I have created a mongodb collection where the schema is as follows.

{user_id, listings: <Array of Documents>, ...}

In the listing documents I have a field named status which has two values publish and draft.

In mongo shell I am trying to bring back only the user listings with status:publish So I run the bellow command and I keep getting back also documents with status:draft that are part of the user with published listings.

db.properties.find({'listings.status':'publish'},{'listings.status':1, _id:0})

Outputs:

[
  { listings: [ { status: 'draft' } ] },
  { listings: [ { status: 'publish' } ] },
  { listings: [ { status: 'draft' }, { status: 'publish' }, { status: 'draft' } ] },
  { listings: [ { status: 'publish' }, { status: 'draft' } ] },
  { listings: [ { status: 'publish' } ] }
]

How can I get an array back of the listings with only status: publish ?

CodePudding user response：

You can use an aggregation pipeline with a $filter:

1. $filter the listings field to contain only items with status publish.
2. $match only documents with $size of listings greater than 0.
3. format

db.collection.aggregate([
  {
    $set: {
      listings: {
        $filter: {
          input: "$listings",
          as: "item",
          cond: {
            $eq: [
              "$$item.status",
              "publish"
            ]
          }
        }
      }
    }
  },
  {$match: {$expr: {$gt: [{$size: "$listings"}, 0]}}},
  {$project: {listings: 1, _id: 0}}
])

Playground example