How to remove mongo specific fields from result (NodeJS, Mongoose) - node.js

Question

I want to remove all Mongo specific fields (like '_id') from query result. Is there a simple method to do this or should I remove fields manually? If yes, then which are that fields and how to do that?
I'm using NodeJS and Mongoose

Answer 1

You can use select() method for remove the field from your query:
Model.find({}).select("-removed_field").then (resp => {
// your code
});
You should specified the "-" before field name, to be remove this field.
If you want remove several fields - you can specified their as array:
Model.find({}).select(["-removed_field1", "-removed_field2" ... ]).then (resp => {
// your code
});
Also you can to select only specified fields, using this method without "-"
Model.find({}).select(["field1", "field2" ... ]).then (resp => {
// your code
});

Answer 2

If you want hide _id property you can use text argument with prefix - which will exclude this or that field from the result, for get sepecifict fields you should pass like this:
Entity.find({ ... }, 'field1 field2', function(err, entity) {
console.log(entity); // { field1: '...', field2: '...' }
});

Answer 3

You can specify a field to be excluded from results by using the optional 2nd parameter projection string of the find method:
Model.find({}, "-a -b").then (res => {
// objects in the res array will all have the
// 'a' and 'b' fields excluded.
});
https://mongoosejs.com/docs/api.html#model_Model.find (see projection)

Answer 4

you can use mongoose instance method two show specific fields from all documents
const userSchema = new mongoose.Schema({
email: {
type: String,
},
name: {
type: String,
maxlength: 128,
index: true,
trim: true,
},
});
userSchema.method({
transform() {
const transformed = {};
const fields = ['name', 'email'];
fields.forEach((field) => {
transformed[field] = this[field];
});
return transformed;
},
});
module.exports = mongoose.model('User', userSchema);

Answer 5

if You want to remove any specific fields like _id, You can try in two ways:
Suppose Here you try to find a user using User Model
User.find({ email: email }, { _id: 0 });
OR
const user = User.find({ email: email });
delete user._doc._id;

Answer 6

OP mentioned "from result", as far as I understood, it means, removing from the query result i.e. query result will contain the field, but will be removed from the query result.
A SO answer here mentions, that to modify a query result (which are immutable), we've to convert the result to Object using toObject() method (making it mutable).
To remove a field from a query result,
let immutableQueryResult = await Col.findById(idToBeSearched)
let mutableQueryResult = immutableQueryResult.toObject()
delete mutableQueryResult.fieldToBeRemoved
console.log(mutableQueryResult)
Another way of getting the mutable result is using the _doc property of the result:
let immutableQueryResult = await Col.findById(idToBeSearched)
let mutableQueryResult = immutableQueryResult._doc // _doc property holds the mutable object
delete mutableQueryResult.fieldToBeRemoved
console.log(mutableQueryResult)