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Query nodes and attributes of parent ways

Overpass API: Get coordinates of postal boundaryCorrect order of nodes in Overpass queryHow can I get a relation with multiple ways as members in addition to the members with Overpass QL?Is there a Service to get only nodes from OSM?Query for WAY that connects two INTERSECTIONSA way to find all footway and street interesections in openstreetmapOverpass / Overpy: Getting Way IDs from NodesOverpass API - Multiple ways for same roadFetching node(nd) reference data of way in Overpass query resultRetrieving OpenStreetMap data (ways and its child nodes) by coordinates using Overpass API?

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}

1

Does the Overpass API provide a way to query a list of nodes and some attributes of their parent ways, and return the data in CSV format?

For example, the following will give me all motorway junctions:

[out:csv(ref, name, ::lat, ::lon)];

// gather results

(

  // query part for: “highway=milestone”

  node["highway"="motorway_junction"]({{bbox}});

);

// print results

out body;

>;

out skel qt;

Is there a way to query the ref=* tag of the roads of which the node is a member?

openstreetmap overpass-api

share|improve this question

asked 8 hours ago

user149408user149408

18017

add a comment | 

1

Does the Overpass API provide a way to query a list of nodes and some attributes of their parent ways, and return the data in CSV format?

For example, the following will give me all motorway junctions:

[out:csv(ref, name, ::lat, ::lon)];

// gather results

(

  // query part for: “highway=milestone”

  node["highway"="motorway_junction"]({{bbox}});

);

// print results

out body;

>;

out skel qt;

Is there a way to query the ref=* tag of the roads of which the node is a member?

openstreetmap overpass-api

share|improve this question

asked 8 hours ago

user149408user149408

18017

add a comment | 

Does the Overpass API provide a way to query a list of nodes and some attributes of their parent ways, and return the data in CSV format?

For example, the following will give me all motorway junctions:

[out:csv(ref, name, ::lat, ::lon)];

// gather results

(

  // query part for: “highway=milestone”

  node["highway"="motorway_junction"]({{bbox}});

);

// print results

out body;

>;

out skel qt;

Is there a way to query the ref=* tag of the roads of which the node is a member?

openstreetmap overpass-api

share|improve this question

asked 8 hours ago

user149408user149408

18017

[out:csv(ref, name, ::lat, ::lon)];

// gather results

(

  // query part for: “highway=milestone”

  node["highway"="motorway_junction"]({{bbox}});

);

// print results

out body;

>;

out skel qt;

share|improve this question

asked 8 hours ago

user149408user149408

18017

share|improve this question

asked 8 hours ago

user149408user149408

18017

asked 8 hours ago

user149408user149408

18017

asked 8 hours ago

user149408user149408

18017

1 Answer
1

active

oldest

votes

2

Yes, that's possible:

[out:csv(_ref, name, ::lat, ::lon)];

node["highway"="motorway_junction"]({{bbox}});



foreach {                  // iterate over nodes



  way(bn)[highway~"^(motorway|trunk)$"][ref] -> .ways; // find ways for each node

  if (ways.count(ways) > 0) {                    // check that at least 1 way was found

    convert result                               // assemble result

             ::id   = id(),                      // use node id

             ::geom = center(geom()),            // and node position

             _ref    = ways.u(t["ref"]),         // parent ref tag

             ::     = ::;                        // all remaining tags of the node

    out geom;                                    // print node

  }

}

share|improve this answer

edited 7 hours ago

user149408

18017

answered 8 hours ago

mmdmmd

2,3651616

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks! I’ve edited the answer to include also trunk roads. There are still issues to be solved where multiple ways with different ref tags are involved: either different spellings (A8 vs. A 8), or the road name actually changes at the junction (A2 becoming S8). The solution to that probably depends on the use case: I would want two rows, one for each road, so it’s probably better to search for the roads and iterate over its junctions.
  
  – user149408
  7 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  To have one single entry for each of the ".ways", you'd need to add another foreach loop and iterate over those ways.
  
  – mmd
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  So I’ve replaced the if with another foreach, and I get short of 200 results (which looks OK). Problem is, how can I refer to tags of the way from the inner foreach?
  
  – user149408
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Try this one: overpass-turbo.eu/s/JY8
  
  – mmd
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  That did the trick. Thanks again!
  
