along with $project it will be more appropriate other wise matching elements will be clubbed together with other elements in document.
db.test.aggregate(
  { "$unwind" : "$shapes" },
  { "$match" : {
     "shapes.color": "red"
  }},
{"$project":{
"_id":1,
"item":1
}}
) 
                 
                                             
                         
                        
manpreet![Tuteehub forum best answer]() Best Answer
                                                
                                                                                                        3 years ago
                                                    Best Answer
                                                
                                                                                                        3 years ago
                                                
                                            
Retrieve only the queried element in an object array in MongoDB collection
Suppose you have the following documents in my collection:
{ "_id":ObjectId("562e7c594c12942f08fe4192"), "shapes":[ { "shape":"square", "color":"blue" }, { "shape":"circle", "color":"red" } ] }, { "_id":ObjectId("562e7c594c12942f08fe4193"), "shapes":[ { "shape":"square", "color":"black" }, { "shape":"circle", "color":"green" } ] }Do query:
db.test.find({"shapes.color": "red"}, {"shapes.color": 1})Or
db.test.find({shapes: {"$elemMatch": {color: "red"}}}, {"shapes.color": 1})Returns matched document (Document 1), but always with ALL array items in shapes:
{ "shapes": [ {"shape": "square", "color": "blue"}, {"shape": "circle", "color": "red"} ] }However, I'd like to get the document (Document 1) only with the array that contains color=red:
{ "shapes": [ {"shape": "circle", "color": "red"} ] }How can I do this?