how to ignore other fields while doing match a field in array in mongodb?

Question

collection is

{ _id: 1, results: [ { product: "abc", score: 10 }, { product: "xyz", score: 5 } ] }
{ _id: 2, results: [ { product: "abc", score: 8 }, { product: "xyz", score: 7 } ] }
{ _id: 3, results: [ { product: "abc", score: 7 }, { product: "xyz", score: 8 } ] }

query is

db.survey.find(
   { results: { $elemMatch: { product: "xyz", score: { $gte: 8 } } } }
)

it returns

{ "_id" : 3, "results" : [ { "product" : "abc", "score" : 7 }, { "product" : "xyz", "score" : 8 } ] }

but i also wanted to ignore { "product" : "abc", "score" : 7 } when returning the json.i tried other match aswel still it returns with other records.how to handle?Kindly help thanks

Does this answer your question? Retrieve only the queried element in an object array in MongoDB collection — Cuong Le Ngoc
– Cuong Le Ngoc, Commented Apr 17, 2020 at 8:51

Kevin Smith · Accepted Answer · 2020-04-17 09:11:44Z

1

You need to project the matching results with a positional operator $, else it will always return the full document that matched the query. (hence why you're getting the full results).

db.survey.find(
   { results: { $elemMatch: { product: "xyz", score: { $gte: 8 } } } },
   { "results.$": 1 }
)

answered Apr 17, 2020 at 9:11

Kevin Smith

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