Getting the minimum of two values in SQL

Question

I have two variables, one is called PaidThisMonth, and the other is called OwedPast. They are both results of some subqueries in SQL. How can I select the smaller of the two and return it as a value titled PaidForPast?

The MIN function works on columns, not variables.

Answer 1

Use a CASE statement.

Example B in this page should be close to what you're trying to do:
http://msdn.microsoft.com/en-us/library/ms181765.aspx

Here's the code from the page:

USE AdventureWorks;
GO
SELECT   ProductNumber, Name, 'Price Range' = 
      CASE 
         WHEN ListPrice =  0 THEN 'Mfg item - not for resale'
         WHEN ListPrice < 50 THEN 'Under $50'
         WHEN ListPrice >= 50 and ListPrice < 250 THEN 'Under $250'
         WHEN ListPrice >= 250 and ListPrice < 1000 THEN 'Under $1000'
         ELSE 'Over $1000'
      END
FROM Production.Product
ORDER BY ProductNumber ;
GO

Answer 2

Building on the brilliant logic / code from mathematix and scottyc, I submit:

DECLARE @a INT, @b INT, @c INT = 0

WHILE @c < 100
    BEGIN
        SET @c += 1
        SET @a = ROUND(RAND()*100,0)-50
        SET @b = ROUND(RAND()*100,0)-50
        SELECT @a AS a, @b AS b,
            @a - ( ABS(@a-@b) + (@a-@b) ) / 2 AS MINab,
            @a + ( ABS(@b-@a) + (@b-@a) ) / 2 AS MAXab,
            CASE WHEN (@a <= @b AND @a = @a - ( ABS(@a-@b) + (@a-@b) ) / 2)
            OR (@a >= @b AND @a = @a + ( ABS(@b-@a) + (@b-@a) ) / 2)
            THEN 'Success' ELSE 'Failure' END AS Status
    END

Although the jump from scottyc's MIN function to the MAX function should have been obvious to me, it wasn't, so I've solved for it and included it here: SELECT @a + ( ABS(@b-@a) + (@b-@a) ) / 2. The randomly generated numbers, while not proof, should at least convince skeptics that both formulae are correct.

Answer 3

The solutions using CASE, IIF, and UDF are adequate, but impractical when extending the problem to the general case using more than 2 comparison values. The generalized solution in SQL Server 2008+ utilizes a strange application of the VALUES clause:

SELECT
PaidForPast=(SELECT MIN(x) FROM (VALUES (PaidThisMonth),(OwedPast)) AS value(x))

Credit due to this website: http://sqlblog.com/blogs/jamie_thomson/archive/2012/01/20/use-values-clause-to-get-the-maximum-value-from-some-columns-sql-server-t-sql.aspx

Answer 4

For MySQL or PostgreSQL, a better way is to use the LEAST and GREATEST functions.

SELECT GREATEST(A.date0, B.date0) AS date0, 
       LEAST(A.date1, B.date1, B.date2) AS date1
FROM A, B
WHERE B.x = A.x

Both are described here: http://dev.mysql.com/doc/refman/5.0/en/comparison-operators.html