Mongoose, Select a specific field with find

Question

I'm trying to select only a specific field with

exports.someValue = function(req, res, next) {
    //query with mongoose
    var query = dbSchemas.SomeValue.find({}).select('name');

    query.exec(function (err, someValue) {
        if (err) return next(err);
        res.send(someValue);
    });
};

But in my json response i'm receiving also the _id, my document schema only has two fiels, _id and name

[{"_id":70672,"name":"SOME VALUE 1"},{"_id":71327,"name":"SOME VALUE 2"}]

Why???

score 296 · Accepted Answer · answered Jun 22 '14 at 06:09

296

The _id field is always present unless you explicitly exclude it. Do so using the - syntax:

exports.someValue = function(req, res, next) {
    //query with mongoose
    var query = dbSchemas.SomeValue.find({}).select('name -_id');

    query.exec(function (err, someValue) {
        if (err) return next(err);
        res.send(someValue);
    });
};

Or explicitly via an object:

exports.someValue = function(req, res, next) {
    //query with mongoose
    var query = dbSchemas.SomeValue.find({}).select({ "name": 1, "_id": 0});

    query.exec(function (err, someValue) {
        if (err) return next(err);
        res.send(someValue);
    });
};

answered Jun 22 '14 at 06:09

Neil Lunn

148,042
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5

I think `.select` just is a filter to choose field after you get all of that, my recommendation is to use `.find({}, 'name -_id')` – hong4rc Mar 02 '19 at 16:15
5

@Hongarc `.find({}, 'name -_id')` seems not working? – Shanika Ediriweera May 29 '19 at 01:53
is there is a way to get the value direct without an obj like [ 'value1', 'value2', 'value3' ] instead of [ 'name':'value1', 'name':'value2', 'name':'value3' ] – M.Amer Jul 14 '19 at 14:21
I get an error using this because apparently I can't use a mixe of inclusions and rejections. I figured I should pass as argument to the `.select()` only this id to exclude it : `.select({ "_id":0 })` – Oriun Jan 19 '21 at 15:49

Aakash · Answer 2 · 2021-11-26T08:13:31.963

82

There is a shorter way of doing this now:

exports.someValue = function(req, res, next) {
    //query with mongoose
    dbSchemas.SomeValue.find({}, 'name', function(err, someValue){
      if(err) return next(err);
      res.send(someValue);
    });
    //this eliminates the .select() and .exec() methods
};

In case you want most of the Schema fields and want to omit only a few, you can prefix the field name with a - (minus sign). For ex "-name" in the second argument will not include name field in the doc whereas the example given here will have only the name field in the returned docs.

edited Nov 26 '21 at 08:13

answered Sep 17 '15 at 12:52

Aakash

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For those who want to filter multiple fields, don't seperate them with comma, just plain space: `blogItemModel.find({}, 'title intro_image intro_text publish_date', function(err, blog_items){..` – Starwave Oct 10 '15 at 18:37
1

for those who want to use where clause in above **dbSchema.Somevalue.find({userEmail:'test@test.com'},'userEmail -_id', function(err, someValue)** – user2180794 May 26 '16 at 07:15
1

multiple fields would be ['name', 'field2', 'field3', 'etc'] instead of just 'name' – Eray T Dec 24 '18 at 19:05

score 34 · Answer 3 · edited Nov 30 '17 at 15:05

34

There's a better way to handle it using Native MongoDB code in Mongoose.

exports.getUsers = function(req, res, next) {

    var usersProjection = { 
        __v: false,
        _id: false
    };

    User.find({}, usersProjection, function (err, users) {
        if (err) return next(err);
        res.json(users);
    });    
}

http://docs.mongodb.org/manual/reference/method/db.collection.find/

Note:

var usersProjection

The list of objects listed here will not be returned / printed.

edited Nov 30 '17 at 15:05

Shea Hunter Belsky

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answered Oct 13 '15 at 16:35

ElvinD

667
1
7
10

How do you do this with findOne? – zero_cool Apr 06 '18 at 01:36
questionnaire needs on name json array. with your method other key are there than they also will come like age,etc – Ratan Uday Kumar Aug 22 '18 at 08:45
Awesome, just what i was looking for – Shiva Teja Jun 01 '19 at 11:31

Nishant Shah · Answer 4 · 2021-06-04T07:47:24.643

33

Tip: 0 means ignore & 1 means show.

Example 1:

User.find({}, { createdAt: 0, updatedAt: 0, isActive: 0, _id : 1 }).then(...)

Example 2:

User.findById(id).select("_id, isActive").then(...)

Example 3:

User.findById(id).select({ _id: 1, isActive: 1, name: 1, createdAt: 0 }).then(...)

edited Jun 04 '21 at 07:47

answered Apr 11 '20 at 17:28

Nishant Shah

2,022
22
20

The most suitable solution for safety without any additional functions! – J A S K I E R Dec 16 '20 at 20:14
Seems you can no longer include mix of 0 and 1 – helloworld Oct 17 '21 at 13:16

Ratan Uday Kumar · Answer 5 · 2018-08-22T08:49:43.187

20

DB Data

[
  {
    "_id": "70001",
    "name": "peter"
  },
  {
    "_id": "70002",
    "name": "john"
  },
  {
    "_id": "70003",
    "name": "joseph"
  }
]

Query

db.collection.find({},
{
  "_id": 0,
  "name": 1
}).exec((Result)=>{
    console.log(Result);
})

Output:

[
  {
    "name": "peter"
  },
  {
    "name": "john"
  },
  {
    "name": "joseph"
  }
]

Working sample playground

link

edited Aug 22 '18 at 08:49

answered Jun 25 '17 at 16:34

Ratan Uday Kumar

5,738
6
35
54

this answer helped me, it explains in a easy to understanding way, +1 to you. – Nivethan Apr 29 '22 at 03:42

score 16 · Answer 6 · edited Nov 19 '19 at 16:36

Exclude

Below code will retrieve all fields other than password within each document:

const users = await UserModel.find({}, {
  password: 0 
});
console.log(users);

Output

[
  {
    "_id": "5dd3fb12b40da214026e0658",
    "email": "example@example.com"
  }
]

Include

Below code will only retrieve email field within each document:

const users = await UserModel.find({}, {
  email: 1
});
console.log(users);

Output

[
  {
    "email": "example@example.com"
  }
]

Ashh · Answer 7 · 2018-09-10T12:15:37.230

5

The precise way to do this is it to use .project() cursor method with the new mongodb and nodejs driver.

var query = await dbSchemas.SomeValue.find({}).project({ name: 1, _id: 0 })

edited Sep 10 '18 at 12:15

answered Aug 19 '18 at 01:22

Ashh

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4

`.project()` works only with aggregate. If you want to use it with find use second argument in `find({},{ name: 1, _id: 0 })` method. – Sham Sep 10 '18 at 08:55
No with the new mongodb node driver you have to use this way. But with mongoose you can use second argument. – Ashh Sep 10 '18 at 09:07
1

ah, it is in nodejs driver. – Sham Sep 10 '18 at 12:14
Don't understand why, but in new versions @Anthony is right...Lower versions you can do it this way...very messy – ackuser Sep 13 '18 at 16:10

score 4 · Answer 8 · answered Sep 07 '20 at 02:39

4

I found a really good option in mongoose that uses distinct returns array all of a specific field in document.

User.find({}).distinct('email').then((err, emails) => { // do something })

answered Sep 07 '20 at 02:39

Steve F.

71
1
2

Mongoose, Select a specific field with find

8 Answers8

Exclude

Include

Linked

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