Have you tried:
select UserName, Sum(ColB) from Table group by UserName;
or
select UserName, Sum(ColB) from Table group by UserName where
UserName="Emily";
Mike
At 11:43 AM 1/12/2011, Nicholas Moreno wrote:
My issue is actually in Excel. I'm hoping someone could help me...
I need to to
My issue is actually in Excel. I'm hoping someone could help me...
I need to total the values in column B for "Emily". Is there a way other
than =SUM (B1+B2+B4+B7)?
--
Emily | 1
-
Emily | 5
-
Greg | 2
-
Bob | 7
-
Emily | 4
---
In the last episode (Jan 12), Feighen Oosterbroek said:
> I know that this is a bit of a vague question, but over a period of days
> mysqldump will take on one day 2min to complete and on the following day
> 25min to complete, with the resulting sql file being maybe 200M bigger.
> The dataset isn'
On 1/12/2011 10:26, mysql wrote:
Hi listers
I have a mysql web application. in this application it would be fine to
be able to track the database entries i have visited, because often
later on i grat my head: which entry did i see this already in?
So i would need a way to find out which entries i
Hi,
I could envision a way of doing this when accessing your tables
only via Stored Procedures instead of using them directly.
With regards,
Martijn Tonies
Upscene Productions
http://www.upscene.com
Download Database Workbench for Oracle, MS SQL Server, Sybase SQL
Anywhere, MySQL, InterBase, N
Hi listers
I have a mysql web application. in this application it would be fine to
be able to track the database entries i have visited, because often
later on i grat my head: which entry did i see this already in?
So i would need a way to find out which entries in which table i have
visited la
Hi Ryan. That's a common issue for reporting. This post might have you
an idea where to generate those inexistent dates (time slots), just
forget about the specific aggregates and partitioning done in there:
http://gpshumano.blogs.dri.pt/2009/09/28/finding-for-each-time-interval-how-many-records-a
Thank you, that did the trick.
Simon
On 11 January 2011 12:09, Steve Meyers wrote:
> On 1/11/11 9:31 AM, Simon Wilkinson wrote:
>
>> select users.id from users where users.id in (select newletters.user_id
>> from
>> newletters left join articles on newletters.id = articles.newsletter_id
>> wher
