Question

我有一张桌子

具有列邮政编码的人

与一个zipcode表相关，该表包含zip码作为id和纬度和经度列。

我发现了一系列使用纬度和经度计算球体上的距离（基本上是拉链码）的函数。我对sql很新，并且想知道如何在存储过程中使用这些函数。

纬度函数：

        ALTER Function [dbo].[LatitudePlusDistance](@StartLatitude Float, @Distance Float) Returns Float
As
Begin
    Return (Select @StartLatitude + Sqrt(@Distance * @Distance / 4766.8999155991))
End

经度函数：

    ALTER FUNCTION [dbo].[LongitudePlusDistance]
(
    @StartLongitude float,
    @StartLatitude float,
    @Distance float
)
    RETURNS Float
    AS
begin

RETURN (select @startLongitude + sqrt(@Distance * @Distance/(4784.39411916406*Cos(2*@StartLatitude/114.591559026165)*Cos(2*@StartLatitude/114.591559026165))))
END
begin

RETURN (select @startLongitude + sqrt(@Distance * @Distance/(4784.39411916406*Cos(2*@StartLatitude/114.591559026165)*Cos(2*@StartLatitude/114.591559026165))))
END`

计算距离函数

    `ALTER Function [dbo].[CalculateDistance]
    (@Longitude1 Decimal(8,5),
    @Latitude1   Decimal(8,5),
    @Longitude2  Decimal(8,5),
    @Latitude2   Decimal(8,5))
Returns Float
As
Begin
Declare @Temp Float

Set @Temp = sin(@Latitude1/57.2957795130823) * sin(@Latitude2/57.2957795130823) + cos(@Latitude1/57.2957795130823) * cos(@Latitude2/57.2957795130823) * cos(@Longitude2/57.2957795130823 - @Longitude1/57.2957795130823)

if @Temp > 1
    Set @Temp = 1
Else If @Temp < -1
    Set @Temp = -1

Return (3958.75586574 * acos(@Temp) )

End`

我试过这样的事情......

    Declare @Longitude Decimal(8,5)
Declare @Latitude Decimal(8,5)

Select  @Longitude = Longitude,
        @Latitude = Latitude
From    ZipCodes
Where   ZipCode = '20013'

    Declare @Distance int

Select  persons.personName, ZipCodes.City,  dbo.CalculateDistance(@Longitude, @Latitude, ZipCodes.Longitude, ZipCodes.Latitude) As Distance
From    persons
        Inner Join ZipCodes
            On persons.zipcode = ZipCodes.ZipCode
Order By dbo.CalculateDistance(@Longitude, @Latitude, ZipCodes.Longitude, ZipCodes.Latitude)
    WHERE dbo.CalculateDistance(@Longitude, @Latitude, ZipCodes.Longitude, ZipCodes.Latitude) As Distance <= @Distance

我试过这只是为了过滤参数化搜索范围之外的人，但它不起作用。说“关键字附近的语法不正确”，其中“

甚至不包括我想要输入zipcode的参数，而不是：

其中zipcode ='20013'

我想要像：

其中zipcode = @zipcode

但它说我需要声明zscalar变量@zipcode并且无论我在哪里尝试这样做...我一直得到同样的错误

感谢您的帮助

Answer 1

您在“where”之前使用“order by”子句是问题之一，并且您不能在条件是另一个问题的地方使用别名。请尝试以下查询：

Declare @Longitude Decimal(8,5)
Declare @Latitude Decimal(8,5)

Select  @Longitude = Longitude,
        @Latitude = Latitude
From    ZipCodes
Where   ZipCode = '20013'

Declare @Distance int
set @Distance = 5

Select  persons.personName, ZipCodes.City,  dbo.CalculateDistance(@Longitude, @Latitude,    ZipCodes.Longitude, ZipCodes.Latitude) As Distance
From    persons Inner Join ZipCodes On persons.zipcode = ZipCodes.ZipCode
WHERE dbo.CalculateDistance(@Longitude, @Latitude, ZipCodes.Longitude, ZipCodes.Latitude) <= @Distance
Order By dbo.CalculateDistance(@Longitude, @Latitude, ZipCodes.Longitude, ZipCodes.Latitude)

zipcode和radius的参数化sql查询无法正常工作

1 个答案: