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Home/ Questions/Q 8342017
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Editorial Team
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Editorial Team
Asked: June 9, 20262026-06-09T05:39:08+00:00 2026-06-09T05:39:08+00:00

I have this simple db, with subset of elements: { _id : ObjectId(5019eb2356d80cd005000000), photo

  • 0

I have this simple db, with subset of elements:

{ "_id" : ObjectId("5019eb2356d80cd005000000"),
    "photo" : "/pub/photos/file1.jpg",
    "comments" : [
        {
            "name" : "mike",
            "message" : "hello to all"
        },
        {
            "name" : "pedro",
            "message" : "hola a todos"
        }
    ]
},
{ "_id" : ObjectId("5019eb4756d80cd005000001"),
    "photo" : "/pub/photos/file2.jpg",
    "comments" : [
        {
            "name" : "luca",
            "message" : "ciao a tutti"
        },
        {
            "name" : "stef",
            "message" : "todos bien"
        },
        {
            "name" : "joice",
            "message" : "vamos a las playa"
        }
    ]
}

when I execute a subset find:
db.photos.find({},{“comments.name”:1})

Im receive this structure:

[
    {
        "_id" : ObjectId("5019eb2356d80cd005000000"),
        "comments" : [
            {
                "name" : "mike"
            },
            {
                "name" : "pedro"
            }
        ]
    },
    {
        "_id" : ObjectId("5019eb4756d80cd005000001"),
        "comments" : [
            {
                "name" : "luca"
            },
            {
                "name" : "stef"
            },
            {
                "name" : "joice"
            }
        ]
    }
]

But I want to get a simple one-dimensional array, like this(or similar):

[
    {
        "name" : "mike"
    },
    {
        "name" : "pedro"
    },
    {
        "name" : "luca"
    },
    {
        "name" : "stef"
    },
    {
        "name" : "joice"
    }
]

I need to implement this query with mongo php official driver, but the language is not important, I just want to understand by what logic can I accomplish this by mongo shell

tnk!

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-09T05:39:09+00:00Added an answer on June 9, 2026 at 5:39 am

    The easiest option would be to use distinct():

    >db.photos.distinct("comments.name");
    [ "mike", "pedro", "joice", "luca", "stef" ]
    

    Here’s another example using JavaScript:

    // Array to save results
    > var names = []
    
    // Find comments and save names
    > db.photos.find({},{"comments.name":1}).forEach(
              function(doc) { doc.comments.forEach(
                  function(comment) {names.push(comment.name)})
              })
    
    // Check the results
    > names
    [ "mike", "pedro", "luca", "stef", "joice" ]
    

    Here’s an example using the new aggregation framework in the upcoming MongoDB 2.2:

    db.photos.aggregate(
      { $unwind : "$comments" },
      { $group : {
         _id: "names",
         names: { $addToSet : "$comments.name" }
      }},
      { $project : {
         '_id' : 0,
         'names' : 1,
      }}
    )
    
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