in the first...
October 8, 2012 at 2:47 pm
SQL Server muss immer the log gescrieben.
October 8, 2012 at 10:48 am
SELECT lastname
FROM HR.Employees
WHERE
lastname LIKE '%[/]%' AND
lastname NOT LIKE '%[/]%[/]%'
October 8, 2012 at 10:26 am
Michael T2 (10/5/2012)
actually that wont work, if shipweight2 and shipweight3 are both null it will update them both. I dont want that
No, if both are NULL, only shipweight2 will UPDATE,...
October 8, 2012 at 7:58 am
Yeah, 217 is basically awful. Typically you want to see ~(20-)30ms, although the "base" time will vary depending on your specific disk configuration, of course.
October 5, 2012 at 4:33 pm
Very close. I would make some slight adjustments as follows:
UPDATE dbo.weights
SET
shipweight2 = CASE WHEN shipweight2 IS NULL THEN @weight ELSE shipweight2 END,
...
October 5, 2012 at 4:12 pm
Eugene Elutin (10/4/2012)
ScottPletcher (10/4/2012)
I don't think such extremely simplistic rules always work well in tables, large or small.
Choosing the best indexes takes a lot of careful consideration of your data...
October 4, 2012 at 12:50 pm
Jeff Moden (10/3/2012)
sqlserver12345 (10/2/2012)
Ex:350 days = 50 0/7 weeks
351 days = 50 1/7 weeks
352...
October 4, 2012 at 8:40 am
Jeff Moden (10/3/2012)
ScottPletcher (10/3/2012)
Customer ids are usually unique and ever-increasing; narrowness is sometimes...
October 4, 2012 at 8:36 am
Jeff Moden (10/3/2012)
Use the clustered index for a more appropriate column
For large tables, the "appropriate column" is usually unique, narrow, and ever-increasing. "Customer Code" columns almost never meet all...
October 3, 2012 at 4:57 pm
--
--
SELECT
col1, col2, days,
CAST(days / 7 AS varchar(5)) + ' ' + CAST(days % 7 AS varchar(1)) + '/7' AS weekdays
FROM...
October 3, 2012 at 4:22 pm
Viewing 15 posts - 7,126 through 7,140 (of 7,608 total)