CODEWITHSUNDEEP
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TJBlackman
TJBlackman

Reputation: 2313

Return specified field using findById | Mongoose MongoDB

When using the Model.findOne() method, I can choose to include a string that limits my return value to just those field, but when I try to do so with the findById method, it appears to not work out.

The example below is a contrived example. I'm using a real _id and it works when using the first example, but I just want to use the findById method because a tiny voice in the back of my mind is telling me that's faster (maybe).

Example: 

let Joe = new PersonModel({
  _id: ObjectId("12345")
  name: 'Joe',
  age: 30,
  password: 'GreatPW'
}).save();

Now find Joe, and return only his name and age. This works!

PersonModel.findOne({_id: Joe._id}, 'name age')
  .then(person => console.log(person))

Now do the same thing, but using findById(). This isn't working for me.

PersonModel.findById(Joe._id, 'name age')
  .then(person => console.log(person))

Upvotes: 3

Views: 2936

Answers (2)

Sergi Nadal
Sergi Nadal

Reputation: 960

I'm not sure but I think you need to write {}, on query and in the projection:

This:

PersonModel.findById(Joe._id, 'name age')

Should be this:

PersonModel.findById({idValueToFind}, {name:1, age:1, _id:0})

Upvotes: 3

N-ate
N-ate

Reputation: 6925

Turn on logging in Mongoose. you'll see that the only difference between those two calls is an extra function call from findById to findOne.

Source code reference: findById (provided by MikaS)

Upvotes: 2

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