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问题:
I am trying to calculate the distance between two positions on a map. I have stored in my data: Longitude, Latitude, X POS, Y POS.
I have been previously using the below snippet.
DECLARE @orig_lat DECIMAL DECLARE @orig_lng DECIMAL SET @orig_lat=53.381538 set @orig_lng=-1.463526 SELECT *, 3956 * 2 * ASIN( SQRT( POWER(SIN((@orig_lat - abs(dest.Latitude)) * pi()/180 / 2), 2) + COS(@orig_lng * pi()/180 ) * COS(abs(dest.Latitude) * pi()/180) * POWER(SIN((@orig_lng - dest.Longitude) * pi()/180 / 2), 2) )) AS distance --INTO #includeDistances FROM #orig dest
I don't however trust the data coming out of this, it seems to be giving slightly inaccurate results.
Some sample data in case you need it
Latitude Longitude Distance 53.429108 -2.500953 85.2981833133896
Could anybody help me out with my code, I don't mind if you want to fix what I already have if you have a new way of achieving this that would be great.
Please state what unit of measurement your results are in.
回答1:
Since you're using SQL Server 2008, you have the geography data type available, which is designed for exactly this kind of data:
DECLARE @source geography = 'POINT(0 51.5)' DECLARE @target geography = 'POINT(-3 56)' SELECT @source.STDistance(@target)
Gives
---------------------- 538404.100197555 (1 row(s) affected)
Telling us it is about 538 km from (near) London to (near) Edinburgh.
Naturally there will be an amount of learning to do first, but once you know it it's far far easier than implementing your own Haversine calculation; plus you get a LOT of functionality.
If you want to retain your existing data structure, you can still use STDistance, by constructing suitable geography instances using the Point method:
DECLARE @orig_lat DECIMAL(12, 9) DECLARE @orig_lng DECIMAL(12, 9) SET @orig_lat=53.381538 set @orig_lng=-1.463526 DECLARE @orig geography = geography::Point(@orig_lat, @orig_lng, 4326); SELECT *, @orig.STDistance(geography::Point(dest.Latitude, dest.Longitude, 4326)) AS distance --INTO #includeDistances FROM #orig dest
回答2:
The below function gives distance between two geocoordinates in miles
create function [dbo].[fnCalcDistanceMiles] (@Lat1 decimal(8,4), @Long1 decimal(8,4), @Lat2 decimal(8,4), @Long2 decimal(8,4)) returns decimal (8,4) as begin declare @d decimal(28,10) -- Convert to radians set @Lat1 = @Lat1 / 57.2958 set @Long1 = @Long1 / 57.2958 set @Lat2 = @Lat2 / 57.2958 set @Long2 = @Long2 / 57.2958 -- Calc distance set @d = (Sin(@Lat1) * Sin(@Lat2)) + (Cos(@Lat1) * Cos(@Lat2) * Cos(@Long2 - @Long1)) -- Convert to miles if @d <> 0 begin set @d = 3958.75 * Atan(Sqrt(1 - power(@d, 2)) / @d); end return @d end
The below function gives distance between two geocoordinates in kilometres
CREATE FUNCTION dbo.fnCalcDistanceKM(@lat1 FLOAT, @lat2 FLOAT, @lon1 FLOAT, @lon2 FLOAT) RETURNS FLOAT AS BEGIN RETURN ACOS(SIN(PI()*@lat1/180.0)*SIN(PI()*@lat2/180.0)+COS(PI()*@lat1/180.0)*COS(PI()*@lat2/180.0)*COS(PI()*@lon2/180.0-PI()*@lon1/180.0))*6371 END
The below function gives distance between two geocoordinates in kilometres using Geography data type which was introduced in sql server 2008
DECLARE @g geography; DECLARE @h geography; SET @g = geography::STGeomFromText('LINESTRING(-122.360 47.656, -122.343 47.656)', 4326); SET @h = geography::STGeomFromText('POINT(-122.34900 47.65100)', 4326); SELECT @g.STDistance(@h);
Usage:
select [dbo].[fnCalcDistanceKM](13.077085,80.262675,13.065701,80.258916)
Reference: Ref1,Ref2
回答3:
As you're using SQL 2008 or later, I'd recommend checking out the GEOGRAPHY data type. SQL has built in support for geospatial queries.
e.g. you'd have a column in your table of type GEOGRAPHY which would be populated with a geospatial representation of the coordinates (check out the MSDN reference linked above for examples). This datatype then exposes methods allowing you to perform a whole host of geospatial queries (e.g. finding the distance between 2 points)
回答4:
Create Function [dbo].[DistanceKM] ( @Lat1 Float(18), @Lat2 Float(18), @Long1 Float(18), @Long2 Float(18) ) Returns Float(18) AS Begin Declare @R Float(8); Declare @dLat Float(18); Declare @dLon Float(18); Declare @a Float(18); Declare @c Float(18); Declare @d Float(18); Set @R = 6367.45 --Miles 3956.55 --Kilometers 6367.45 --Feet 20890584 --Meters 6367450 Set @dLat = Radians(@lat2 - @lat1); Set @dLon = Radians(@long2 - @long1); Set @a = Sin(@dLat / 2) * Sin(@dLat / 2) + Cos(Radians(@lat1)) * Cos(Radians(@lat2)) * Sin(@dLon / 2) * Sin(@dLon / 2); Set @c = 2 * Asin(Min(Sqrt(@a))); Set @d = @R * @c; Return @d; End GO
Usage:
select dbo.DistanceKM(37.848832506474, 37.848732506474, 27.83935546875, 27.83905546875)
Outputs:
0,02849639
You can change @R parameter with commented floats.
回答5:
In addition to the previous answers, here is a way to calculate the distance inside a SELECT:
CREATE FUNCTION Get_Distance ( @La1 float , @Lo1 float , @La2 float, @Lo2 float ) RETURNS TABLE AS RETURN -- Distance in Meters SELECT GEOGRAPHY::Point(@La1, @Lo1, 4326).STDistance(GEOGRAPHY::Point(@La2, @Lo2, 4326)) AS Distance GO
Usage:
select Distance from Place P1, Place P2, outer apply dbo.Get_Distance(P1.latitude, P1.longitude, P2.latitude, P2.longitude)
Scalar functions also work but they are very inefficient when computing large amount of data.
I hope this might help someone.
回答6:
It looks like Microsoft invaded brains of all other respondents and made them write as complicated solutions as possible. Here is the simplest way without any additional functions/declare statements:
SELECT geography::Point(LATITUDE_1, LONGITUDE_1, 4326).STDistance(geography::Point(LATITUDE_2, LONGITUDE_2, 4326))
Simply substitute your data instead of LATITUDE_1, LONGITUDE_1, LATITUDE_2, LONGITUDE_2 e.g.:
SELECT geography::Point(53.429108, -2.500953, 4326).STDistance(geography::Point(c.Latitude, c.Longitude, 4326)) from coordinates c
