Aggregate query result in mongodb












0














I have collection with documents like this one:



{  
"_id": 1,
"people": [
{
"name": "Bob",
"age": "15"
},
{
"name": "Alice",
"age": "18"
}
]
}


My query is:



db.groups.aggregate({ $match: { "_id": 1 }}, { $project: { "_id": 0, "people.name": 1 } })


This query returns:



{  
"people": [
{
"name": "Bob"
},
{
"name": "Alice"
}
]
}


But I need the result like:



{ "names": [ "Bob", "Alice" ] }


Which parameters should I add to the .aggregate() function?










share|improve this question


















  • 1




    No need to use $map here... Try this .groups.aggregate({ $match: { "_id": 1 }}, { $project: { "_id": 0, "names": "$people.name" } })
    – Anthony Winzlet
    Nov 23 '18 at 15:16












  • Thanks! It's helped
    – TheYarik
    Nov 23 '18 at 15:22
















0














I have collection with documents like this one:



{  
"_id": 1,
"people": [
{
"name": "Bob",
"age": "15"
},
{
"name": "Alice",
"age": "18"
}
]
}


My query is:



db.groups.aggregate({ $match: { "_id": 1 }}, { $project: { "_id": 0, "people.name": 1 } })


This query returns:



{  
"people": [
{
"name": "Bob"
},
{
"name": "Alice"
}
]
}


But I need the result like:



{ "names": [ "Bob", "Alice" ] }


Which parameters should I add to the .aggregate() function?










share|improve this question


















  • 1




    No need to use $map here... Try this .groups.aggregate({ $match: { "_id": 1 }}, { $project: { "_id": 0, "names": "$people.name" } })
    – Anthony Winzlet
    Nov 23 '18 at 15:16












  • Thanks! It's helped
    – TheYarik
    Nov 23 '18 at 15:22














0












0








0


0





I have collection with documents like this one:



{  
"_id": 1,
"people": [
{
"name": "Bob",
"age": "15"
},
{
"name": "Alice",
"age": "18"
}
]
}


My query is:



db.groups.aggregate({ $match: { "_id": 1 }}, { $project: { "_id": 0, "people.name": 1 } })


This query returns:



{  
"people": [
{
"name": "Bob"
},
{
"name": "Alice"
}
]
}


But I need the result like:



{ "names": [ "Bob", "Alice" ] }


Which parameters should I add to the .aggregate() function?










share|improve this question













I have collection with documents like this one:



{  
"_id": 1,
"people": [
{
"name": "Bob",
"age": "15"
},
{
"name": "Alice",
"age": "18"
}
]
}


My query is:



db.groups.aggregate({ $match: { "_id": 1 }}, { $project: { "_id": 0, "people.name": 1 } })


This query returns:



{  
"people": [
{
"name": "Bob"
},
{
"name": "Alice"
}
]
}


But I need the result like:



{ "names": [ "Bob", "Alice" ] }


Which parameters should I add to the .aggregate() function?







mongodb aggregate






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Nov 23 '18 at 14:39









TheYarik

66




66








  • 1




    No need to use $map here... Try this .groups.aggregate({ $match: { "_id": 1 }}, { $project: { "_id": 0, "names": "$people.name" } })
    – Anthony Winzlet
    Nov 23 '18 at 15:16












  • Thanks! It's helped
    – TheYarik
    Nov 23 '18 at 15:22














  • 1




    No need to use $map here... Try this .groups.aggregate({ $match: { "_id": 1 }}, { $project: { "_id": 0, "names": "$people.name" } })
    – Anthony Winzlet
    Nov 23 '18 at 15:16












  • Thanks! It's helped
    – TheYarik
    Nov 23 '18 at 15:22








1




1




No need to use $map here... Try this .groups.aggregate({ $match: { "_id": 1 }}, { $project: { "_id": 0, "names": "$people.name" } })
– Anthony Winzlet
Nov 23 '18 at 15:16






No need to use $map here... Try this .groups.aggregate({ $match: { "_id": 1 }}, { $project: { "_id": 0, "names": "$people.name" } })
– Anthony Winzlet
Nov 23 '18 at 15:16














Thanks! It's helped
– TheYarik
Nov 23 '18 at 15:22




Thanks! It's helped
– TheYarik
Nov 23 '18 at 15:22












1 Answer
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oldest

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0














The solution is:



db.groups.aggregate({ $match: { "_id": 1 }}, { $project: { "_id": 0, "names": "$people.name" } })





share|improve this answer























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    1 Answer
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    active

    oldest

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    0














    The solution is:



    db.groups.aggregate({ $match: { "_id": 1 }}, { $project: { "_id": 0, "names": "$people.name" } })





    share|improve this answer




























      0














      The solution is:



      db.groups.aggregate({ $match: { "_id": 1 }}, { $project: { "_id": 0, "names": "$people.name" } })





      share|improve this answer


























        0












        0








        0






        The solution is:



        db.groups.aggregate({ $match: { "_id": 1 }}, { $project: { "_id": 0, "names": "$people.name" } })





        share|improve this answer














        The solution is:



        db.groups.aggregate({ $match: { "_id": 1 }}, { $project: { "_id": 0, "names": "$people.name" } })






        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Nov 23 '18 at 15:23

























        answered Nov 23 '18 at 15:17









        TheYarik

        66




        66






























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