use mongoose model.find() to get all entries of only 1 field

Question

I\'ve tried to use different variation of model.find(), but none do what I want.

code below is what I\'m working with, but it displays every single field, and I only wan

Answer 1

You can use the projection parameter to specify which field you would like to include in the result.

db.collection.find(query, projection)

If find() includes a projection argument, the matching documents contain only the projection fields and the _id field. You can optionally exclude the _id field.

The projection parameter takes a document of the following form:

{ field1: <boolean>, field2: <boolean> ... }

The <boolean> value can be any of the following:

1 or true to include the field. The find() method always includes the _id field even if the field is not explicitly stated to return in the projection parameter.

0 or false to exclude the field.

When you include a field you can only explicitly exclude fields in the case of the _id field.

app.get('/api/videos', function (req, res) {
    Video.find({}, {_id:0, iframe:1 }, function (err, docs) {
       res.json(docs);
    });
});

Answer 2

What you are looking for is called projection:

Video.find({}, {iframe: 1}, function (err, docs) {
   res.json(docs);
});

The second parameter to the find function tells which field to return. If you do not want the _id as well, then use: {_id:0, iframe:1}

Like so:

Video.find({}, {_id:0, iframe:1}, function (err, docs) {
   res.json(docs);
});

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However, projection does not give you distinct values. It only returns the fields you want to use (along with repetitions).