Is there a Max function in SQL Server that takes two values like Math.Max in .NET?

Question

I want to write a query like this:

SELECT o.OrderId, MAX(o.NegotiatedPrice, o.SuggestedPrice)
FROM Order o

But this isn't how the MAX function works, right? It is an aggregate function so it expects a single parameter and then returns the MAX of all rows.

Does anyone know how to do it my way?

Answer 1

You'd need to make a User-Defined Function if you wanted to have syntax similar to your example, but could you do what you want to do, inline, fairly easily with a CASE statement, as the others have said.

The UDF could be something like this:

create function dbo.InlineMax(@val1 int, @val2 int)
returns int
as
begin
  if @val1 > @val2
    return @val1
  return isnull(@val2,@val1)
end

... and you would call it like so ...

SELECT o.OrderId, dbo.InlineMax(o.NegotiatedPrice, o.SuggestedPrice) 
FROM Order o

Answer 2

If you're using SQL Server 2008 (or above), then this is the better solution:

SELECT o.OrderId,
       (SELECT MAX(Price)
        FROM (VALUES (o.NegotiatedPrice),(o.SuggestedPrice)) AS AllPrices(Price))
FROM Order o

All credit and votes should go to Sven's answer to a related question, "SQL MAX of multiple columns?"
I say it's the "best answer" because:

1. It doesn't require complicating your code with UNION's, PIVOT's, UNPIVOT's, UDF's, and crazy-long CASE statments.
2. It isn't plagued with the problem of handling nulls, it handles them just fine.
3. It's easy to swap out the "MAX" with "MIN", "AVG", or "SUM". You can use any aggregate function to find the aggregate over many different columns.
4. You're not limited to the names I used (i.e. "AllPrices" and "Price"). You can pick your own names to make it easier to read and understand for the next guy.
5. You can find multiple aggregates using SQL Server 2008's derived_tables like so:
   SELECT MAX(a), MAX(b) FROM (VALUES (1, 2), (3, 4), (5, 6), (7, 8), (9, 10) ) AS MyTable(a, b)

Answer 3

Can be done in one line:

-- the following expression calculates ==> max(@val1, @val2)
SELECT 0.5 * ((@val1 + @val2) + ABS(@val1 - @val2)) 

Edit: If you're dealing with very large numbers you'll have to convert the value variables into bigint in order to avoid an integer overflow.

Answer 4

I don't think so. I wanted this the other day. The closest I got was:

SELECT
  o.OrderId,
  CASE WHEN o.NegotiatedPrice > o.SuggestedPrice THEN o.NegotiatedPrice 
     ELSE o.SuggestedPrice
  END
FROM Order o

Answer 5

Why not try IIF function (requires SQL Server 2012 and later)

IIF(a>b, a, b)

That's it.

(Hint: be careful about either would be null, since the result of a>b will be false whenever either is null. So b will be the result in this case)

Answer 6

DECLARE @MAX INT
@MAX = (SELECT MAX(VALUE) 
               FROM (SELECT 1 AS VALUE UNION 
                     SELECT 2 AS VALUE) AS T1)

Answer 7

The other answers are good, but if you have to worry about having NULL values, you may want this variant:

SELECT o.OrderId, 
   CASE WHEN ISNULL(o.NegotiatedPrice, o.SuggestedPrice) > ISNULL(o.SuggestedPrice, o.NegotiatedPrice)
        THEN ISNULL(o.NegotiatedPrice, o.SuggestedPrice)
        ELSE ISNULL(o.SuggestedPrice, o.NegotiatedPrice)
   END
FROM Order o

Answer 8

Sub Queries can access the columns from the Outer query so you can use this approach to use aggregates such as MAX across columns. (Probably more useful when there is a greater number of columns involved though)

;WITH [Order] AS
(
SELECT 1 AS OrderId, 100 AS NegotiatedPrice, 110 AS SuggestedPrice UNION ALL
SELECT 2 AS OrderId, 1000 AS NegotiatedPrice, 50 AS SuggestedPrice
)
SELECT
       o.OrderId, 
       (SELECT MAX(price)FROM 
           (SELECT o.NegotiatedPrice AS price 
            UNION ALL SELECT o.SuggestedPrice) d) 
        AS MaxPrice 
FROM  [Order]  o

Answer 9

SQL Server 2012 introduced IIF:

SELECT 
    o.OrderId, 
    IIF( ISNULL( o.NegotiatedPrice, 0 ) > ISNULL( o.SuggestedPrice, 0 ),
         o.NegotiatedPrice, 
         o.SuggestedPrice 
    )
FROM 
    Order o

Handling NULLs is recommended when using IIF, because a NULL on either side of your boolean_expression will cause IIF to return the false_value (as opposed to NULL).

