Quick Scalar Tables–#SQLNewBlogger
Another post for me that is simple and hopefully serves as an example for people trying to get blogging as...
2016-07-28
937 reads
Another post for me that is simple and hopefully serves as an example for people trying to get blogging as...
2016-07-28
937 reads
Gathering too much data isn't always helpful, and doesn't necessarily help you make better decisions. However, that's up to you to decide how much is too much.
2016-07-28
83 reads
2016-07-28
1,193 reads
Today Steve Jones notes that the ability to search if invaluable as the amount of information, or even objects, we deal with grows.
2016-07-27
98 reads
The first time I heard someone mention drift at Redgate, it made sense to me. After all, in the context...
2016-07-27 (first published: 2016-07-22)
4,415 reads
Releasing software is stressful, and as Steve Jones notes today, can cause bankruptcy if you don't have a solid process.
2016-07-26
105 reads
2016-07-25
82 reads
Today Steve Jones wonders why we don't unit test more, especially for databases.
2016-07-25
129 reads
One of the things that you need with a Continuous Integration server is that ability to build your software on...
2016-07-22 (first published: 2016-07-18)
1,494 reads
2016-07-22
1,265 reads
A RAG pipeline that answers questions in the demo is not the same thing...
By Steve Jones
“A multitude of bad ideas is necessary for one good idea” – from Excellent...
Yesterday I gave a talk for MSSQLTips called “Building a DBA Agent for Your...
Comments posted to this topic are about the item Adding new column with DEFAULT
I've got a web application for my side business. I've added a SQL Server...
Comments posted to this topic are about the item Looking for New Blood
Which number will the COUNT() return after executing the following statements:
DROP TABLE IF EXISTS #test; CREATE TABLE #test (id INT) INSERT INTO #test (id) SELECT * FROM GENERATE_SERIES(1, 3) AS gs ; ALTER TABLE #test ADD flag BIT CONSTRAINT DF_#test_flag DEFAULT 0; go UPDATE #test SET flag = 0 WHERE id = 1 UPDATE #test SET flag = 1 WHERE id = 2 INSERT INTO #test (id) VALUES (4) SELECT COUNT(*) FROM #test AS t WHERE flag = 0See possible answers