Internally SQL Server converts the statement into the CASE statement and executes. We can check it using the actual execution plan of a query. Execute the query from Example 6 with an Actual execution plan. In the actual execution plan, look at the properties of Compute Scalar. You can see...
它配备了数据编辑、SQL查询和数据模型工具，并支持所有SQLite对象类型。 Navicat for SQLite主要功能包...
SQL Copy SELECT [Result] = IIF( 45 > 30, NULL, NULL ); The result of this statement is an error. C. IIF with NULL parameters SQL Copy DECLARE @P INT = NULL, @S INT = NULL; SELECT [Result] = IIF( 45 > 30, @P, @S ); Here is the result set. Copy Result --- ...
问SQL嵌套IIF函数EN我尝试用嵌套的IIF函数在SQL中制作一个复杂的公式。在这个公式中有很多IFF。但不知...
SQL DECLARE@aINT=45, @bINT=40;SELECT[Result] =IIF( @a > @b,'TRUE','FALSE'); Here's the result set. Result --- TRUE B. IIF with NULL constants SQL SELECT[Result] =IIF(45>30,NULL,NULL); The result of this statement is an error. C. IIF...
For more information, seeIIF Function Query Hints in SQL Server Analysis Services 2008andExecution Plans and Plan Hints for MDX IIF Function and CASE Statement. Examples The following query shows a simple use ofIIFinside a calculated measure to return one of two different string values when the ...
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The result of this statement is an error. C. IIF with NULL parameters DECLARE @P INT = NULL, @S INT = NULL; SELECT IIF ( 45 > 30, @p, @s ) AS Result; Here is the result set. Result --- NULL (1 row(s) affected) See Also...
The result of this statement is an error. C. IIF with NULL parameters DECLARE @P INT = NULL, @S INT = NULL; SELECT IIF ( 45 > 30, @p, @s ) AS Result; Here is the result set. Result --- NULL (1 row(s) affected) See Also...