How to get only virtual attributes. Node.JS + Mongoose + MongoDB

Question

Consider I have a Mongoose schema defined

var Schema_MySchema = new mongoose.Schema({
Field1: String, Field2: String
});

I have added virtual attribute & setting options in the following way:

Schema_MySchema.virtual('USERNAME').get(function () {
    return this.Field1;
});

Schema_MySchema.set('toJSON', { virtuals: true });

Using the schema object to retrieve the data from MongoDB as follows:

var Mod_Obj = mongoose.model('SchemaName', Schema_MySchema);
var Model_Instance = new Mod_Obj();
Model_Instance.find({}, function (err, docs) {
   /* 
      Here I get a object docs with following structure:
      [{ _id: ObjectId("511XXXdff9c4c419000008"), 
          Field1: 'SOMEvalueFromMongoDB_1', 
          Field2: 'SOMEvalueFromMongoDB_2',
          USERNAME: 'SOMEvalueFromMongoDB_1'
      }]
  */
});

Now I want to remove actual attribute returned from MongoDB, say I want to remove Field1 from docs

I tried following ways to remove:

1. delete docs.Field1;

2. var Json_Obj = docs.toJSON(); delete Json_Obj.Field1;

3. var Json_Obj = docs.toObject({ virtuals: true }); delete Json_Obj.Field1;

4. delete docs[0].Field1;

5. delete docs[0]['Field1'];

All the ways didn't work. :-(

If execute for testing on simple JSON it worked:

var a = { 'A' : 1, 'B': 2 };
delete a.A;
console.log(a); //prints only object with B attribute only. i.e. { B: 2 }

Could anyone tell me whats wrong here?

Thanks in Advance...

did you try `delete docs['Field1']` and what what was output? — Mithun Satheesh, Feb 08 '13 at 07:51

score 2 · Accepted Answer · edited May 23 '17 at 10:28

I believe you are missing the index there as docs is an array.

if your docs has the following structure

[{ _id: ObjectId("511XXXdff9c4c419000008"), 
      Field1: 'SOMEvalueFromMongoDB_1', 
      Field2: 'SOMEvalueFromMongoDB_2',
      USERNAME: 'SOMEvalueFromMongoDB_1'
  }]

the deletion code should be

delete docs[index]['Field1']

ie. to delete Field1 of 1st element

delete docs[0]['Field1']

UPDATED TO COMMENTS:

please refer to this answer on SO.

Here you will need to do like

var k = docs[index].toObject();
delete k['Field1'];

score 1 · Answer 2 · answered Feb 08 '13 at 08:38

You have to save the doc after field is deleted, Try as follows

var Mod_Obj = mongoose.model('SchemaName', Schema_MySchema);
var Model_Instance = new Mod_Obj();
Model_Instance.find({}, function (err, docs) {
   /* 
      Here I get a object docs with following structure:
      [{ _id: ObjectId("511XXXdff9c4c419000008"), 
          Field1: 'SOMEvalueFromMongoDB_1', 
          Field2: 'SOMEvalueFromMongoDB_2',
          USERNAME: 'SOMEvalueFromMongoDB_1'
      }]
  */
delete docs[0].Field1;
docs[0].save(function(err){
if(err){
console.log(err);
}else{
console.log("Field1 is deleted");
}
});
});

score 1 · Answer 3 · answered Feb 08 '13 at 09:34

Found some solution. Here is the way to do it with Mongoose

schema.options.toObject.transform = function (doc, ret, options) {
  // remove the _id of every document before returning the result
  delete ret._id;
}

Go to Document.toObject([options]) then scroll to topic Transform

score 0 · Answer 4 · answered Apr 15 '21 at 04:27

0

You need update model to return it.

toJSON: {
    virtuals: true,
    transform: function(doc, ret) {
         ret.id = ret._id;
         delete ret._id;
      }
}

answered Apr 15 '21 at 04:27

Nguyễn Văn Linh

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How to get only virtual attributes. Node.JS + Mongoose + MongoDB

4 Answers4