如何在 sql server 2008 中调用标量函数

Question

我创建了一个标量函数，它创建成功，但是当我使用 select 语句调用该函数时，它显示 Invalid object name 'dbo.fun_functional_score'。

我的功能：

 ALTER function [dbo].[fun_functional_score] (@phy_id varchar(20))
 returns  varchar(50)

as
begin 

declare @level_initial int, @level_current int

-- initial functional score
set @level_initial=(SELECT pflag.fun_level
FROM tbl_phy_demographic_details as [phy] 
    inner join tbl_all_purple_flag_level as [pflag] on phy.Demographic_id=pflag.Id 
WHERE phy.Physicion_id=@phy_id
    and pflag.visited_count=(select MAX(visited_count)-1 from tbl_all_purple_flag_level ))


-- current functional score
set @level_current=(SELECT pflag.fun_level
FROM tbl_phy_demographic_details as [phy] 
    inner join tbl_all_purple_flag_level as [pflag] on phy.Demographic_id=pflag.Id 
WHERE phy.Physicion_id=@phy_id
    and pflag.visited_count=(select MAX(visited_count) from tbl_all_purple_flag_level ))


--to calculate functional score
declare @fun_level varchar(20),@result varchar(50)

set  @fun_level=@level_current-@level_initial;

 if @fun_level = 0   set @result='Maintained' 
if @fun_level = '-1'  set @result='Minor Improvement' 
if @fun_level = '-2'  set @result='Moderate Improvement' 
if @fun_level = '-3'  set @result='Significant Improvement' 
if @fun_level =  '-4'  set @result='Substantial Improvement' 
if @fun_level =  '1'  set @result='Minor Reduction' 
if @fun_level =  '2'  set @result='Moderate Reduction' 
if @fun_level =  '3'  set @result='Significant Reduction' 
if @fun_level =  '4'  set @result='Substantial Reduction' 




return @result

end

我用这个选择来调用

 select * from dbo.fun_functional_score('01091400003') as [er]

或

 select * from dbo.fun_functional_score('01091400003') 

两者都显示错误“无效的对象名称'dbo.fun_functional_score'。 "

我犯错的地方。谁能帮帮我...

Answer 1

您的语法适用于返回结果集并且可以像表一样查询的表值函数。对于标量函数做

 select  dbo.fun_functional_score('01091400003') as [er]

Answer 2

您有一个标量值函数，而不是表值函数。 from 子句用于表。直接在列列表中查询值即可。

select dbo.fun_functional_score('01091400003')

Answer 3

由于某种原因，我无法使用我的标量函数，直到我使用括号引用它，如下所示：

select [dbo].[fun_functional_score]('01091400003')

Answer 4

对于标量函数语法是

Select dbo.Function_Name(parameter_name)

Select dbo.Department_Employee_Count('HR')

Answer 5

你可以像这样调用你的标量函数

declare @value <type>;
 exec @value= <dbo.fn_nameOfFunction>  <param1>, <param2>;
 print @value;