Hi James,
What I am doing is like that. I am reading the points from an XML
file. & right now, trying to figure out a math.
Somehow, somewhere there is a problem with this code & I cant seem to
find where.
See, If u can find.
*********************************************Code STarts
Here********************************************************
function testPolyArray2()
{
//********************************************Initilize Lat & Long &
Create Array***********************************
for(var p = 0; p < dataArray.length; p++)
{
ptsg3.push (new GLatLng(dataArray[p][0], dataArray[p][1]));
}
//
********************************************************************************************************************
var newStpt =0;
var yul =0;
var njStart=0;
var newBreak =1;
for(var p = newStpt; p < dataArray.length; p++)
{
for(var o = p+1; o < dataArray.length; o++)
{
var def = dataArray[p][0] + "," + dataArray[p][1]
var lef = dataArray[o][0] + "," + dataArray[o][1]
if(def == lef)
{
for(var yu
=newBreak;yu<o;yu++)
//for(var yu
=0;yu<172;yu++)
{
var str = String(ptsg3[yu]);
str = str.replace("(", " ");
str = str.replace(")", " ");
holdop = str.split(',')
if(o > 296)
{
//alert("Breaking.....");
break;
}
else
{
getDistance[yul] = calcDist(dataArray[o+1][1],dataArray[o+1]
[0],holdop[1],holdop[0])
}
myarray[yul] = getDistance[yul]
+ "--" + holdop[1] + "," +
holdop[0]
//document.getElementById("selected").value =
document.getElementById("selected").value + myarray[yul] + "\n";
yul = yul +1;
}
//alert(o+1);
document.getElementById("selected").value =
document.getElementById("selected").value + "\n***Start Group***\n";
//alert(dataArray[o+1][1] + " , " + dataArray[o+1][0]);
yul =0;
var getMin = getDistance.min();
//alert("The Smallest Point is: " +
getMin + " And the Break
point is:" + o);
newBreak = o;
for(var mj
=0;mj<myarray.length;mj++)
{
//alert(myarray);
if(Left(myarray[mj], 8)
== Left(getMin, 8))
{
var mjSplit =
myarray[mj].split("--");
var findStr =
Left(mjSplit[1],mjSplit[1].search(/,/i));
findStr1 =
Mid(mjSplit[1], mjSplit[1].search(/,/i)+2,
mjSplit[1].length)
var newrt =
String("(" + trimAll(findStr1) + ", " +
trimAll(findStr) + ")");
var newmj =
FNArrayGetSearchLinearRowI(newrt,ptsg3);
//alert("Start Slicing
from: " + newmj + " to :" + o + "Then
From: " + njStart + " to: " + eval(newmj+1));
var str4 =
ptsg3.slice(njStart,newmj+1);
var str5 =
ptsg3.slice(newmj,o);
var newStr2 = str5 +
"," + str4
var mySplitResult = newStr2.split("),");
//alert(mySplitResult.length);
for(var
kl
=0;kl<mySplitResult.length-1;kl++)
{
var
abcd =
mySplitResult[kl].replace("(","");
var
mySplit2 = abcd.split(",");
ptsgNew2.push(new GLatLng(mySplit2[0],mySplit2[1]));
}
njStart = ptsgNew2.length + 1;
getDistance.length =0;
myarray.length
=0;
}
}
newStpt = o;
}
}
//break;
}
poly2 = new GPolygon(ptsgNew2,"#000000",1,1,"#0000FF",0.5,
{clickable:false});
map.addOverlay(poly2);
map.setCenter(new GLatLng(dataArray[dataArray.length-1][0],
dataArray[dataArray.length-1][1]), 11);
}
**************************************************************************************************************************
On Oct 15, 2:33 pm, jameslove <[EMAIL PROTECTED]> wrote:
> Hi Sunny,
> Why don't know where the second polygon starts? Are you talking about
> a series of polygons which are contained within one series of points
> but you need to break them out into a series of polygons? If this is
> the case, I'd look for differences in proximity. If the polygons are
> sufficiently far enough apart, you could differentiate them that way.
> This seems so unique you'll need to figure out the math yourself,
> since finding an existing algorithm seems doubtful.
>
> -James Love
> Vernon, BC
> Canadawww.jameslove.com
>
> On Oct 15, 11:25 am, Sunny <[EMAIL PROTECTED]> wrote:
>
> > Hi,
>
> > Do anyone Know, How we can reorder the polygons based on points.
> > Like If you have coordinates for three polygons.But you dont know,
> > from where the second polygon starts. You loop through the points &
> > See, where the first polygon ends., when It ends, you take the next
> > point, where first one ends & again search where this one ends.
> > Some algorithm to arrange these points may help.
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