Help W/ Distance Calculation Query

Mar 28, 2007

I'm trying to run a dyncamic query that returns all records within a specific distance of a certain point. The longitude and latitude of each record is stored in the database. The query is constructed from two dynamic variables $StartLatitude and $StartLongitude with represent the starting point.

SELECT UserID, ACOS(SIN($StartLatitude * PI() / 180) * SIN(Latitude * PI() / 180) + COS($StartLatitude * PI() / 180) * COS(Latitude * PI() / 180) * COS(($StartLongitude - Longitude) * PI() / 180)) * 180 / PI() * 60 * 1.1515 AS Distance
FROM HPN_Painters
HAVING (Distance <= 150)

It runs fine until I add the 'HAVING (Distance <= 150)' clause, in which I recieve the error: Invalid column name 'Distance' It seems that Distance cannot be referenced in the HAVING clause.

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Great Circle Distance Calculation

Oct 15, 2007

Great Circle distance calculation
Is there any stored procedure or application that implements Great Circle distance calculation 

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Great Circle Distance Calculation

Oct 15, 2007

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Hello Friends
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Thank you for your timeSara

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this is a sfar as I have got with the sql version


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VS 2005 BI Tools

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I have a table with the following creation sql script

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      [COURSE_OFFERING_ID] [int]
      [STUDENT_ID] [int]

[Code] ....

Below is the sql insert for the above table’s data


[Code] ...

I was trying to calculate GPA and Commulative GPA (CGPA) for this student and the formula of GPA is SUM(Honor)/SUM(CREDIT) and the formula of the CGPA is the Average of the SUM(Honor)/SUM(CREDIT) grouped by semester and I wrote the below query which is calculating the GPA correctly but wrongly calculating the CGPA.

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select x.STUDENT_ID,AVG(x.gpa)as

[Code] ....

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104 2.25 2.25
104 3 3
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104 2.25 2.575
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Sep 19, 2005


I have a query which gives me the following results:

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33 8345000701 0 1116ALPS Premium Due
56 1000000701 0 1116Regular Premium Income
63 6836000701 1 516ALPS Group Holding
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The Column bDebit has either value '0' or value '1' in it ('0' being debit - positive amount, and '1' - credit, negative amount).

I would like it to show the net amount for each account. Therefore in plain English I would like to take all GLCodes that are the same (eg 6843000701) and sum all amounts that have debit value of '0' and subtract all amounts that have debit value '1'. Therefore I would only see '6843000701' code once, and the amount would be '0' becase 600 - 600 = 0.

The current query is:
SELECT dbo.tbGLTransactions.lLedgerCode, dbo.tbGLTransactions.sGLCode, dbo.tbGLTransactions.bDebit, SUM(curAmount)As TotalSum, dbo.tbLedgerCode.sGLDesc
FROM dbo.tbGLTransactions
INNER JOIN dbo.tbLedgerCode
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WHERE dbo.tbGLTransactions.lGLExtractRun = '452'
Group By dbo.tbGLTransactions.lLedgerCode, dbo.tbGLTransactions.sGLCode, dbo.tbGLTransactions.bDebit, dbo.tbLedgerCode.sGLDesc
Order By dbo.tbGLTransactions.bDebit, dbo.tbLedgerCode.sGLDesc

Is someone able to help me as to how i need to modify this query to get the desired result?


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I hope I was enough clear with my poor english.
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Jan 12, 2008

I have a user defined function, I want to determine the distance between the 2 points. I have it working but i'm having a problem getting to print.