  – user149408
  6 hours ago
  

  
  
  
  

add a comment | 

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2

Yes, that's possible:

[out:csv(_ref, name, ::lat, ::lon)];

node["highway"="motorway_junction"]({{bbox}});



foreach {                  // iterate over nodes



  way(bn)[highway~"^(motorway|trunk)$"][ref] -> .ways; // find ways for each node

  if (ways.count(ways) > 0) {                    // check that at least 1 way was found

    convert result                               // assemble result

             ::id   = id(),                      // use node id

             ::geom = center(geom()),            // and node position

             _ref    = ways.u(t["ref"]),         // parent ref tag

             ::     = ::;                        // all remaining tags of the node

    out geom;                                    // print node

  }

}

share|improve this answer

edited 7 hours ago

user149408

18017

answered 8 hours ago

mmdmmd

2,3651616

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks! I’ve edited the answer to include also trunk roads. There are still issues to be solved where multiple ways with different ref tags are involved: either different spellings (A8 vs. A 8), or the road name actually changes at the junction (A2 becoming S8). The solution to that probably depends on the use case: I would want two rows, one for each road, so it’s probably better to search for the roads and iterate over its junctions.
  
  – user149408
  7 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  To have one single entry for each of the ".ways", you'd need to add another foreach loop and iterate over those ways.
  
  – mmd
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  So I’ve replaced the if with another foreach, and I get short of 200 results (which looks OK). Problem is, how can I refer to tags of the way from the inner foreach?
  
  – user149408
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Try this one: overpass-turbo.eu/s/JY8
  
  – mmd
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  That did the trick. Thanks again!
  
  – user149408
  6 hours ago
  

  
  
  
  

add a comment | 

2

Yes, that's possible:

[out:csv(_ref, name, ::lat, ::lon)];

node["highway"="motorway_junction"]({{bbox}});



foreach {                  // iterate over nodes



  way(bn)[highway~"^(motorway|trunk)$"][ref] -> .ways; // find ways for each node

  if (ways.count(ways) > 0) {                    // check that at least 1 way was found

    convert result                               // assemble result

             ::id   = id(),                      // use node id

             ::geom = center(geom()),            // and node position

             _ref    = ways.u(t["ref"]),         // parent ref tag

             ::     = ::;                        // all remaining tags of the node

    out geom;                                    // print node

  }

}

share|improve this answer

edited 7 hours ago

user149408

18017

answered 8 hours ago

mmdmmd

2,3651616

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks! I’ve edited the answer to include also trunk roads. There are still issues to be solved where multiple ways with different ref tags are involved: either different spellings (A8 vs. A 8), or the road name actually changes at the junction (A2 becoming S8). The solution to that probably depends on the use case: I would want two rows, one for each road, so it’s probably better to search for the roads and iterate over its junctions.
  
  – user149408
  7 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  To have one single entry for each of the ".ways", you'd need to add another foreach loop and iterate over those ways.
  
  – mmd
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  So I’ve replaced the if with another foreach, and I get short of 200 results (which looks OK). Problem is, how can I refer to tags of the way from the inner foreach?
  
  – user149408
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Try this one: overpass-turbo.eu/s/JY8
  
  – mmd
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  That did the trick. Thanks again!
  
  – user149408
  6 hours ago
  

  
  
  
  

add a comment | 

Yes, that's possible:

[out:csv(_ref, name, ::lat, ::lon)];

node["highway"="motorway_junction"]({{bbox}});



foreach {                  // iterate over nodes



  way(bn)[highway~"^(motorway|trunk)$"][ref] -> .ways; // find ways for each node

  if (ways.count(ways) > 0) {                    // check that at least 1 way was found

    convert result                               // assemble result

             ::id   = id(),                      // use node id

             ::geom = center(geom()),            // and node position

             _ref    = ways.u(t["ref"]),         // parent ref tag

             ::     = ::;                        // all remaining tags of the node

    out geom;                                    // print node

  }

}

share|improve this answer

edited 7 hours ago

user149408

18017

answered 8 hours ago

mmdmmd

2,3651616

[out:csv(_ref, name, ::lat, ::lon)];

node["highway"="motorway_junction"]({{bbox}});



foreach {                  // iterate over nodes



  way(bn)[highway~"^(motorway|trunk)$"][ref] -> .ways; // find ways for each node

  if (ways.count(ways) > 0) {                    // check that at least 1 way was found

    convert result                               // assemble result

             ::id   = id(),                      // use node id

             ::geom = center(geom()),            // and node position

             _ref    = ways.u(t["ref"]),         // parent ref tag

             ::     = ::;                        // all remaining tags of the node

    out geom;                                    // print node

  }

}

share|improve this answer

edited 7 hours ago

user149408

18017

answered 8 hours ago

mmdmmd

2,3651616

edited 7 hours ago

user149408

18017

edited 7 hours ago

user149408

18017

answered 8 hours ago

mmdmmd

2,3651616

answered 8 hours ago

mmdmmd

2,3651616