Viewing 15 posts - 5,881 through 5,895 (of 8,731 total)
Hi Luis,
First of all, I'm sorry for the comments made that didn't help you on your task.
Second, you shouldn't keep something that is fast enough right now. You should...
September 3, 2014 at 1:32 pm
ROW_NUMBER won't work on SQL Server 2000.
You could use a running total solution and the fastest is the quirky update. You can find different type of solutions in the following...
September 3, 2014 at 11:42 am
Eirikur Eiriksson (9/3/2014)
Koen Verbeeck (9/3/2014)
TomThomson (9/3/2014)
Luis Cazares (9/3/2014)
xsevensinzx (9/3/2014)
Koen Verbeeck (9/3/2014)
xsevensinzx (9/3/2014)
September 3, 2014 at 11:27 am
Oh my, so much work when ROUND function has a truncation option.
DECLARE @MyPay decimal(12,9);
SET @MyPay = 258.235543210;
SELECT ROUND(@MyPay,2,1)
September 3, 2014 at 11:23 am
xsevensinzx (9/3/2014)
Koen Verbeeck (9/3/2014)
xsevensinzx (9/3/2014)
September 3, 2014 at 8:41 am
HowardW (9/3/2014)
Grant Fritchey (9/3/2014)
September 3, 2014 at 8:17 am
TomThomson (9/2/2014)
Sean Lange (9/2/2014)
Lynn Pettis (9/2/2014)
AARRRGGGGGGG!!!!!!!!!!! Someone fire the man. He has no business in this business.
In my experience nearly 95% of the people in this business...
September 2, 2014 at 6:11 pm
Lynn's code is explained in here:
http://www.sqlservercentral.com/articles/comma+separated+list/71700/
September 2, 2014 at 1:38 pm
Alan.B (9/2/2014)
1. A is joined to B to create AB.
2. AB is joined to C to create ABC
Let's say you had 5...
September 2, 2014 at 1:13 pm
David Burrows (9/2/2014)
I do not oracle
You're lucky 😀
September 2, 2014 at 8:46 am
JackTimber (9/1/2014)
I am having some problem understanding joins (Specially left,right and outer apply)when it comes to more then 3 tables.
Like how are the result set formed , which...
September 2, 2014 at 8:43 am
Eirikur Eiriksson (9/1/2014)
Lynn Pettis (8/22/2014)
And Mr. Celko is back with his baseball bat. :Whistling:In more than one thread. :Whistling:
September 1, 2014 at 5:12 pm
CELKO (9/1/2014)
there's no 1NF violation since each value of the column is a scalar value.
NO! The column itself has to be scalar, not each value in a list, not...
September 1, 2014 at 4:54 pm
CELKO (9/1/2014)
I have the following code and I want to pass more than one value:
DECLARE @myvendedor VARCHAR (255);
SET @my_vendor = '87,30';
Please read any book on RDBMS. In...
September 1, 2014 at 4:06 pm
You're welcome.
I hope that you've learned something from this. The main lesson would be that you don't need to count yourself because SQL Server will do it for you....
September 1, 2014 at 1:50 pm
Viewing 15 posts - 5,881 through 5,895 (of 8,731 total)