Is it possible to flatten MongoDB result query?
I have a deeply nested collection in my MongoDB collection.
When I run the following query:
db.countries.findOne({},{'data.country.neighbor.name':1,'_id':0})
I end up with this nested result here:
{"data" : {
"country" : [
{
"neighbor" : [
{
"name" : "Austria"
},
{
"name" : "Switzerland"
}
]
},
{
"neighbor" : {
"name" : "Malaysia"
}
},
{
"neighbor" : [
{
"name" : "Costa Rica"
},
{
"name" : "Colombia"
}
]
}
]
}}
Now, this is what I want:
['Austria', 'Switzerland', 'Malaysia', 'Costa Rica', 'Colombia']
or this:
{'name':['Austria', 'Switzerland', 'Malaysia', 'Costa Rica', 'Colombia']}
or anything else similar... Is this possible?
You can use $project & $unwind & $group of aggregation framework to get the result closer to your requirement.
> db.countries.aggregate({$project:{a:'$data.country.neighbor.name'}},
{$unwind:'$a'},
{$unwind:'$a'},
{$group:{_id:'a',res:{$addToSet:'$a'}}})
{
"result" : [
{
"_id" : "a",
"res" : [
"Colombia",
"Malaysia",
"Switzerland",
"Costa Rica",
"Austria"
]
}
],
"ok" : 1
}
$unwind used twice since the name array is nested deep. And It will only work if the neighbor attribute is an array. In your example one neighbor field (Malaysia) is not an array
Done it much simpler way, maybe it is recent
db.countries.aggregate({$unwind:'$data.country.neighbor.name'})
It's pretty straightforward under the new aggregation framework. The $project and $unwind operation are right for the purpose.
Follow WeChat
Success story sharing
Want to stay one step ahead of the latest teleworks?
Subscribe Now相似问题
- Random record from MongoDB
- Find objects between two dates MongoDB
- How to query MongoDB with "like"
- Update MongoDB field using value of another field
- Query for documents where array size is greater than 1
- How do I drop a MongoDB database from the command line?
- How can I list all collections in the MongoDB shell?
- Merge/flatten an array of arrays
- Conditional $sum in MongoDB
{ "result" : [ ], "ok" : 1 }:/