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



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


return ( @Miles )


set @Ans = SIN(@lat1 / @DegToRad) * SIN(@lat2 / @DegToRad) + COS(@lat1 / @DegToRad ) * COS( @lat2 / @DegToRad ) * COS(ABS(@long2 - @long1 )/@DegToRad)

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set @Miles = CEILING(@Miles)

return ( @Miles )


EXEC Distance '39.943762', '-78.122265', '32.334709', '-96.633546'
PRINT @RC /* in miles */

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DECLARE @Latitude NUMERIC(9, 6), @Longitude NUMERIC(9, 6)

DECLARE @MyLatitude NUMERIC(9, 6), @MyLongitude NUMERIC(9, 6)
Set @Latitude = 42.329596;
Set @Longitude = -83.709286;
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Question: How do we calculate the distance in miles between the 2 points.

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@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
SELECT @s1_char = SUBSTRING(@s1, @i, 1), @c = @i, @cv0 = CAST(@i AS binary(2)), @j = 1
WHILE @j <= @s2_len
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
SELECT @cv1 = @cv0, @i = @i + 1

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I am trying to use the haversine function to find the distance betweentwo points on a sphere, specifically two zip codes in my database. I'mneither horribly familiar with SQL syntax nor math equations :), so Iwas hoping I could get some help. Below is what I'm using and it is,as best as I can figure, the correct formula. It is not however,giving me correct results. Some are close, others don't seem right atall. Any ideas?SET @lat1 = RADIANS(@lat1)SET @log1 = RADIANS(@log1)SET @lat2 = RADIANS(@lat2)SET @log2 = RADIANS(@log2)SET @Dlat = ABS(@lat2 - @lat1)SET @Dlog = ABS(@log2 - @log1)SET @R = 3956 /*Approximate radius of earth in miles*/SET @A = SQUARE(SIN(@Dlat/2)) + COS(@lat1) * COS(@lat2) *SQUARE(SIN(@Dlog/2))SET @C = 2 * ATN2(SQRT(@A), SQRT(1 - @A))/*SET @C = 2 * ASIN(min(SQRT(@A))) Alternative calculation*/SET @distance = @R * @Cthnx,cjrsumner

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Given the following example;

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2 45 1.5465432300
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Mar 28, 2007

This function computes the great circle distance in Kilometers using the Haversine formula distance calculation.

If you want it in miles, change the average radius of Earth to miles in the function.

create function dbo.F_GREAT_CIRCLE_DISTANCE
@Latitude1 float,
@Longitude1 float,
@Latitude2 float,
@Longitude2 float
returns float

Computes the Great Circle distance in kilometers
between two points on the Earth using the
Haversine formula distance calculation.

Input Parameters:
@Longitude1 - Longitude in degrees of point 1
@Latitude1 - Latitude in degrees of point 1
@Longitude2 - Longitude in degrees of point 2
@Latitude2 - Latitude in degrees of point 2

declare @radius float

declare @lon1 float
declare @lon2 float
declare @lat1 float
declare @lat2 float

declare @a float
declare @distance float

-- Sets average radius of Earth in Kilometers
set @radius = 6371.0E

-- Convert degrees to radians
set @lon1 = radians( @Longitude1 )
set @lon2 = radians( @Longitude2 )
set @lat1 = radians( @Latitude1 )
set @lat2 = radians( @Latitude2 )

set @a = sqrt(square(sin((@lat2-@lat1)/2.0E)) +
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set @distance =
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return @distance


Edit: corrected spelling


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Hi All,
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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:

Fitzroy, 0xE6100000010C292499D53BE642C0A7406667511F6240
Footscray, 0xE6100000010C89B7CEBF5DE642C02D23F59ECA1C6240

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

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


I then looked at formatting this as in Australia we go by KM so after some searching I found two solutions one for Miles and the other KM

So I changed Select statement to look like this

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

My result is then


When I go to google maps and do a direction request between the locations provided above it says 10.2km (depending on traffic)

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

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Two things I don't like about query:

1. I have to do a UNION to another query that retrieves everything that is in the same city in order to have complete results.
2. very slow to retrieve results (> 1 minute)

Sample DDL: 2 tables
create table dim_lead
date_created datetime,
[contact_first_name] varchar(20),
[contact_last_name] varchar(20),
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[Code] .....

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