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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?


F
Faraz

MongoDB 2.2's new $elemMatch projection operator provides another way to alter the returned document to contain only the first matched shapes element:

db.test.find(
    {"shapes.color": "red"}, 
    {_id: 0, shapes: {$elemMatch: {color: "red"}}});

Returns:

{"shapes" : [{"shape": "circle", "color": "red"}]}

In 2.2 you can also do this using the $ projection operator, where the $ in a projection object field name represents the index of the field's first matching array element from the query. The following returns the same results as above:

db.test.find({"shapes.color": "red"}, {_id: 0, 'shapes.$': 1});

MongoDB 3.2 Update

Starting with the 3.2 release, you can use the new $filter aggregation operator to filter an array during projection, which has the benefit of including all matches, instead of just the first one.

db.test.aggregate([
    // Get just the docs that contain a shapes element where color is 'red'
    {$match: {'shapes.color': 'red'}},
    {$project: {
        shapes: {$filter: {
            input: '$shapes',
            as: 'shape',
            cond: {$eq: ['$$shape.color', 'red']}
        }},
        _id: 0
    }}
])

Results:

[ 
    {
        "shapes" : [ 
            {
                "shape" : "circle",
                "color" : "red"
            }
        ]
    }
]

any solution if I want it to return every elements that matches it instead of just the first?
I'm afraid I am using Mongo 3.0.X :-(
@charliebrownie Then use one of the other answers that use aggregate.
This also works: db.test.find({}, {shapes: {$elemMatch: {color: "red"}}});
Is this an error: $$shape.color? the double $$ in the condition of the $filter.
S
Stennie

The new Aggregation Framework in MongoDB 2.2+ provides an alternative to Map/Reduce. The $unwind operator can be used to separate your shapes array into a stream of documents that can be matched:

db.test.aggregate(
  // Start with a $match pipeline which can take advantage of an index and limit documents processed
  { $match : {
     "shapes.color": "red"
  }},
  { $unwind : "$shapes" },
  { $match : {
     "shapes.color": "red"
  }}
)

Results in:

{
    "result" : [
        {
            "_id" : ObjectId("504425059b7c9fa7ec92beec"),
            "shapes" : {
                "shape" : "circle",
                "color" : "red"
            }
        }
    ],
    "ok" : 1
}

@JohnnyHK: In this case, $elemMatch is another option. I actually got here by way of a Google Group question where $elemMatch wouldn't work because it only returns the first match per document.
Thanks, I wasn't aware of that limitation so that's good to know. Sorry for deleting my comment you're responding to, I decided to post another answer instead and didn't want to confuse people.
@JohnnyHK: No worries, there are now three useful answers for the question ;-)
For other searchers, in addition to this I also tried adding { $project : { shapes : 1 } } - which seemed to work and would be helpful if the enclosing documents were large and you just wanted to view the shapes key values.
@calmbird I updated the example to include an initial $match stage. If you're interested in a more efficient feature suggestion I would watch/upvote SERVER-6612: Support projecting multiple array values in a projection like the $elemMatch projection specifier in the MongoDB issue tracker.
N
Niels van der Rest

Caution: This answer provides a solution that was relevant at that time, before the new features of MongoDB 2.2 and up were introduced. See the other answers if you are using a more recent version of MongoDB.

The field selector parameter is limited to complete properties. It cannot be used to select part of an array, only the entire array. I tried using the $ positional operator, but that didn't work.

The easiest way is to just filter the shapes in the client.

If you really need the correct output directly from MongoDB, you can use a map-reduce to filter the shapes.

function map() {
  filteredShapes = [];

  this.shapes.forEach(function (s) {
    if (s.color === "red") {
      filteredShapes.push(s);
    }
  });

  emit(this._id, { shapes: filteredShapes });
}

function reduce(key, values) {
  return values[0];
}

res = db.test.mapReduce(map, reduce, { query: { "shapes.color": "red" } })

db[res.result].find()

a
anvarik

Another interesing way is to use $redact, which is one of the new aggregation features of MongoDB 2.6. If you are using 2.6, you don't need an $unwind which might cause you performance problems if you have large arrays.

db.test.aggregate([
    { $match: { 
         shapes: { $elemMatch: {color: "red"} } 
    }},
    { $redact : {
         $cond: {
             if: { $or : [{ $eq: ["$color","red"] }, { $not : "$color" }]},
             then: "$$DESCEND",
             else: "$$PRUNE"
         }
    }}]);

$redact "restricts the contents of the documents based on information stored in the documents themselves". So it will run only inside of the document. It basically scans your document top to the bottom, and checks if it matches with your if condition which is in $cond, if there is match it will either keep the content($$DESCEND) or remove($$PRUNE).

