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。但不知何故，Microsoft query不会接受以下语句：派生表就是一个由查询结果生成的临时表。他是在外部查询的 FROM 中定义的。派生表的存在范围只是在外部查询中，只要外部查询结束了，派生表也就不存在了。派生表一定要写在 FROM 后面范围内，用（）括起来。后面跟着派生表...
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...
The result of this statement is an error.C. IIF with NULL parametersSQL Копирај DECLARE @P INT = NULL, @S INT = NULL; SELECT [Result] = IIF( 45 > 30, @P, @S ); Here's the result set.Копирај Result --- NULL Next steps...
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 ...
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...
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SQL DECLARE@aINT=45, @bINT=40;SELECT[Result] =IIF( @a > @b,'TRUE','FALSE'); Here is 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...
Because the date test appeared late in the WHERE clause, I assumed that there wasn't an index over the date(s). (Don't ask me why. It is always a mistake to assume such things.) My apologies for my carelessness. :w00t: I took out the DATEADD for simplicity of explanation and just...
Returns one of two values, depending on whether the Boolean expression evaluates to true or false in SQL Server 2012.Transact-SQL Syntax ConventionsSyntaxCopy IIF ( boolean_expression, true_value, false_value ) Argumentsboolean_expression A valid Boolean expression. If this argument is not a ...
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...