Answer 10

I would go with the solution provided by kcrumley Just modify it slightly to handle NULLs

create function dbo.HigherArgumentOrNull(@val1 int, @val2 int)
returns int
as
begin
  if @val1 >= @val2
    return @val1
  if @val1 < @val2
    return @val2

 return NULL
end

EDIT Modified after comment from Mark. As he correctly pointed out in 3 valued logic x > NULL or x < NULL should always return NULL. In other words unknown result.

Answer 11

Its as simple as this:

CREATE FUNCTION InlineMax
(
    @p1 sql_variant,
    @p2 sql_variant
)  RETURNS sql_variant
AS
BEGIN
    RETURN CASE 
        WHEN @p1 IS NULL AND @p2 IS NOT NULL THEN @p2 
        WHEN @p2 IS NULL AND @p1 IS NOT NULL THEN @p1
        WHEN @p1 > @p2 THEN @p1
        ELSE @p2 END
END;

Answer 12

Oops, I just posted a dupe of this question...

The answer is, there is no built in function like Oracle's Greatest, but you can achieve a similar result for 2 columns with a UDF, note, the use of sql_variant is quite important here.

create table #t (a int, b int) 

insert #t
select 1,2 union all 
select 3,4 union all
select 5,2

-- option 1 - A case statement
select case when a > b then a else b end
from #t

-- option 2 - A union statement 
select a from #t where a >= b 
union all 
select b from #t where b > a 

-- option 3 - A udf
create function dbo.GREATEST
( 
    @a as sql_variant,
    @b as sql_variant
)
returns sql_variant
begin   
    declare @max sql_variant 
    if @a is null or @b is null return null
    if @b > @a return @b  
    return @a 
end


select dbo.GREATEST(a,b)
from #t

kristof

Posted this answer:

create table #t (id int IDENTITY(1,1), a int, b int)
insert #t
select 1,2 union all
select 3,4 union all
select 5,2

select id, max(val)
from #t
    unpivot (val for col in (a, b)) as unpvt
group by id

Answer 13

Here's a case example that should handle nulls and will work with older versions of MSSQL. This is based on the inline function in one one of the popular examples:

case
  when a >= b then a
  else isnull(b,a)
end

Answer 14

I probably wouldn't do it this way, as it's less efficient than the already mentioned CASE constructs - unless, perhaps, you had covering indexes for both queries. Either way, it's a useful technique for similar problems:

SELECT OrderId, MAX(Price) as Price FROM (
   SELECT o.OrderId, o.NegotiatedPrice as Price FROM Order o
   UNION ALL
   SELECT o.OrderId, o.SuggestedPrice as Price FROM Order o
) as A
GROUP BY OrderId

Answer 15

Here is an IIF version with NULL handling (based on of Xin's answer):

IIF(a IS NULL OR b IS NULL, ISNULL(a,b), IIF(a > b, a, b))

The logic is as follows, if either of the values is NULL, return the one that isn't NULL (if both are NULL, a NULL is returned). Otherwise return the greater one.

Same can be done for MIN.

IIF(a IS NULL OR b IS NULL, ISNULL(a,b), IIF(a < b, a, b))

Answer 16

SELECT o.OrderId,   
--MAX(o.NegotiatedPrice, o.SuggestedPrice)  
(SELECT MAX(v) FROM (VALUES (o.NegotiatedPrice), (o.SuggestedPrice)) AS value(v)) as ChoosenPrice  
FROM Order o

Answer 17

You can do something like this:

select case when o.NegotiatedPrice > o.SuggestedPrice 
then o.NegotiatedPrice
else o.SuggestedPrice
end

Answer 18

SELECT o.OrderID
CASE WHEN o.NegotiatedPrice > o.SuggestedPrice THEN
 o.NegotiatedPrice
ELSE
 o.SuggestedPrice
END AS Price

Answer 19

CREATE FUNCTION [dbo].[fnMax] (@p1 INT, @p2 INT)
RETURNS INT
AS BEGIN

    DECLARE @Result INT

    SET @p2 = COALESCE(@p2, @p1)

    SELECT
        @Result = (
                   SELECT
                    CASE WHEN @p1 > @p2 THEN @p1
                         ELSE @p2
                    END
                  )

    RETURN @Result

END

Answer 20

For the answer above regarding large numbers, you could do the multiplication before the addition/subtraction. It's a bit bulkier but requires no cast. (I can't speak for speed but I assume it's still pretty quick)

SELECT 0.5 * ((@val1 + @val2) + ABS(@val1 - @val2))

Changes to

SELECT @val1*0.5+@val2*0.5 + ABS(@val1*0.5 - @val2*0.5)

at least an alternative if you want to avoid casting.

