SQL Server Geography - sql

Is there any possible way to improve the below query:
DECLARE #radiusInMeters FLOAT = 400;
DECLARE #dgeog geography = geography::Point(given_latitude, given_longitude, 4326).STBuffer(#radiusInMeters);
select [fdx].latitude, [fdx].longitude
from [dbo].[fdx]
where #dgeog.STIntersects(geography::STGeomFromText('POINT(' + convert(varchar(20), [fdx].longitude) + ' ' + convert(varchar(20), [fdx].latitude) + ')', 4326)
) = 1

kcung and Hasan BINBOGA are correct, you need a spatial index.
Look at your query:
#dgeog.STIntersects(xxxx) = 1
This requires [xxxx] to be a geography data type. In order for [xxxx] to be a geography data type, the STGeomFromText function must be applied to the row. And because this is the only part of your WHERE clause, the function must be applied to all rows.
If the table fdx is particularly large, this means that the CLR function will have to be applied over and over again. This is not (in SQL-Server terms) a fast process.
Try this, if you can:
ALTER dbo.fdx ADD Point AS (GEOGRAPHY::Point(Latitude, Longitude, 4326)) PERSISTED
GO
CREATE SPATIAL INDEX SIndex_FDX ON dbo.fdx (Point)
USING GEOGRAPHY_GRID
WITH (
GRIDS = (LEVEL_1 = HIGH,LEVEL_2 = HIGH,LEVEL_3 = HIGH,LEVEL_4 = HIGH),
CELLS_PER_OBJECT = 1
)
GO
DECLARE #Latitude DECIMAL(15,10) = 0
DECLARE #Longitude DECIMAL(15,10) = 0
DECLARE #Radius FLOAT = 400
DECLARE #g GEOGRAPHY = GEOGRAPHY::Point(#Latitude, #Longitude, 4326).STBuffer(#Radius)
SELECT * FROM dbo.fdx WHERE Point.STIntersects(#g) = 1
A note: You should convert your lat/long pairs into decimals before using them to compute the geography column. There is an implicit conversion from float to decimal to string when you use a float as an input that will trim your coordinates down to 4 decimal places. If you explicitly convert first, that will not be an issue.
Also, if you have any null lat/long values in dbo.fdx, you need to filter them in the WHERE clause as a null value will cause your spatial index not to work properly.

You can create spatial index :
https://msdn.microsoft.com/en-us/library/bb934196.aspx

Related

How do I calculate distance between two latitude, longitude points in miles using standard SQL without trigonometry?

How do I calculate distance between two latitude, longitude points in miles using standard SQL without trigonometry?
Without trig you get incorrect results. Find an explanation with answer, formula and example here:
https://jonisalonen.com/2014/computing-distance-between-coordinates-can-be-simple-and-fast
Approximation without precision and trig
DECLARE #SqDegreeLatInMiles AS REAL = 4774.81
DECLARE #SqDegreeLongInMiles AS REAL = 2809
-- VA, Zip=23452
DECLARE #Point_A_Lat AS REAL = 36.8366
DECLARE #Point_A_Long AS REAL = 76.0952
-- TX, Zip=75225
DECLARE #Point_B_Lat AS REAL = 32.8644
DECLARE #Point_B_Long AS REAL = 96.7946
DECLARE #DistanceInMiles AS REAL
SET #DistanceInMiles = SQRT (#SqDegreeLatInMiles * POWER(#Point_A_Lat - #Point_B_Lat,2) + #SqDegreeLongInMiles * POWER(#Point_A_Long - #Point_B_Long,2))
PRINT #DistanceInMiles

