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Given a Geometry object in OpenLayers 3. How would one go about getting its center?

Older versions of OpenLayers provided a getCentroid method. There was also a getBounds workaround. But these appear to be removed in OpenLayers 3.

swiss_knight
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Joseph Ravenwolfe
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4 Answers4

22

You can get its extent center using ol.extent.getCenter. In my case I have a vector layer and I want to get the center of a feature after I click it.

So

create a simple click interaction and add it to the map

 var select = new ol.interaction.Select();
 map.addInteraction(select);

For each click...

select.on('select', function(e) {

Get the first feature that is selected, from the "selected" array. Then get its geometry, and then its extent.

Use that extent to find its center, using ol.extent.getCenter

    var aa = e.selected[0].getGeometry().getExtent();
    var oo = ol.extent.getCenter(aa);
    console.log("The center is :  "+oo); // voila!!!!

 });

The same code worked for lines, points, and polygons.

PS. The ol.extent.getCenter is stable, used in OL version 3.9.0 and version 3.10.1 and can you can find it here

slevin
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    For lines, this will get you the center of the extent occupied by line geometry. To get a point that is actually on the line, you can use e.selected[0].getGeometry().getClosestPoint(oo) – jOshT Jul 07 '17 at 05:57
21

Bit of a workaround but you can use the getExtent method on the geometry to set the extent on the map. I presume the center of the view will then be the center of the geometry;

Test = new ol.geom.Geometry();
map.getView().fitExtent(Test.getExtent, map.getSize());
var CenterOfGeom = map.getView().getCenter()

If you do not want to change the view (which I can imagine), you can think of a function to calculate the center of the extent. This will be easy in certain types of coordinate systems (EPSG:3857, which is meter based), but more difficult in others (EPSG:4326, based on lon/lat coords). A function which could calculate this center (in EPSG:3857) would be as following;

function getCenterOfExtent(Extent){
var X = Extent[0] + (Extent[2]-Extent[0])/2;
var Y = Extent[1] + (Extent[3]-Extent[1])/2;
return [X, Y];
}

Hope this helps!

Tim.Lucas
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    I think it should be

    `var x = extent[0] + (extent[2] - extent[0]) / 2;

    var y = extent[1] + (extent[3] - extent[1]) / 2;`

     – Gerrit Fölster Feb 11 '15 at 07:49
  • Yes you are right, small mistake there, changed my awnser – Tim.Lucas Feb 12 '15 at 14:29
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    Something so trivial should just be shipped with OL3. +1 for the agnostic getCenterOfExtent – Joel Mar 25 '15 at 16:07
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    @Joel Hey, now it is. OL version 3.9.0 has ol.extent.getCenter and its stable. For example, I did var aa = e.selected[0].getGeometry().getExtent(); var oo = ol.extent.getCenter(aa); and gave me the center of points, lines and polygons. Check it out here – slevin Oct 21 '15 at 17:22
  • That is actually quite good :) – Joel Oct 22 '15 at 10:46
1

You can also get the center with:

var center = e.feature.getGeometry().getCenter();
Bwyss
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    In which version of OpenLayers 3 does this work? In 3.15.1, I get a Uncaught TypeError: feature.getGeometry(...).getCenter is not a function. – Dirk May 20 '16 at 10:22
  • The getCenter() method is only applicable to circle geometries ol.geom.Circle. Is it possible that you have a different type of geometry? – Bwyss May 22 '16 at 14:21
  • Yep, I tried it with a LineString, good to know that this only works with circles. For LineStrings, however, one can use the code in @slevin's answer which will return the point "in the middle of the line". – Dirk May 23 '16 at 08:21
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OpenLayers v 6.1.1

var center = e.feature.getGeometry().getExtent().getCenter()

lord5et
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