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My MongoDB document is 

{
"_id" : ObjectId("59f9850e44b0e37ffba68899"),
"code":32,
"sweet" : {
    "reddy" : 0.0,
    "sekhar" : 1.0,
    "praveen" : 1.0,
    "all" : 1.0
},
"salt" : {
    "reddy" : 0.0,
    "sekhar" : 1.0,
    "praveen" : 1.0,
    "all" : 1.0
},
"pepper" : {
    "reddy" : 0.0,
    "sekhar" : 1.0,
    "praveen" : 1.0,
    "all" : 1.0
},
"All" : {
    "reddy" : 0.0,
    "sekhar" : 1.0,
    "praveen" : 1.0,
    "all" : 1.0
}
}

How can i get the only one particular Object field values without writing the new case classe for Objects.

I have tried to get them for particular Object following queries in mongo command prompt

db.test.find({code:{$in:[32]},{new: 0 }}).pretty();

db.test.find({code:{$in:[32]},{new: 1}}).pretty();

db.test.find({code:{$in:[32]},{"new": 1}}).pretty();

db.test.find({code:{$in:[32]},{"new": 0}}).pretty();
cchantep
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newbie
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  • Do you mean `db.test.find({ code: 32 },{ "pepper": 1 })`? Not really clear what you are asking. Perhaps, "show your expected result" – Neil Lunn Nov 01 '17 at 08:41
  • This is my expected output { "_id" : ObjectId("59f981dd44b0e37ffba68898"), "pepper" : { "reddy" : 0, "sekhar" : 1, "praveen" : 1, "all" : 1 } } – newbie Nov 01 '17 at 09:40
  • db.test.find({ code: 32 },{ "pepper": 1 }) it's working fine Thank you – newbie Nov 01 '17 at 09:42

0 Answers0