Find Data Within 10m Distance Of Coordinates?

Nov 22, 2013

I am trying to write a piece of SQL which gives me a list of enquiries within 10 metre distance of a enquiry.

The idea is to identify possible duplicates.

Table: enquiry

Primary key: enquiry_number

Co-ordinates data fields: enquiry.enquiry_easting and enquiry.enquiry_northing.

I will need to self-search on the same table to find possible enquiries within 10m distance.

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What Data Type To Save GPS Coordinates In A Data Table.

Sep 20, 2006

I am working on a program in VB 2005 in which i want to store and retrieve GPS coordinates. I am not sure which data type is the best to use to enter Latitude & Longitude numbers and maintain their proper integrity.

Like LAT ( N38 28.025' ) and LONG (W105 52.098' )

The numbers will be entered by the user and that format can be maintained, but how to re-enter & or insert them into the database using the same format is my real question.

I hope I have explained this right. The numbers in BOLD are what I need to maintain.

Thanks for any help in advance.

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Apr 29, 2015

I have the two following locations.

They're both towns in Australia , State of Victoria

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Footscray,-37.799736, 144.899734

After running geography::Point(Latitude, Longitude , 4326) on the latitude and longitude provided for each location, my Geography column for each row is populated with the following:

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Footscray, 0xE6100000010C89B7CEBF5DE642C02D23F59ECA1C6240

In my SQL Query, I have the following which works out the distance between both towns. Geo being my Geography column

DECLARE @s geography = 0xE6100000010C292499D53BE642C0A7406667511F6240 -- Fitzroy
DECLARE @t geography = 0xE6100000010C89B7CEBF5DE642C02D23F59ECA1C6240 -- Footscray
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The result I get is

6954.44911927616

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So I changed Select statement to look like this

select @s.STDistance(@t)/1000 -- format to KM

My result is then

6.95444911927616

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Now I'm new to this spatial data within SQL, why would I get a different result from google maps?

Also I would like to round this number so its easier to use within my where statement so I'm using Ceiling as shown here:

SELECT CEILING(@s.STDistance(@t)/1000)

Is ceiling the correct way to go?

Reason I need to round this is because we are allowing the end user to search by radius so if they pass in 50km I will then say

Where CEILING(@s.STDistance(@t)/1000) < 50

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Hi
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+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++




Code Snippetcreate function dbo.Distance( @lat1 float , @long1 float , @lat2 float , @long2 float)
returns float

as

begin

declare @DegToRad as float
declare @Ans as float
declare @Miles as float

set @DegToRad = 57.29577951
set @Ans = 0
set @Miles = 0

if @lat1 is null or @lat1 = 0 or @long1 is null or @long1 = 0 or @lat2 is
null or @lat2 = 0 or @long2 is null or @long2 = 0

begin

return ( @Miles )

end

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return ( @Miles )

end

DECLARE @RC float
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CREATE FUNCTION edit_distance(@s1 nvarchar(3999), @s2 nvarchar(3999))
RETURNS int
AS
BEGIN
DECLARE @s1_len int, @s2_len int, @i int, @j int, @s1_char nchar, @c int, @c_temp int,
@cv0 varbinary(8000), @cv1 varbinary(8000)
SELECT @s1_len = LEN(@s1), @s2_len = LEN(@s2), @cv1 = 0x0000, @j = 1, @i = 1, @c = 0
WHILE @j <= @s2_len
SELECT @cv1 = @cv1 + CAST(@j AS binary(2)), @j = @j + 1
WHILE @i <= @s1_len
BEGIN
SELECT @s1_char = SUBSTRING(@s1, @i, 1), @c = @i, @cv0 = CAST(@i AS binary(2)), @j = 1
WHILE @j <= @s2_len
BEGIN
SET @c = @c + 1
SET @c_temp = CAST(SUBSTRING(@cv1, @j+@j-1, 2) AS int) +
CASE WHEN @s1_char = SUBSTRING(@s2, @j, 1) THEN 0 ELSE 1 END
IF @c > @c_temp SET @c = @c_temp
SET @c_temp = CAST(SUBSTRING(@cv1, @j+@j+1, 2) AS int)+1
IF @c > @c_temp SET @c = @c_temp
SELECT @cv0 = @cv0 + CAST(@c AS binary(2)), @j = @j + 1
END
SELECT @cv1 = @cv0, @i = @i + 1
END
RETURN @c
END

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@Longitude1 float,
@Latitude2 float,
@Longitude2 float
)
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/*
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declare @lon2 float
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declare @lat2 float

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