Question

My fetched DBObject looks like following

{ "_id" : { "$oid" : "50c28ac1de86acf0bdfbeca0"} , 
"schedule" : { "startAt" : 1354926785198 , "endAt" : 1391155200000 , "repeatForever" : true , "interval" : 3600} , "destination" : "Storage-East"}

I want to extract JSON string sans “_id” so that I can deserialize it back to my Java object. If I use following I can remove the ‘_id’ field and it is fine to get the Java object back from JSON String. Is there any other elegant way of doing this ?

dbObj.removeField("_id");
String jsonString = dbObj.toString();

// Now readValue from the json string 

thanks.

Answer 1

Instead of removing the data afterwords, just use results projections. You simply remove the _id by using the result projections in a find statement:

//find all documents where destination equals Storage-East, but exclude _id from result set
db.inventory.find( { "destination": 'Storage-East' }, { _id:0 } )

You can find the documentation http://docs.mongodb.org/manual/core/read-operations/#result-projections.