How do I return rows with a specific value first?

Question

I want my query to return the rows of the table where a column contains a specific value first, and then return the rest of the rows alphabetized.

If I have a table something like this example:

 - Table: Users
 - id - name -  city
 - 1    George  Seattle
 - 2    Sam     Miami
 - 3    John    New York
 - 4    Amy     New York
 - 5    Eric    Chicago
 - 6    Nick    New York

And using that table I want to my query to return the rows which contain New York first, and then the rest of the rows alphabetized by city. Is this possible to do using only one query?

Answer 1

On SQL Server, Oracle, DB2, and many other database systems, this is what you can use:

ORDER BY CASE WHEN city = 'New York' THEN 1 ELSE 2 END, city

Answer 2

If your SQL dialect is intelligent enough to treat boolean expressions as having a numeric value, then you can use:

SELECT *
FROM `Users`
ORDER BY (`city` = 'New York') DESC, `city`

Answer 3

My answer may be old and not required but someone may need different approach,hence posting it here.

I had same requirement implemented this, worked for me.

Select * from Users
ORDER BY
(CASE WHEN city = 'New York' THEN 0 ELSE 1 END), city
GO

PS

this is for SQL