Error Handling for numbers of delimiters when extracting substrings

Situation: I have a column where each cell can have up to 5 delimiters. However, it's possible that there are none.
Objective: How do i handle errors such as :
Invalid length parameter passed to the LEFT or SUBSTRING function.
in the case that it cannot find the specified delimiter.
Query:
declare #text VARCHAR(111) = 'abc-def-geeee-ifjf-zzz'
declare #start1 as int
declare #start2 as int
declare #start3 as int
declare #start4 as int
declare #start_index_reverse as int
set #start1 = CHARINDEX('-',#text,1)
set #start2 = CHARINDEX('-',#text,charindex('-',#text,1)+1)
set #start3 = CHARINDEX('-',#text,charindex('-',#text,CHARINDEX('-',#text,1)+1)+1)
set #start4 = CHARINDEX('-',#text,charindex('-',#text,CHARINDEX('-',#text,CHARINDEX('-',#text,1)+1)+1)+1)
set #start_index_reverse = CHARINDEX('-',REVERSE(#text),1)
select
LEFT(#text,#start1-1) AS Frst,
SUBSTRING(#text,#start1+1,#start2-#start1-1) AS Scnd,
SUBSTRING(#text,#start2+1,#start3-#start2-1) AS Third,
SUBSTRING(#text,#start3+1,#start4-#start3-1)AS Third,
RIGHT(#text,#start_index_reverse-1) AS Lst
In this case my variable includes 5 delimiters and so my query works but if i removed one '-' it would break.
XML support in SQL Server brings about some unintentional but useful tricks. Converting this string to XML allows for some parsing that is far less messy than native string handling, which is very far from awesome.
DECLARE #test varchar(111) = 'abc-def-ghi-jkl-mnop'; -- try also with 'abc-def'
;WITH n(x) AS
(
SELECT CONVERT(xml, '<x>' + REPLACE(#test, '-', '</x><x>') + '</x>')
)
SELECT
Frst = x.value('/x[1]','varchar(111)'),
Scnd = x.value('/x[2]','varchar(111)'),
Thrd = x.value('/x[3]','varchar(111)'),
Frth = x.value('/x[4]','varchar(111)'),
Ffth = x.value('/x[5]','varchar(111)')
FROM n;
For a table it's almost identical:
DECLARE #foo TABLE ( col varchar(111) );
INSERT #foo(col) VALUES('abc-def-ghi-jkl-mnop'),('abc'),('def-ghi');
;WITH n(x) AS
(
SELECT CONVERT(xml, '<x>' + REPLACE(col, '-', '</x><x>') + '</x>')
FROM #foo
)
SELECT
Frst = x.value('/x[1]','varchar(111)'),
Scnd = x.value('/x[2]','varchar(111)'),
Thrd = x.value('/x[3]','varchar(111)'),
Frth = x.value('/x[4]','varchar(111)'),
Ffth = x.value('/x[5]','varchar(111)')
FROM n;
Results (sorry about the massive size, seems this doesn't handle 144dpi well):
add a test before your last select
then you should decide how to handle the other case (when one of start is 0)
You can also refer to this link about splitting a string in sql server
which is uses a loop and can handle any number of delimiters
if #start1>0 and #start2>0 and #start3>0 and #start4>0
select LEFT(#text,#start1-1) AS Frst,
SUBSTRING(#text,#start1+1,#start2-#start1-1) AS Scnd,
SUBSTRING(#text,#start2+1,#start3-#start2-1) AS Third,
SUBSTRING(#text,#start3+1,#start4-#start3-1)AS Third,
RIGHT(#text,#start_index_reverse-1) AS Lst

Calculating distance between two points (Latitude, Longitude)

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.
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
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
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
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.
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)
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.

