create function statusGrade
    (@status as varchar(11))
returns varchar(11)
as
    -- Process
    begin
        declare @mark as char(1)
        set @mark = ('A''B''C')

        if (@mark = 'A' or @mark = 'B' or @mark = 'C')
            set @status = 'success'
            return @status

        if (@mark = 'D' or @mark = 'E')
            set @status = 'fail'
            return @status
    end

选择statusgrade（'a'）<-这是我的输入
尝试执行时出现错误：
“statusgrade”不是可识别的内置函数名。