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Point in Polygon ( myWithin(@point, @polygon) )
Posted by: AndroNick Nad
Date: May 09, 2010 03:30AM

My function is 6 times faster than here http://forums.mysql.com/read.php?23,201913,201913

CREATE FUNCTION myWithin(p POINT, poly POLYGON) RETURNS INT(1) DETERMINISTIC
BEGIN
DECLARE n INT DEFAULT 0;
DECLARE pX DECIMAL(9,6);
DECLARE pY DECIMAL(9,6);
DECLARE ls LINESTRING;
DECLARE poly1 POINT;
DECLARE poly1X DECIMAL(9,6);
DECLARE poly1Y DECIMAL(9,6);
DECLARE poly2 POINT;
DECLARE poly2X DECIMAL(9,6);
DECLARE poly2Y DECIMAL(9,6);
DECLARE i INT DEFAULT 0;
DECLARE result INT(1) DEFAULT 0;
SET pX = X(p);
SET pY = Y(p);
SET ls = ExteriorRing(poly);
SET poly2 = EndPoint(ls);
SET poly2X = X(poly2);
SET poly2Y = Y(poly2);
SET n = NumPoints(ls);
WHILE i<n DO
SET poly1 = PointN(ls, (i+1));
SET poly1X = X(poly1);
SET poly1Y = Y(poly1);
IF ( ( ( ( poly1X <= pX ) && ( pX < poly2X ) ) || ( ( poly2X <= pX ) && ( pX < poly1X ) ) ) && ( pY > ( poly2Y - poly1Y ) * ( pX - poly1X ) / ( poly2X - poly1X ) + poly1Y ) ) THEN
SET result = !result;
END IF;
SET poly2X = poly1X;
SET poly2Y = poly1Y;
SET i = i + 1;
END WHILE;
RETURN result;
End;

Usage:

SET @point = PointFromText('POINT(5 5)') ;
SET @polygon = PolyFromText('POLYGON((0 0,10 0,10 10,0 10))') ;
SELECT myWithin(@point, @polygon) AS result ;
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Edited 1 time(s). Last edit at 05/09/2010 03:24AM by AndroNick Nad.

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