Node.js 从mongodb查找聚合获取true或false（仅值而非数组或对象）

我正在尝试创建一个函数，该函数返回用户是否是最喜欢的

我在用户模型上使用MongoDB聚合，带有lookout with Favorites模型（其中所有Favorites用户都已保存），带有管道和$project

下面是MongoDB聚合函数查询

await User.aggregate([
            {
                $match: {
                    _id: ObjectId(_id),
                },
            },
            {
                $lookup: {
                    from: 'userportfolios',
                    localField: '_id',
                    foreignField: 'user',
                    as: 'user_portfolios',
                },
            },
            {
                $lookup: {
                    from: 'favourites',
                    pipeline: [
                        {
                            $match: {
                                user: ObjectId(data.user._id),
                                favourites: {
                                    $elemMatch: {
                                        $eq: ObjectId(_id),
                                    },
                                },
                            },
                        },
                        {
                            $project: {
                                _id: 0,
                                is_favourite: {
                                    $cond: [
                                        { $ifNull: ['$is_favourite', false] },
                                        { $literal: true },
                                        { $literal: false },
                                    ],
                                },
                            },
                        },
                    ],
                    as: 'is_favourite',
                },
            },
        ]);

我得到了这个结果

 [
    {
      "_id": "602b5078e8eb5339e0134351",
      "is_active": true,
      "is_admin": false,
      "is_block": false,
      "user_type": "company",
      "firstName": "Sagar",
      "lastName": "Davara",
      "mobileNumber": "*******",
      "email": "s*********@gmail.com",
      "password": "b7370338e54b1cb051be0dd54eba2f7f",
      "verificationCode": "210768",
      "createdAt": "2021-02-16T04:56:24.091Z",
      "updatedAt": "2021-02-16T04:56:24.091Z",
      "user_portfolios": [],
      "is_favourite": [
        {
          "is_favourite": false
        }
      ]
    }
  ]

但是我想得到

是你最喜欢的：true

而不是一个对象数组 我需要mongodb的输出

 [
    {
      "_id": "602b5078e8eb5339e0134351",
      "is_active": true,
      "is_admin": false,
      "is_block": false,
      "user_type": "company",
      "firstName": "Sagar",
      "lastName": "Davara",
      "mobileNumber": "6353764283",
      "email": "sagardspeed3@gmail.com",
      "password": "b7370338e54b1cb051be0dd54eba2f7f",
      "verificationCode": "210768",
      "createdAt": "2021-02-16T04:56:24.091Z",
      "updatedAt": "2021-02-16T04:56:24.091Z",
      "user_portfolios": [],
      "is_favourite": false
    }
  ]

---

只需在

$lookup

阶段后添加以下阶段：

{
$addFields：{
is_Favorite:{$arrayElemAt:[“$is_Favorite.is_Favorite”，0]}
}
}