Database Design Contest at the 2012 AITP Region 3 Student Conference
Had a great Friday in Lafayette, LA on the campus of the University of Louisiana-Lafayette. 31 different teams submitted database...
2012-10-14
976 reads
Had a great Friday in Lafayette, LA on the campus of the University of Louisiana-Lafayette. 31 different teams submitted database...
2012-10-14
976 reads
Problem Statement
SQL Server has got in-built functions to convert the given string into LOWER() or UPPER() format but it does...
2012-10-14
1,315 reads
I received an email from my boss on the morning supervisors approve timesheets. He stated that they are complaining about...
2012-10-14
2,075 reads
Chennai SQL Serve User Group Meet for Oct'2012 happend today at Microsoft Traning Centre located in Radhakrishnan Road, Chennai.
Its pleasure seeing enthusiastic...
2012-10-13
906 reads
The San Francisco SQL Server User Group – Oct 10 2012 Thanks you everyone for attending session, here is the presentation...
2012-10-13
325 reads
2012-10-13
14 reads
The San Francisco SQL Server User Group – Oct 10 2012 Thanks you everyone for attending session, here is the presentation...
2012-10-13
1,007 reads
2012-10-13
9 reads
2012-10-13
5 reads
2012-10-13
6 reads
By ReviewMyDB
My T-SQL Tuesday #202 entry for Marlon Ribunal's invitation on memorable SQL Server outages....
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...
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