Return object without its key in MongoDb
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
add a comment |
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
add a comment |
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
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
mongodb mongodb-query
edited Nov 13 '18 at 21:33
Cody Patterson
asked Nov 13 '18 at 21:21
Cody PattersonCody Patterson
747
747
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
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: "$first.anotherKey",
])
I like your answer, but unfortunately I'm using 3.2 and$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
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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: "$first.anotherKey",
])
I like your answer, but unfortunately I'm using 3.2 and$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
add a comment |
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: "$first.anotherKey",
])
I like your answer, but unfortunately I'm using 3.2 and$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
add a comment |
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",
])
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: "$first.anotherKey",
])
edited Nov 13 '18 at 21:34
answered Nov 13 '18 at 21:22
micklmickl
14k51536
14k51536
I like your answer, but unfortunately I'm using 3.2 and$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 $project, modified my answer
– mickl
Nov 13 '18 at 21:37
Thank you. This solved the issue.
– Cody Patterson
Nov 13 '18 at 21:44
add a comment |
I like your answer, but unfortunately I'm using 3.2 and$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
add a comment |
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