SQL Compare

SQLServerCentral Editorial

20 Years of Compare

  • Editorial

When we started SQLServerCentral, there were originally 7 of us. We all decided to "invest" $50 to get the site going. With this seed money, we paid for a VM that hosted both SQL Server and IIS. This was enough money to run the site for 6+ months, and we set about building an online […]

(1)

You rated this post out of 5. Change rating

2019-09-04

279 reads

External Article

Managing multiple database versions

  • Article

Sometimes, it's necessary to have different versions of a database running in production. But how do you establish a single source of truth in source control? Alex Yates from DLM Consultants shows how to use object filters in SQL Compare to build multiple database versions from the same source.

2016-12-19

3,159 reads

Blogs

T-SQL Tuesday #202: 100 Hours

By

It’s time for T-SQL Tuesday again and this is a great prompt to start...

T-SQL Tuesday

By

T-SQL Tuesday is a monthly blog party hosted by a different community member each...

Exploring DiskANN: Part 2: PQ, SSDs, caching and beam search

By

How I used AI for this postChatGPT to generate images based on info specifically...

Read the latest Blogs

Forums

Performance Regression After Upgrading from SQL Server 2016 to 2022

By abdalah.mehdoini

Bonjour à tous, La semaine dernière, nous avons effectué une mise à niveau de...

Adding new column with DEFAULT

By Thomas Franz

Comments posted to this topic are about the item Adding new column with DEFAULT

How do I connect to a SQL Server database on my ISP?

By Doctor Who 2

I've got a web application for my side business. I've added a SQL Server...

Visit the forum

Question of the Day

Adding new column with DEFAULT

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 = 0
 

See possible answers