IIF() is one of the new built-in logical function introduced as a Part of Sql Server 2012. IIF() is the shorthand way of writing CASE statement or IF-ELSE statement. In-fact if we see the execution plan the IIF() function internally translates to a CASE statement.

IIF() function takes three arguments, first argument should be a Boolean expression otherwise it raises an exception. If Boolean expression evaluates to TRUE then Second argument will be the result otherwise Third argument will be the result. This functions result’s Datatype will be the one with highest precedence from the Datatypes of Second and Third Argument.

Syntax: IIF ( boolean_expression, true_value, false_value )

Now let us understand this function with examples:

Example 1: Comparing two Integer numbers

SELECT IIF(1 > 10, 'TRUE', 'FALSE' )

Result: FALSE

Example 2: Comparing two Integer Variables

DECLARE @Marks INT = 60, @Minimum INT = 35
SELECT IIF(@Marks >= @Minimum, 'PASS', 'FAIL' )

Result: PASS

Example 3: Comparing Strings

DECLARE @NAME Varchar(50) = 'Basav'
SELECT IIF(@NAME IN ('Basav','Kalpana'), 
                    'Likes SQL', 'Likes ORACLE' )

Result: Likes SQL

Example 4: Result’s Datatype will be the one with highest precedence from the Datatypes of Second and Third Argument.

SELECT IIF(1 > 10, 1.5, 40)

Result: 40.0

Example 5: Nested IIF() Statement.

DECLARE @Percentage AS NUMERIC(5,2) = 71
SELECT IIF(@Percentage >= 70, 'Distinction', 
        IIF(@Percentage>=35 AND @Percentage<70, 'Pass', 'Fail'))

Result: Distinction

Note: As IIF() statement internally translates to an CASE statement, so the Max Nesting level is 10.

Example 6: IIF() with both return value arguments value as NULL constant

SELECT IIF(1 > 2, NULL, NULL)

Result:
Msg 8133, Level 16, State 1, Line 1
At least one of the result expressions in a CASE specification must be an expression other than the NULL constant

Example 7: IIF() with on of the return value argument value as NULL constant

SELECT IIF(1 > 2, NULL, 'NO')

Result: NO

Example 8: IIF() with both return value arguments are variables with value NULL

DECLARE @A INT = NULL, @B INT = NULL
SELECT IIF(1 > 2, @A, @B)

Result: NULL

You may also like to read other LOGICAL FUNCTION CHOOSE() introduced in SQL SERVER 2012

Please correct me, if my understanding is wrong. Comments are always welcome.

You may like to read the below new built-in function’s introduced in Sql Server 2012:

New Built in Functions introduced in Sql Server
CONVERSION FUNCTIONS
PARSE	TRY_PARSE
TRY_CONVERT
STRING FUNCTIONS
CONCAT	FORMAT
LOGICAL FUNCTIONS
CHOOSE	IIF
DATE AND TIME FUNCTIONS
EOMONTH
DATEFROMPARTS	DATETIMEFROMPARTS
SMALLDATETIMEFROMPARTS	DATETIME2FROMPARTS
TIMEFROMPARTS	DATETIMEOFFSETFROMPARTS