How do I fetch records matching output of aggregate function in mongoDB?

Question

I have queried mongodb by using aggregate function and got two fields in the output. The result of my db.Collection.aggregate(..) looks like the below:

{ 
  "_id" : NumberLong(203440), 
  "date" : ISODate("2013-05-11T00:00:00Z")
}
{ 
  "_id" : NumberLong(203520), 
  "date" :ISODate("2013-01-05T00:00:00Z") 
} 
{ 
  "_id" : NumberLong(203970),
  "date": ISODate("2013-01-11T00:00:00Z") 
} 
{ 
  "_id" : NumberLong(203660),
  "date" : ISODate("2013-01-11T00:00:00Z")
} 
{ 
  "_id" : NumberLong(203360), 
  "date" : ISODate("2013-01-11T00:00:00Z")
}

How do I get the records in the collection for which these two fields are true?(in a single query)

i.e If each record in my collection has the fields data,_id, x, y, z, a , b and c, how do I fetch list of records for which the date and _id are equal to the above result of aggregate?

The value in the aggregation result "_id" is actually the field x in the document. — SHILPA AR
– SHILPA AR, Commented Sep 7, 2015 at 2:23
Please show the original document and the aggregation pipeline you have tried so far. — Blakes Seven
– Blakes Seven, Commented Sep 7, 2015 at 2:26

SHILPA AR · Accepted Answer · 2015-10-10 14:33:47Z

0

In the aggregation command, using the below in the $group part helped me.

"allFields": { "$first": "$$CURRENT" }

In my response, I got the entire document in "allFields" field.

answered Oct 10, 2015 at 14:33

SHILPA AR

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