2012-01-24 (first published: 2010-07-28)
9,767 reads
2012-01-24 (first published: 2010-07-28)
9,767 reads
2012-01-19 (first published: 2010-07-21)
9,119 reads
2012-01-17 (first published: 2010-07-14)
8,961 reads
2012-01-12 (first published: 2010-07-07)
8,793 reads
Leibniz breaks out of a buyer's dilemma that might defeat a subtler mind.
2012-01-10 (first published: 2010-06-30)
9,239 reads
Centuries ago, the philosopher Leibniz had a dream that all decisions could be made by machine. Today, his vision comes to life, more or less.
2012-01-05 (first published: 2010-06-23)
9,072 reads
2012-01-03 (first published: 2010-06-16)
9,169 reads
2011-12-29 (first published: 2010-06-09)
8,896 reads
2011-12-27 (first published: 2010-06-02)
11,230 reads
2011-12-22 (first published: 2010-05-26)
11,694 reads
By Steve Jones
Superheroes and saints never make art. Only imperfect beings can make art because art...
One feature that I have been waiting for years! The new announcement around optimize...
Following on from my last post about Getting Started With KubeVirt & SQL Server,...
Comments posted to this topic are about the item The AI Bubble and the...
Hi, in a simple oledb source->derived column->oledb destination data flow, 2 of my...
hi, i noticed the sqlhealth extended event is on by default , and it...
I am currently working with Sql Server 2022 and AdventureWorks database. First of all, let's set the "Read Committed Snapshot" to ON:
use master; go alter database AdventureWorks set read_committed_snapshot on with no_wait; goThen, from Session 1, I execute the following code:
--Session 1 use AdventureWorks; go create table ##t1 (id int, f1 varchar(10)); go insert into ##t1 values (1, 'A');From another session, called Session 2, I open a transaction and execute the following update:
--Session 2 use AdventureWorks; go begin tran; update ##t1 set f1 = 'B' where id = 1;Now, going back to Session 1, what happens if I execute this statement?
--Session 1 select f1 from ##t1 where id = 1;See possible answers