In the example above, first $match returns the whole shapes array, and $redact strips it down to the expected result.

Note that {$not:"$color"} is necessary, because it will scan the top document as well, and if $redact does not find a color field on the top level this will return false that might strip the whole document which we don't want.


perfect answer. As you mentioned $unwind will consume lot of RAM. So this will be better when compared.
I have a doubt. In the example, "shapes" is an array. Will "$redact" scan all the objects in the "shapes" array ?? How this will be good with respect to performance??
not all of it, but the result of your first match. That is the reason why you put $match as your first aggregate stage
okkk.. if an index created on "color" field, even then it will scan all the objects in the "shapes" array??? Which could be the efficient way of matching multiple objects in an array???
Brilliant! I do not understand how $eq works here. I left it off originally and this didn't work for me. Somehow, it looks in the array of shapes to find the match, but the query never specifies which array to look in. Like, if the documents had shapes and, for example, sizes; would $eq look in both arrays for matches? Is $redact just looking for anything within the document that matches the 'if' condition?
E
Egor Neliuba

Better you can query in matching array element using $slice is it helpful to returning the significant object in an array.

db.test.find({"shapes.color" : "blue"}, {"shapes.$" : 1})

$slice is helpful when you know the index of the element, but sometimes you want whichever array element matched your criteria. You can return the matching element with the $ operator.


Does it return all documents that contain shapes.color : blue or just the first one?
V
Viral Patel
 db.getCollection('aj').find({"shapes.color":"red"},{"shapes.$":1})

OUTPUTS

{

   "shapes" : [ 
       {
           "shape" : "circle",
           "color" : "red"
       }
   ]
}

thanks for the query, but it is just returning the first one even though the condition is matching for multiple elements in the array, any suggestion?
J
Jinxcat

The syntax for find in mongodb is

    db.<collection name>.find(query, projection);

and the second query that you have written, that is

    db.test.find(
    {shapes: {"$elemMatch": {color: "red"}}}, 
    {"shapes.color":1})

in this you have used the $elemMatch operator in query part, whereas if you use this operator in the projection part then you will get the desired result. You can write down your query as

     db.users.find(
     {"shapes.color":"red"},
     {_id:0, shapes: {$elemMatch : {color: "red"}}})

This will give you the desired result.


This works for me. However, It appears that "shapes.color":"red" in the query parameter (the first parameter of the find method) is not necessary. You can replace it with {} and get the same results.
@ErikOlson Your suggestion is right in the above case, where we need to find all the document that with red color and to apply the projection on them only. But let's say if somebody requires to find out all the document that have color blue but it should return only those element of that shapes array that have color red. In this case the above query can be referenced by somebody else also..
This seems to be the easiest, but I can't make it work it. It only returns the first matching subdocument.
@MahmoodHussain This answer is almost 7 years old, so may be version issue. Can you check latest documentation. I will try to run similar on latest version and share my findings. Can you explain what exactly you are trying to achieve ?
@Vicky Patient.find( { user: req.user._id, _id: req.params.patientId, "tests.test": req.params.testId, }, { "tests.$": 1, name: 1, } ) .populate({ path: "tests", populate: { path: "test", model: "Test", }, }) .exec((err, patient) => { if (err || !patient) { return res.status(404).send({ error: { message: err } }); } return res.send({ patient }); }); But then populate is throwing an error
C
Community

Thanks to JohnnyHK.

Here I just want to add some more complex usage.

// Document 
{ 
"_id" : 1
"shapes" : [
  {"shape" : "square",  "color" : "red"},
  {"shape" : "circle",  "color" : "green"}
  ] 
} 

{ 
"_id" : 2
"shapes" : [
  {"shape" : "square",  "color" : "red"},
  {"shape" : "circle",  "color" : "green"}
  ] 
} 


// The Query   
db.contents.find({
    "_id" : ObjectId(1),
    "shapes.color":"red"
},{
    "_id": 0,
    "shapes" :{
       "$elemMatch":{
           "color" : "red"
       } 
    }
}) 


//And the Result

{"shapes":[
    {
       "shape" : "square",
       "color" : "red"
    }
]}

a
azhar

You just need to run query

db.test.find(
{"shapes.color": "red"}, 
{shapes: {$elemMatch: {color: "red"}}});

output of this query is

{
    "_id" : ObjectId("562e7c594c12942f08fe4192"),
    "shapes" : [ 
        {"shape" : "circle", "color" : "red"}
    ]
}

as you expected it'll gives the exact field from array that matches color:'red'.