Answer 21

In its simplest form...

CREATE FUNCTION fnGreatestInt (@Int1 int, @Int2 int )
RETURNS int
AS
BEGIN

    IF @Int1 >= ISNULL(@Int2,@Int1)
        RETURN @Int1
    ELSE
        RETURN @Int2

    RETURN NULL --Never Hit

END

Answer 22

For SQL Server 2012:

SELECT 
    o.OrderId, 
    IIF( o.NegotiatedPrice >= o.SuggestedPrice,
         o.NegotiatedPrice, 
         ISNULL(o.SuggestedPrice, o.NegiatedPrice) 
    )
FROM 
    Order o

Answer 23

In SQL Server 2012 or higher, you can use a combination of IIF and ISNULL (or COALESCE) to get the maximum of 2 values.
Even when 1 of them is NULL.

IIF(col1 >= col2, col1, ISNULL(col2, col1)) 

Or if you want it to return 0 when both are NULL

IIF(col1 >= col2, col1, COALESCE(col2, col1, 0)) 

Example snippet:

-- use table variable for testing purposes
declare @Order table 
(
  OrderId int primary key identity(1,1),
  NegotiatedPrice decimal(10,2),
  SuggestedPrice decimal(10,2)
);

-- Sample data
insert into @Order (NegotiatedPrice, SuggestedPrice) values
(0, 1),
(2, 1),
(3, null),
(null, 4);

-- Query
SELECT 
     o.OrderId, o.NegotiatedPrice, o.SuggestedPrice, 
     IIF(o.NegotiatedPrice >= o.SuggestedPrice, o.NegotiatedPrice, ISNULL(o.SuggestedPrice, o.NegotiatedPrice)) AS MaxPrice
FROM @Order o

Result:

OrderId NegotiatedPrice SuggestedPrice  MaxPrice
1       0,00            1,00            1,00
2       2,00            1,00            2,00
3       3,00            NULL            3,00
4       NULL            4,00            4,00

Answer 24

Here is @Scott Langham's answer with simple NULL handling:

SELECT
      o.OrderId,
      CASE WHEN (o.NegotiatedPrice > o.SuggestedPrice OR o.SuggestedPrice IS NULL) 
         THEN o.NegotiatedPrice 
         ELSE o.SuggestedPrice
      END As MaxPrice
FROM Order o

Answer 25

select OrderId, (
    select max([Price]) from (
        select NegotiatedPrice [Price]
        union all
        select SuggestedPrice
    ) p
) from [Order]

Answer 26

In Presto you could use use

SELECT array_max(ARRAY[o.NegotiatedPrice, o.SuggestedPrice])

Answer 27

 -- Simple way without "functions" or "IF" or "CASE"
 -- Query to select maximum value
 SELECT o.OrderId
  ,(SELECT MAX(v)
   FROM (VALUES (o.NegotiatedPrice), (o.SuggestedPrice)) AS value(v)) AS MaxValue
  FROM Order o;

Answer 28

Expanding on Xin's answer and assuming the comparison value type is INT, this approach works too:

SELECT IIF(ISNULL(@A, -2147483648) > ISNULL(@B, -2147483648), @A, @B)

This is a full test with example values:

DECLARE @A AS INT
DECLARE @B AS INT

SELECT  @A = 2, @B = 1
SELECT  IIF(ISNULL(@A, -2147483648) > ISNULL(@B, -2147483648), @A, @B)
-- 2

SELECT  @A = 2, @B = 3
SELECT  IIF(ISNULL(@A, -2147483648) > ISNULL(@B, -2147483648), @A, @B)
-- 3

SELECT  @A = 2, @B = NULL
SELECT  IIF(ISNULL(@A, -2147483648) > ISNULL(@B, -2147483648), @A, @B)
-- 2    

SELECT  @A = NULL, @B = 1
SELECT  IIF(ISNULL(@A, -2147483648) > ISNULL(@B, -2147483648), @A, @B)
-- 1