Return object without its key in MongoDb










1















I am using MongoDb 3.2



Say I have this document in a MongoDb collection:




"_id" : ObjectId("5bad65b9777f6df3ce840fd1"),
"entryCode" : "1234",
"first" :
"someKey" : "x",
"anotherKey" : "y"
,
"second" :
"someKey" : "u",
"anotherKey" : "v",




When I execute the following query:
db.collection.find(entryCode:"1234",_id:0, first:1)



I get this returned result:




"first" :
"someKey" : "x",
"anotherKey" : "y"




However, what I'd like to return is this:




"someKey" : "x",
"anotherKey" : "y"



Notice I do not want the key first to be a part of the returned value. Just the object value within. What Mongo query can I use?










share|improve this question




























    1















    I am using MongoDb 3.2



    Say I have this document in a MongoDb collection:




    "_id" : ObjectId("5bad65b9777f6df3ce840fd1"),
    "entryCode" : "1234",
    "first" :
    "someKey" : "x",
    "anotherKey" : "y"
    ,
    "second" :
    "someKey" : "u",
    "anotherKey" : "v",




    When I execute the following query:
    db.collection.find(entryCode:"1234",_id:0, first:1)



    I get this returned result:




    "first" :
    "someKey" : "x",
    "anotherKey" : "y"




    However, what I'd like to return is this:




    "someKey" : "x",
    "anotherKey" : "y"



    Notice I do not want the key first to be a part of the returned value. Just the object value within. What Mongo query can I use?










    share|improve this question


























      1












      1








      1








      I am using MongoDb 3.2



      Say I have this document in a MongoDb collection:




      "_id" : ObjectId("5bad65b9777f6df3ce840fd1"),
      "entryCode" : "1234",
      "first" :
      "someKey" : "x",
      "anotherKey" : "y"
      ,
      "second" :
      "someKey" : "u",
      "anotherKey" : "v",




      When I execute the following query:
      db.collection.find(entryCode:"1234",_id:0, first:1)



      I get this returned result:




      "first" :
      "someKey" : "x",
      "anotherKey" : "y"




      However, what I'd like to return is this:




      "someKey" : "x",
      "anotherKey" : "y"



      Notice I do not want the key first to be a part of the returned value. Just the object value within. What Mongo query can I use?










      share|improve this question
















      I am using MongoDb 3.2



      Say I have this document in a MongoDb collection:




      "_id" : ObjectId("5bad65b9777f6df3ce840fd1"),
      "entryCode" : "1234",
      "first" :
      "someKey" : "x",
      "anotherKey" : "y"
      ,
      "second" :
      "someKey" : "u",
      "anotherKey" : "v",




      When I execute the following query:
      db.collection.find(entryCode:"1234",_id:0, first:1)



      I get this returned result:




      "first" :
      "someKey" : "x",
      "anotherKey" : "y"




      However, what I'd like to return is this:




      "someKey" : "x",
      "anotherKey" : "y"



      Notice I do not want the key first to be a part of the returned value. Just the object value within. What Mongo query can I use?







      mongodb mongodb-query






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Nov 13 '18 at 21:33







      Cody Patterson

















      asked Nov 13 '18 at 21:21









      Cody PattersonCody Patterson

      747




      747






















          1 Answer
          1






          active

          oldest

          votes


















          3














          You need $replaceRoot operator, try:



          db.col.aggregate([
          $match: entryCode:"1234" ,

          $replaceRoot:
          newRoot: "$first"


          ])


          EDIT: In MongoDB 3.2 you can only use $project and specify fields explicitly:



          db.col.aggregate([
          $match: entryCode:"1234" ,

          $project:
          _id: 0,
          someKey: "$first.someKey",
          anotherKey: "$first.anotherKey",


          ])





          share|improve this answer

























          • I like your answer, but unfortunately I'm using 3.2 and $replaceRoot is only available in 3.4. I've edited my question to clarify.

            – Cody Patterson
            Nov 13 '18 at 21:34











          • @CodyPatterson okay so then you have to use $project, modified my answer

            – mickl
            Nov 13 '18 at 21:37











          • Thank you. This solved the issue.

