Expanding The Scope of Bridge Tables
Timothy Claason continues talking about database design in a new article. This one talks about expanding the functionality that Bridge Tables provide in your schema.
2010-03-23
8,220 reads
Timothy Claason continues talking about database design in a new article. This one talks about expanding the functionality that Bridge Tables provide in your schema.
2010-03-23
8,220 reads
In building a database, typically we want a well normalized design. However there are cases for considering options for denormalization in complex systems. Timothy Claason gives you some thoughts on the subject.
2010-03-15
11,434 reads
Is your application scalable under increased activity? Timothy Claason brings us a methodology for testing how your application will deal with database load.
2010-03-10
12,665 reads
2010-03-02
10,674 reads
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,...
By DesertDBA
I haven’t posted in a while (well, not here at least since I’ve been...
Comments posted to this topic are about the item Refactoring SQL Code, which is...
Comments posted to this topic are about the item The Read Committed Snapshot Isolation...
Comments posted to this topic are about the item Working with JSON/JSONB Data in...
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