Extra parenthesis changing result of formula in SQL Server 2008

In all other languages (arithmetic engines in general) putting an extra set of parenthesis around operators of same priority does not impact results. But recently in a testing project I noticed that MS SQL server changes the results in those cases. Please take a look at the query below, and let me know if you have any idea (or a setting in SQL Server administration) or any links to MSDN article explaining the behavior.
select (0.55 * 287.61 / 0.66) calc_no_parens
,(0.55 * (287.61 / 0.66)) calc_parens
,round(0.55 * 287.61 / 0.66,2) no_paren_round
,round(0.55 * (287.61 / 0.66),2) paren_round;
Results
Column Record 1
calc_no_parens 239.6750000
calc_parens 239.67499985
no_paren_round 239.6800000
paren_round 239.67000000
To me, first two of them should return 239.675, and round should give 239.68.
You will get the desired result if you declare each value as Float.
DECLARE #Float1 float, #Float2 float, #Float3 float;
SET #Float1 = 0.55;
SET #Float2 = 287.61;
SET #Float3 = 0.66;
select (#Float1 * #Float2 / #Float3) calc_no_parens
,(#Float1* (#Float2/ #Float3)) calc_parens
,round(#Float1 * #Float2/ #Float3,2) no_paren_round
,round(#Float1* (#Float2/ #Float3),2) paren_round;
Output
calc_no_parens calc_parens no_paren_round paren_round
239.675 239.675 239.68 239.68
You may want to see this article: So-called "exact" numerics are not at all exact!
I can see what is happening, but I don't think there is a fix.
SQL calculates and stores each part of the function as a SQL data type (in this case it's a floating point number).
287.61/0.66 produces 435.7727272727272727272727272... which SQL will store as a floating point number to some degree of accuracy, however it isn't exact (after all, it's a floating point number).
For more info on floating point numbers: How is floating point stored? When does it matter?
Habib's answer made me thinking this has to be with decimal data types my columns are using. After a bit of research, I found this
Precision, Scale, and Length (Transact-SQL)
As you can see in that article, division operation significantly changes the both scale and precision of resulting decimal. Then I tried an variation of my query, this time adding extra parenthesis around Multiplication operation.
select distinct (0.55 * 287.61 / 0.66) calc_no_parens
,(0.55 * (287.61 / 0.66)) calc_parens_div
,((0.55 * 287.61) / 0.66) calc_parens_mult
,round(0.55 * 287.61 / 0.66,2) no_paren_round
,round(0.55 * (287.61 / 0.66),2) paren_round
,round((0.55 * 287.61) / 0.66,2) paren_round2;
Results
Column Record 1
calc_no_parens 239.6750000
calc_parens_div 239.67499985
calc_parens_mult 239.6750000
no_paren_round 239.6800000
paren_round 239.67000000
paren_round2 239.6800000
So as long as division is the last operator in the formula we get correct answers. Its not a fix to the problem, but a learning to self in any future testing projects.
When you use numbers SQL try to convert them dynamically:
{
SELECT
0.55*(287.61 / 0.66) PrecisionError,
0.55* (CONVERT(NUMERIC(24,12), 287.61) / CONVERT(NUMERIC(24,12), 0.66)) NotPrecisionError
DECLARE #V SQL_VARIANT
SET #V = 0.55*(287.61 / 0.66)
SELECT
Value = #V
,[TYPE] = CONVERT(SYSNAME, sql_variant_property(#V, 'BaseType')) + '(' +
CONVERT(VARCHAR(10), sql_variant_property(#V, 'Precision')) + ',' +
CONVERT(VARCHAR(10), sql_variant_property(#V, 'Scale')) + ')'
SET #V = 0.55 * (CONVERT(NUMERIC(24,14), 287.61) / CONVERT(NUMERIC(24,14), 0.66))
SELECT
Value = #V
,[TYPE] = CONVERT(SYSNAME, sql_variant_property(#V, 'BaseType')) + '(' +
CONVERT(VARCHAR(10), sql_variant_property(#V, 'Precision')) + ',' +
CONVERT(VARCHAR(10), sql_variant_property(#V, 'Scale')) + ')'
}
RESULTS
PrecisionError NotPrecisionError
239.67499985 239.6750000000000
Value TYPE
239.67499985 numeric(14,8)
Value TYPE
239.6750000000000 numeric(38,13)

