Question

我是Sql函数的新手，想要一些关于为什么我的函数没有像我想的那样表现的指针..

我的数据库有一个UserLogin表，其中包含所有用户电子邮件等。然后它有一个Customer，Teachers和Advisors表，这3个表中的任何一个ID都引用UserLogin表中的ID。这个结构值得商榷，但我已经解决了这个问题。

我的功能的目的是找到一个用户名，给出他的ID。

CREATE FUNCTION GetUserLoginName
(
-- Add the parameters 
@id int
)
RETURNS varchar
AS
BEGIN
Declare @firstName varchar
declare @lastName varchar

-- If an advisor exists with ID, find names in Advisor table.
if (exists(
select *
from Advisor
where ID = @id))
Begin

SET @firstName = (
select u.FirstName
from dbo.Advisor as u
WHERE u.ID = @id)

SET @lastName = (
select u.LastName
from dbo.Advisor as u
WHERE u.ID = @id)
End

-- Else if a customer exists with ID, find names in customer table.
else if (exists(
select *
from Customer
where ID = @id))
Begin

SET @firstName = (
select u.FirstName
from dbo.Customer as u
WHERE u.ID = @id)

SET @lastName = (
select u.LastName
from dbo.Customer as u
WHERE u.ID = @id)
End

-- Else if a Teacher exists with ID, find names in Teacher table.
else if (exists(
select *
from Teacher
where ID = @id))
Begin

SET @firstName = (
select u.FirstName
from dbo.Teacher as u
WHERE u.ID = @id)

SET @lastName = (
select u.LastName
from dbo.Teacher as u
WHERE u.ID = @id)
End

RETURN (@firstName + ' ' + @lastName)

END

该功能正在返回名称为“Karl Pedersen”的“K”......

Answer 1

除了定义varchars的大小之外，您还可以使此函数更简单。这些方面的东西。

C

REATE FUNCTION GetUserLoginName
(
    -- Add the parameters 
    @id int
)
RETURNS varchar(30)
AS
BEGIN
    Declare @FullName varchar(30)

    select @FullName = x.FullName
    from
    (
        select a.FirstName + ' ' + a.LastName as FullName
        from dbo.Advisor as a
        WHERE a.ID = @id

        union all

        select c.FirstName + ' ' + c.LastName
        from dbo.Customer as c
        WHERE c.ID = @id

        union all

        select t.FirstName + ' ' + t.LastName
        from dbo.Teacher as t
        WHERE t.ID = @id
    ) x

    return @FullName

end

几个表上的SQL Scalar函数

1 个答案: