<div dir="ltr"><div>Hi,</div><div><br></div><div>Something like accumulating ids of obstacles for each polygon, ordered by height, and taking the first element of the array:<br></div><div><br></div><div>select <a href="http://p.id">p.id</a>, (array_agg(<a href="http://o.id">o.id</a> order by height_m))[1] as heighest_id<br></div><div>from polyobstacles p join obstacles o on st_contains(p.geom, o.geom);</div><div><br></div><div>Nicolas<br></div></div><br><div class="gmail_quote"><div dir="ltr" class="gmail_attr">On Mon, 9 Mar 2020 at 10:21, <<a href="mailto:paul.malm@lfv.se">paul.malm@lfv.se</a>> wrote:<br></div><blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex">
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<p class="MsoNormal"><span lang="EN-US">Hi,<u></u><u></u></span></p>
<p class="MsoNormal"><span lang="EN-US">I have 2 tables:<u></u><u></u></span></p>
<p class="MsoNormal"><span lang="EN-US">obstacles (point layer) with a column height_m (INTEGER) and a
<u></u><u></u></span></p>
<p class="MsoNormal"><span lang="EN-US">polyobstacles (polygon layer)<u></u><u></u></span></p>
<p class="MsoNormal"><span lang="EN-US">the point layer has obstacles all over the map and the polygon layer has some polygons containing some of the obstacle points.<u></u><u></u></span></p>
<p class="MsoNormal"><span lang="EN-US">Now I would like to select the highest obstacle in each polygon. If there is several points with the same highest height a random obstacle of those highest shall be selected.<u></u><u></u></span></p>
<p class="MsoNormal"><span lang="EN-US">Is there someone that can get me on the track for such a query, I’m losing myself in subqueries?<u></u><u></u></span></p>
<p class="MsoNormal"><span lang="EN-US">Kind regards,<u></u><u></u></span></p>
<p class="MsoNormal"><span lang="EN-US">Paul<u></u><u></u></span></p>
<p class="MsoNormal"><span lang="EN-US"><u></u> <u></u></span></p>
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<b><span style="font-size:9pt;color:rgb(51,51,51)" lang="EN-US"> Paul Malm</span></b><span style="font-size:9pt;color:rgb(51,51,51)" lang="EN-US"><br>
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