Viewing 15 posts - 736 through 750 (of 1,248 total)
You're welcome. Hopefully it pointed you in the right direction.
February 13, 2015 at 8:54 am
I had to update an existing date table to show whether a particular day was a bank holiday in England or Wales or not. The table already had a...
February 13, 2015 at 4:38 am
Sangeeth878787 (2/13/2015)
February 13, 2015 at 3:01 am
This will be a lot easier if you've got a calendar table. This one[/url] is a good place to start. You might have to modify it to show...
February 13, 2015 at 2:25 am
It's funny. When it snows in my area (usually get 1 snow/ice storm per year, of no more than 4 to 6 inches), we get all of the people...
February 11, 2015 at 8:52 am
Try this one[/url] first. There's many out there so a little searching might find one that does exactly what you need. Working days vary from country to country...
February 11, 2015 at 3:43 am
Have you got a calendar table? That will make the calculation of working days much easier.
February 11, 2015 at 3:26 am
Stuart Davies (2/10/2015)
Ed Wagner (2/9/2015)
SQLRNNR (2/9/2015)
Sir KneighfCriminal (had to look that one up)
Smooth
Plane
February 10, 2015 at 1:13 am
Brandie Tarvin (2/5/2015)
Oh, and let's not forget where it all starts.
My cousin's daughter once said 'I know its a boys' job, but I want to be a scientist when I...
February 5, 2015 at 8:55 am
Edvard Korsbæk (2/4/2015)
I have this table:
[dbo].[Login_details](
[login_data_id] [int] IDENTITY(1,1) NOT NULL,
[login_no] [int] NULL,
[logindate] [datetime] NULL,
[logoutdate] [datetime] NULL,
CONSTRAINT [PK_Login_details] PRIMARY KEY CLUSTERED
(
[login_data_id] ASC
)
This query:
Select count(*) from dbo.Login_details group by...
February 4, 2015 at 2:03 am
Sqlraider (2/2/2015)
Iwas Bornready (2/2/2015)
February 3, 2015 at 1:16 am
Viewing 15 posts - 736 through 750 (of 1,248 total)