t
turivishal

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
    }
  }
)

can you pls describe that this accomplishes with an input and output set?
a
ashishSober

Likewise you can find for the multiple

db.getCollection('localData').aggregate([
    // Get just the docs that contain a shapes element where color is 'red'
  {$match: {'shapes.color': {$in : ['red','yellow'] } }},
  {$project: {
     shapes: {$filter: {
        input: '$shapes',
        as: 'shape',
        cond: {$in: ['$$shape.color', ['red', 'yellow']]}
     }}
  }}
])

This answer is indeed the prefered 4.x way: $match to cut down the space, then $filter to keep what you want, overwriting the input field (use output of $filter on field shapes to $project back on to shapes. Style note: best not to use the field name as the as argument because that can lead to confusion later with $$shape and $shape. I prefer zz as the as field because it really stands out.
S
Suraj Rao
db.test.find( {"shapes.color": "red"}, {_id: 0})

Welcome to Stack Overflow! Thank you for the code snippet, which might provide some limited, immediate help. A proper explanation would greatly improve its long-term value by describing why this is a good solution to the problem, and would make it more useful to future readers with other similar questions. Please edit your answer to add some explanation, including the assumptions you've made.
K
KARTHIKEYAN.A

Use aggregation function and $project to get specific object field in document

db.getCollection('geolocations').aggregate([ { $project : { geolocation : 1} } ])

result:

{
    "_id" : ObjectId("5e3ee15968879c0d5942464b"),
    "geolocation" : [ 
        {
            "_id" : ObjectId("5e3ee3ee68879c0d5942465e"),
            "latitude" : 12.9718313,
            "longitude" : 77.593551,
            "country" : "India",
            "city" : "Chennai",
            "zipcode" : "560001",
            "streetName" : "Sidney Road",
            "countryCode" : "in",
            "ip" : "116.75.115.248",
            "date" : ISODate("2020-02-08T16:38:06.584Z")
        }
    ]
}

A
Ali

Although the question was asked 9.6 years ago, this has been of immense help to numerous people, me being one of them. Thank you everyone for all your queries, hints and answers. Picking up from one of the answers here.. I found that the following method can also be used to project other fields in the parent document.This may be helpful to someone.

For the following document, the need was to find out if an employee (emp #7839) has his leave history set for the year 2020. Leave history is implemented as an embedded document within the parent Employee document.

db.employees.find( {"leave_history.calendar_year": 2020}, 
    {leave_history: {$elemMatch: {calendar_year: 2020}},empno:true,ename:true}).pretty()


{
        "_id" : ObjectId("5e907ad23997181dde06e8fc"),
        "empno" : 7839,
        "ename" : "KING",
        "mgrno" : 0,
        "hiredate" : "1990-05-09",
        "sal" : 100000,
        "deptno" : {
                "_id" : ObjectId("5e9065f53997181dde06e8f8")
        },
        "username" : "none",
        "password" : "none",
        "is_admin" : "N",
        "is_approver" : "Y",
        "is_manager" : "Y",
        "user_role" : "AP",
        "admin_approval_received" : "Y",
        "active" : "Y",
        "created_date" : "2020-04-10",
        "updated_date" : "2020-04-10",
        "application_usage_log" : [
                {
                        "logged_in_as" : "AP",
                        "log_in_date" : "2020-04-10"
                },
                {
                        "logged_in_as" : "EM",
                        "log_in_date" : ISODate("2020-04-16T07:28:11.959Z")
                }
        ],
        "leave_history" : [
                {
                        "calendar_year" : 2020,
                        "pl_used" : 0,
                        "cl_used" : 0,
                        "sl_used" : 0
                },
                {
                        "calendar_year" : 2021,
                        "pl_used" : 0,
                        "cl_used" : 0,
                        "sl_used" : 0
                }
        ]
}

R
Rafiq

if you want to do filter, set and find at the same time.

let post = await Post.findOneAndUpdate(
          {
            _id: req.params.id,
            tasks: {
              $elemMatch: {
                id: req.params.jobId,
                date,
              },
            },
          },
          {
            $set: {
              'jobs.$[i].performer': performer,
              'jobs.$[i].status': status,
              'jobs.$[i].type': type,
            },
          },
          {
            arrayFilters: [
              {
                'i.id': req.params.jobId,
              },
            ],
            new: true,
          }
        );