Mongodb query object without field

Question

I'm trying to query a full mongodb field but without an intern field, like this document example:

{
  _id: ObjectId('x8234728x8z8381'),
  Example: {
       field1: 'first',
       field2: 'second',
       field3: {
          field 3.1: 'third'
       }
  }
}

I would like to return all "Example" field, but without "$field2". Is it possible in mongodb?

Ouch! You named your fields with a $? It's doable, but why use $ and then have to worry about possible consequences? — rickhg12hs
– rickhg12hs, Commented Jun 12, 2022 at 16:40
It was just an example to mock my real data hahaha, when I put this data I didn't pay attention to it ahahah — Victor Resende
– Victor Resende, Commented Jun 12, 2022 at 16:43
It's also better not to use images for code/errors/back traces/stack traces/etc. If you put them in a code block, it's easier for us to copy/paste. Code images are like answer repellent. :-) — rickhg12hs
– rickhg12hs, Commented Jun 12, 2022 at 16:46

Gibbs · Accepted Answer · 2022-06-12 16:57:08Z

1

db.collection.aggregate({
  "$project": {
    "Example.field2": 0
  }
})

You just ignore it in projection.

answered Jun 12, 2022 at 16:57

Gibbs

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Comments

rickhg12hs · Accepted Answer · 2022-06-12 16:56:32Z

0

You can use find and then just not "project" the fields you want to eliminate.

db.collection.find({
  "_id": ObjectId("62a6144faf59381072fa83d6")
},
{
  "Example.field2": false
})

Try it on mongoplayground.net.

answered Jun 12, 2022 at 16:56

rickhg12hs

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