Error converting data type varchar to numeric CAST not working

I know this has been beaten like a dead horse. However no matter how I slice it, cast it or convert it I have the same issue.
Error converting data type varchar to numeric.
SELECT property_id, property_case_number, property_address, property_city,
property_state, property_zip, property_lon, property_lat
FROM property
WHERE (property_active = 1)
AND
(property_county = (SELECT property_county FROM property AS property_1
WHERE (property_id = 9165)))
AND
(property_id <> 9165)
AND
property_lon IS NOT Null
AND
property_lat IS NOT Null
AND
dbo.LatLonRadiusDistance(
CONVERT(DECIMAL(15,12),(select property_lat from property where property_id = 9165)),
CONVERT(DECIMAL(15,12),(select property_lon from property where property_id = 9165)),
property_lat,property_lon) <= '5'
I run into this issue as soon as I add dbo.LatLonRadiusDistance at the end.
dbo.LatLonRadiusDistance compares lat & lon distance in miles.
FUNCTION [dbo].[LatLonRadiusDistance]
(
#lat1Degrees decimal(15,12),
#lon1Degrees decimal(15,12),
#lat2Degrees decimal(15,12),
#lon2Degrees decimal(15,12)
)
RETURNS decimal(9,4)
AS
BEGIN
DECLARE #earthSphereRadiusKilometers as decimal(10,6)
DECLARE #kilometerConversionToMilesFactor as decimal(7,6)
SELECT #earthSphereRadiusKilometers = 6366.707019
SELECT #kilometerConversionToMilesFactor = .621371
-- convert degrees to radians
DECLARE #lat1Radians decimal(15,12)
DECLARE #lon1Radians decimal(15,12)
DECLARE #lat2Radians decimal(15,12)
DECLARE #lon2Radians decimal(15,12)
SELECT #lat1Radians = (#lat1Degrees / 180) * PI()
SELECT #lon1Radians = (#lon1Degrees / 180) * PI()
SELECT #lat2Radians = (#lat2Degrees / 180) * PI()
SELECT #lon2Radians = (#lon2Degrees / 180) * PI()
-- formula for distance from [lat1,lon1] to [lat2,lon2]
RETURN ROUND(2 * ASIN(SQRT(POWER(SIN((#lat1Radians - #lat2Radians) / 2) ,2)
+ COS(#lat1Radians) * COS(#lat2Radians) * POWER(SIN((#lon1Radians - #lon2Radians) / 2), 2)))
* (#earthSphereRadiusKilometers * #kilometerConversionToMilesFactor), 4)
END
I'm sure it's something to do with
(select property_lat from property where property_id = 9165)
But no matter how I cast or convert it doesn't change things.
And if I run the function by itself it doesn't give an error.
Anyone have any insights?
here is a sample row
8462 023-125514 15886 W MOHAVE ST GOODYEAR AZ 85338-0000 -112.400297000000 33.429041000000
property_lat & property_lon are varchar(50)
Most likely you are expecting boolean short circuit to rescue the order of evaluating your WHERE clause. This is a known fallacy: boolean operator short circuit is not guaranteed in SQL. See On SQL Server boolean operator short-circuit for a discussion and proof that boolean short circuit can be skipped by query optimizer. Similar topic is T-SQL functions do no imply a certain order of execution. The gist of it is that SQL is a declarative language, not an imperative one.
In your case probably your cast and converts will be called for properties with IDs different from property_id = 9165 and property_active=1 and may attempt to cast string values that are not numerics to a numeric, hence the exception you see. Is difficult to give a precise diagnosis since so much information is missing from your problem description (like the exact definition of all object involved, including all tables, indexes, column types etc).
Your best avenue is to upgrade to SQL Server 2008 and use the built in geography type which has built-in support for STDistance:
This is a close approximate to the geodesic distance. The deviation of
STDistance() on common earth models from the exact geodesic distance
is no more than .25%.
After playing with the query I got it working.
SELECT [property_id], [property_case_number], [property_address], [property_city],
[property_state], [property_zip], [property_lon],
[property_lat]
FROM property
WHERE ([property_active] = 1)
AND
([property_county] = (SELECT b.property_county FROM property b WHERE (b.property_id = #prop_id)))
AND
([property_id] <> #prop_id)
AND
[property_lon] IS NOT Null
AND
[property_lat] IS NOT Null
AND
dbo.LatLonRadiusDistance(
(SELECT c.property_lat FROM property c WHERE (c.property_id = #prop_id)),
(SELECT d.property_lon FROM property d WHERE (d.property_id = #prop_id)),
CAST([property_lat] as FLOAT),
CAST([property_lon] as FLOAT)) <= 5
adding the [] seems to have skirted the issue I was having.