SQL Server RTM - June 1
The next version of SQL Server 2016 will be released on June 1, 2016, which means you can start planning those upgrades.
The next version of SQL Server 2016 will be released on June 1, 2016, which means you can start planning those upgrades.
How do you go about transferring a disk-based workload to the respective memory-optimized design? How do you process memory-optimized tables? How important for performance are natively-compiled Stored Procedures? Artemakis Artemiou comes up with a step-by-step guide to implementing an in-memory OLTP solution .
The way in which SQL Server chooses to join your tables in a query can dramatically affect performance. In this article, Jason Brimhall explains how a hash match works and shows some performance numbers.
There are some obvious advantages to having the Query Store, but what is the performance impact that it is likely to have on a busy OLTP database server? It is early days, of course and until we get more experience we have to rely on Microsoft's estimate of a performance impact of 3-5% on average. However, this will depend on a number of factors such as usage an the way it is configured. Enrico explores some of these factors in order to give a clearer picture of what you should expect.
In this piece, Steve Jones shows how you can enforce a requirement of a Primary Key (PK) with a unit test, but also allow exceptions where needed.
A simple change might solve some of those tempdb issues various customers experience.
Aaron Bertrand explores yet another scenario where a date/time function seems to cause the optimizer to behave unexpectedly.
One of the great things about IT is that we have so many tools to help us do our jobs. We can even build tools as needed. So why don't more IT people take advantage of this?
By Steve Jones
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In SQL Server 2025, I have a table (dbo.UserPermission) that contains this data:
UserID UserPermissions 15 23 37What is returned when I run this code:
select bit_count(UserPermissions) as PermissionCount from dbo.UserPermission where UserID = 3;See possible answers