问题
Aggregate, $unwind and $group is not my solution as they make query very slow, there for I am looking to get my record by db.collection.find() method.
The problem is that I need more then one value from sub array. For example from the following example I want to get the "type" : "exam" and "type" : "quiz" elements.
{
"_id" : 22,
"scores" : [
{
"type" : "exam",
"score" : 75.04996547553947
},
{
"type" : "quiz",
"score" : 10.23046475899236
},
{
"type" : "homework",
"score" : 96.72520512117761
},
{
"type" : "homework",
"score" : 6.488940333376703
}
]
}
I am looking something like
db.students.find(
// Search criteria
{ '_id': 22 },
// Projection
{ _id: 1, scores: { $elemMatch: { type: 'exam', type: 'quiz' } }}
)
The result should be like
{ "_id": 22, "scores" : [ { "type" : "exam", "type" : "quiz" } ] }
But this over ride the type: 'exam' and returns only type: 'quiz'. Have anybody any idea how to do this with db.find()?
回答1:
This is not possible directly using find
and elemMatch
because of following limitation of elemMatch
and mongo array fields
.
The $elemMatch operator limits the contents of an field from the query results to contain only the first element matching the $elemMatch condition. ref. from $elemMacth
and mongo array field limitations as below
Only one positional $ operator may appear in the projection document.
The query document should only contain a single condition on the array field being projected. Multiple conditions may override each other internally and lead to undefined behavior. ref from mongo array field limitations
So either you tried following this to find out only exam
or quiz
db.collectionName.find({"_id":22,"scores":{"$elemMatch":{"type":"exam"}}},{"scores.$.type":1}).pretty()
is shows only exam
scores array.
Otherwise you should go through aggregation
来源:https://stackoverflow.com/questions/29672035/extract-two-sub-array-values-in-mongodb-by-elemmatch