How to ignore duplicate "_id" and insert unique "_id" in mongodb - mongodb

Question

I have a JSON structure at first insertion as
collection.insert(query.query, function(err, docs) {
callback(err,docs);
dbCon.close();
});
JSON Structure:
"employees":[
{"_id":1, "lastName":"Doe"},
{"_id":2, "lastName":"Smith"},
{"_id":3,"lastName":"Jones"}
]
When I first insert this JSON into mongodb, it gets inserted without any errors.
When I try insert again on the same database, with below JSON
"employees":[
{"_id":2, "lastName":"Smith"},
{"_id":5, "lastName":"Peter"},
{"_id":6,"``lastName":"James"}
]
Now the mondo db is throwing duplicate key error.
Is there any way where I can omit the "_id":2 and insert "_id":5 and "_id":6 into mongo?
Please help me..

Answer 1

You can perform an upsert in your case :
db.people.update(
{ name: "Andy" },
{
name: "Andy",
rating: 1,
score: 1
},
{ upsert: true }
)
If the first query gets a blank collection, that it will insert the data in this case if a Recrod with name Andy is not there, it will insert one.

Answer 2

In the first time, you insert a document
{"_id":2, "lastName":"Smith"}.
In the second time , you insert a document {"_id":2, "lastName":"Smith"}.
Their ids are the same ,not unique.
Suggest create id automatically by mongdb.
You can add the following code:
collection.findOne({"_id":2}.exec(function(err,result){
if(!result){
collection.insert(XXX);
}
}));