[Talk-GB] overpass query involving pubs

Roland Olbricht roland.olbricht at gmx.de
Fri Oct 27 04:40:38 UTC 2017


> I am confused by this.
> I understood ( x; y; ); to be the union of x and y.  Rather than x 
> passing its results onto y. What am i missing.

It is both at the same time. Given an example like

(way[foo=bar];node(w););

the node(w)-part should depend on the way[foo=bar] part.
And we want both (the ways and the nodes they depend on) in the result.

The rest is a trade-off what is more frequent.

Best regards,

Roland



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