            – Cody Patterson
            Nov 13 '18 at 21:44










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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3














          You need $replaceRoot operator, try:



          db.col.aggregate([
          $match: entryCode:"1234" ,

          $replaceRoot:
          newRoot: "$first"


          ])


          EDIT: In MongoDB 3.2 you can only use $project and specify fields explicitly:



          db.col.aggregate([
          $match: entryCode:"1234" ,

          $project:
          _id: 0,
          someKey: "$first.someKey",
          anotherKey: "$first.anotherKey",


          ])





          share|improve this answer

























          • I like your answer, but unfortunately I'm using 3.2 and $replaceRoot is only available in 3.4. I've edited my question to clarify.

            – Cody Patterson
            Nov 13 '18 at 21:34











          • @CodyPatterson okay so then you have to use $project, modified my answer

            – mickl
            Nov 13 '18 at 21:37











          • Thank you. This solved the issue.

            – Cody Patterson
            Nov 13 '18 at 21:44















          3














          You need $replaceRoot operator, try:



          db.col.aggregate([
          $match: entryCode:"1234" ,

          $replaceRoot:
          newRoot: "$first"


          ])


          EDIT: In MongoDB 3.2 you can only use $project and specify fields explicitly:



          db.col.aggregate([
          $match: entryCode:"1234" ,

          $project:
          _id: 0,
          someKey: "$first.someKey",
          anotherKey: "$first.anotherKey",


          ])





          share|improve this answer

























          • I like your answer, but unfortunately I'm using 3.2 and $replaceRoot is only available in 3.4. I've edited my question to clarify.

            – Cody Patterson
            Nov 13 '18 at 21:34











          • @CodyPatterson okay so then you have to use $project, modified my answer

            – mickl
            Nov 13 '18 at 21:37











          • Thank you. This solved the issue.

            – Cody Patterson
            Nov 13 '18 at 21:44













          3












          3








          3







          You need $replaceRoot operator, try:



          db.col.aggregate([
          $match: entryCode:"1234" ,

          $replaceRoot:
          newRoot: "$first"


          ])


          EDIT: In MongoDB 3.2 you can only use $project and specify fields explicitly:



          db.col.aggregate([
          $match: entryCode:"1234" ,

          $project:
          _id: 0,
          someKey: "$first.someKey",
          anotherKey: "$first.anotherKey",


          ])





          share|improve this answer















          You need $replaceRoot operator, try:



          db.col.aggregate([
          $match: entryCode:"1234" ,

          $replaceRoot:
          newRoot: "$first"


          ])


          EDIT: In MongoDB 3.2 you can only use $project and specify fields explicitly:



          db.col.aggregate([
          $match: entryCode:"1234" ,

          $project:
          _id: 0,
          someKey: "$first.someKey",
          anotherKey: "$first.anotherKey",


          ])






          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Nov 13 '18 at 21:34

























          answered Nov 13 '18 at 21:22









          micklmickl

          14k51536




          14k51536












          • I like your answer, but unfortunately I'm using 3.2 and $replaceRoot is only available in 3.4. I've edited my question to clarify.

            – Cody Patterson
            Nov 13 '18 at 21:34











          • @CodyPatterson okay so then you have to use $project, modified my answer

            – mickl
            Nov 13 '18 at 21:37











          • Thank you. This solved the issue.

            – Cody Patterson
            Nov 13 '18 at 21:44

















          • I like your answer, but unfortunately I'm using 3.2 and $replaceRoot is only available in 3.4. I've edited my question to clarify.

            – Cody Patterson
            Nov 13 '18 at 21:34











          • @CodyPatterson okay so then you have to use $project, modified my answer

            – mickl
            Nov 13 '18 at 21:37











          • Thank you. This solved the issue.

            – Cody Patterson
            Nov 13 '18 at 21:44
















          I like your answer, but unfortunately I'm using 3.2 and $replaceRoot is only available in 3.4. I've edited my question to clarify.

          – Cody Patterson
          Nov 13 '18 at 21:34





          I like your answer, but unfortunately I'm using 3.2 and $replaceRoot is only available in 3.4. I've edited my question to clarify.

          – Cody Patterson
          Nov 13 '18 at 21:34













          @CodyPatterson okay so then you have to use $project, modified my answer

          – mickl
          Nov 13 '18 at 21:37





          @CodyPatterson okay so then you have to use $project, modified my answer

          – mickl
          Nov 13 '18 at 21:37













          Thank you. This solved the issue.

          – Cody Patterson
          Nov 13 '18 at 21:44





          Thank you. This solved the issue.

          – Cody Patterson
          Nov 13 '18 at 21:44



















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