Query an ArrayPosition in MongoDB

Question

I have a collection like this:

> db.nodes.find()  
{ "_id" : ObjectId("534d44e182bee8420ace927f"), "id" : "59598841", "created_by" : "JOSM", "geo" : { "type" : "Point", "coordinates" : [ 9.7346094, 52.371738 ] } }  
{ "_id" : ObjectId("534d44e182bee8420ace9280"), "id" : "59598842", "created_by" : "JOSM", "geo" : { "type" : "Point", "coordinates" : [ 9.7343616, 52.3718121 ] } }  
{ "_id" : ObjectId("534d44e182bee8420ace9281"), "id" : "59598845", "created_by" : "JOSM", "geo" : { "type" : "Point", "coordinates" : [ 9.7331504, 52.372057 ] } }  
{ "_id" : ObjectId("534d44e182bee8420ace9282"), "id" : "59835778", "created_by" : "JOSM", "geo" : { "type" : "Point", "coordinates" : [ 9.7354137, 52.3711697 ] } }  
{ "_id" : ObjectId("534d44e182bee8420ace9283"), "id" : "60409270", "created_by" : "JOSM", "geo" : { "type" : "Point", "coordinates" : [ 9.7354388, 52.3735999 ] } }

Now I want to query the coordinates-array to find the document with the greatest lon-value. How can I do that, I have no idea :(

Tschüss, Andre

Does the answer provided not answer your question? It should. — Neil Lunn
– Neil Lunn, Commented Apr 19, 2014 at 5:56

Neil Lunn · Accepted Answer · 2014-04-16 07:56:18Z

2

So actually getting the "lon" which is the first value, of the array may not seem immediately apparent, but is quite simple with aggregate:

db.nodes.aggregate([
    { "$project": {
        "_id": {
          "_id": "$_id",
          "id": "$id",
          "created_by": "$created_by",
          "geo": "$geo",
        },
        "coordinates": "$geo.coordinates"
    }},
    { "$unwind": "$coordinates" },
    { "$group": {
        "_id": "$_id",
        "lon": { "$first": "$coordinates" }
    }},
    { "$sort": { "lon": 1 } },
    { "$limit": 1 },   
    { "$project": {
        "_id": "$_id._id",
          "id": "$_id.id",
          "created_by": "$_id.created_by",
          "geo": "$_id.geo",
    }}
])

Which gives the whole document with the higest value. Or if you just want the value:

db.nodes.aggregate([
    { "$unwind": "$geo.coordinates" },
    { "$group": {
        "_id": "$_id",
        "lon": { "$first": "$geo.coordinates" }
    }},
    { "$group": {
        "_id": null,
        "lon": { "$max": "$lon" }
    }}
])

edited Apr 16, 2014 at 7:56

answered Apr 16, 2014 at 7:42

Neil Lunn

151k36 gold badges356 silver badges327 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

anvarik Over a year ago

Good answer Neil. Do you happen to know whether is it possible to $group let's say to the fifth element of the array? $first and $last is obvious

Neil Lunn Over a year ago

@anvarik possible yes. Practical? Sadly no. There is a JIRA sitting around for this but it is still a long way off implementation. Mostly syntax issues. But for the possible, you can always ask the question or I have answered at least two that do this kind of thing.

Lalit Agarwal · Accepted Answer · 2014-04-16 08:05:09Z

0

Try using the aggregation framework

db.nodes.aggregate(
 { $unwind: "geo.coordinate" },
 { $group: { _id: { id: "$id"}, lon: { $first: "geo.coordinate" } } },
 { $group: { _id: null, maxval: { $max: "$lon" } } }
)

For more info on aggregation look here: http://docs.mongodb.org/manual/reference/operator/aggregation/

edited Apr 16, 2014 at 8:05

answered Apr 16, 2014 at 7:33

Lalit Agarwal

2,3641 gold badge16 silver badges18 bronze badges

3 Comments

anvarik Over a year ago

he is asking greatest lon-value, not the greatest value. Your answer might give larger lat-values as well

Neil Lunn Over a year ago

And aggreeing with @anvarik there is a way to get each of those values out of the array as is shown in the answer I gave.

Lalit Agarwal Over a year ago

@NeiLunn Agree with you answer. My bad!! I didn't understand the question properly.

Collectives™ on Stack Overflow

Query an ArrayPosition in MongoDB

2 Answers 2

2 Comments

3 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

2 Comments

3 Comments

Your Answer

Sign up or log in

Post as a guest

Related