Rocky Mountain Tech Trifecta v3.0 Keynote
I had the main room keynote at the recent Rocky Mountain Tech Trifecta v3.0 in Denver. Jeff Certain & Ben Hoelting...
2011-03-09
1,018 reads
I had the main room keynote at the recent Rocky Mountain Tech Trifecta v3.0 in Denver. Jeff Certain & Ben Hoelting...
2011-03-09
1,018 reads
Last week was a great week all around. It started with Canada's first SQL Saturday in Vancouver - packed with over 300 attendees watching five tracks of local speakers...
2011-03-09
13 reads
Last week was a great week all around. It started with Canada's first SQL Saturday in Vancouver - packed with over 300 attendees watching five tracks of local speakers...
2011-03-09
15 reads
Yes, this looks familiar. Didn’t I write this last week? Well it’s new and the same. We have another pre-con available on the Friday, 3/25, before SQL Saturday 67....
2011-03-09
32 reads
Yes, this looks familiar. Didn’t I write this last week? Well it’s new and the same. We have another pre-con...
2011-03-09
982 reads
Last week was a great week all around. It started with Canada's first SQL Saturday in Vancouver - packed with over...
2011-03-09
1,042 reads
I follow the blog of Johnny Long (twitter) from time-to-time, especially as he chronicled his work in Uganda. I know...
2011-03-09
810 reads
For my data analysis and trending, I wanted to find a simple distribution across quartiles.
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Microsoft has released the RTM version of Visual Studio 2010 SP1, which is available on the MSDN Subscribers web site.
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Comments posted to this topic are about the item Let's Talk Certifications
Comments posted to this topic are about the item The Hidden Security Risks of...
Comments posted to this topic are about the item The Empty Aggregate
I have this table in a SQL Server 2025 database:
CREATE TABLE [dbo].[CustomerOrder] ( [OrderID] [int] NULL, [CustomerID] [int] NULL, [total] [money] NULL ) ON [PRIMARY] GOWhat is returned from this code? (answers are for the sum and then the count)
SELECT SUM(total) AS sum,
COUNT(total) AS count
FROM dbo.CustomerOrder;
See possible answers