CODEWITHSUNDEEP
mongodbmongooseaggregation-framework
monad98
monad98

Reputation: 291

mongodb aggregation: group by _id

After aggregation I got this result. The _id field is each class's id and I applied {$group:{_id:"$_id", .....

[{ _id: 54c977314f5293b74ea54f96, subject: 'math' }, score: 73.5335 },
{ _id: 54c977314f5293b74ea54f96, subject: 'science' }, score: 56.2192 },
{ _id: 54c977314f5293b74ea54f96, subject: 'history' }, score: 82.8821 },
{ _id: 54c974bdff0d993b4ecf34ce, subject: 'math' }, score: 68.2598 },
{ _id: 54c974bdff0d993b4ecf34ce, subject: 'science' }, score: 77.8712 },
{ _id: 54c974bdff0d993b4ecf34ce, subject: 'history' }, score: 71.9218 },
                                 ...                                  ]

The problem is _id value is not string type. so when I send JSON to client, it says malformed JSON.

How can I get string type _id? Thanks in advance.

Upvotes: 2

Views: 1607

Answers (2)

Larry Battle
Larry Battle

Reputation: 9178

Don't form the JSON string yourself. Use JSON.stringify( output_from_aggregation ).

However, if you don't want do that, then do this.

  • Make sure to form strings wrapped in double quotes ("), and not single quotes ('). Strings in single quotes are invalid in JSON.
  • You need to wrap the _id value in a string. It's an invalid hex number since it's too large to parse.

Old code:

...
{ _id: 54c974bdff0d993b4ecf34ce, subject: 'science' }, score: 77.8712 },
...

New Code:

...
{ _id: "54c974bdff0d993b4ecf34ce", subject: "science" }, score: 77.8712 },
...

More info:

Upvotes: 2

Dev
Dev

Reputation: 13753

In your case, _id is automatically generated and is of BSON (Binary JSON) type. You can manually choose _id as a String and write its value. For example

db.collection.insert( { _id: "10", item: "box", qty: 20 } )

you can see documentation for more detail on _id field http://docs.mongodb.org/manual/reference/method/db.collection.insert/

Upvotes: 1

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