arrays - MongoDB:如何获取查找集合的对象值

Question

我是 MongoDB 新手。我想找到对象值来查找集合。

以下是集合:

语言:

{
  _id: 15,
  language:English,
  status: Active
},
{
  _id: 20,
  language:Spanish,
  status: Active
},

类别:

这里15, 20是name对象中的语言ID

{
  _id: 1,
  name: { 
          15: Office,
          20: Oficina 
        },
  status: Active
},
{
  _id: 2,
  name: { 
          15: Restaurant,
          20: Restaurante 
        },
  status: Active
 },

属性:

{
  _id: 1,
  name: { 
          15: Lake View,
          20: Vista al lago 
        },
  cat_id : 1
  status: Active
},
{
  _id: 2,
  name: { 
          15: The Apple,
          20: La manzana 
        },
  cat_id : 1
  status: Active
 },
{
  _id: 3,
  name: { 
          15: Blue Monday,
          20: Lunes azul 
        },
  cat_id : 2
  status: Active
 },

查询:

var langId = 15;
db.Property.aggregate([
  {
    $lookup:
      {
        from: 'Category',
        localField: 'cat_id',
        foreignField: '_id',
        as: 'catdetails',
      },
  },
  {
    $project: {
      _id: 1,
      name: `$name.${langId}`,
      cat_id: 1,
      status: 1,
      'catdetails._id': 1,
      'catdetails.name': `$catdetails.name.${langId}`,
      'catdetails.status': 1,
    },
  },
]);

结果:

[
  {
    "_id": 1,
    "name": "Lake View",
    "cat_id": 1,
    "status": "Active",
    "catdetails": {
                    "_id": "1",
                    "name":[ "Office" ],
                  }
   },
   {
    "_id": 2,
    "name": "The Apple",
    "cat_id": 2,
    "status": "Active",
    "catdetails": {
                    "_id": "1",
                    "name":[ "Office" ],
                  }
   },
   {
    "_id": 3,
    "name": "Blue Monday",
    "cat_id": 2,
    "status": "Active",
    "catdetails": {
                    "_id": 2,
                    "name":[ "Restaurant" ],
                  }
   },
]

在catdetails对象name的结果中出现这样的数组符号“name”:[“Restaurant”]我想要这个没有像这样的数组符号“name”:“Restaurant”

预期结果:

[
  {
    "_id": 1,
    "name": "Lake View",
    "cat_id": 1,
    "status": "Active",
    "catdetails": {
                    "_id": "1",
                    "name":"Office",
                  }
   },
   {
    "_id": 2,
    "name": "The Apple",
    "cat_id": 2,
    "status": "Active",
    "catdetails": {
                    "_id": "1",
                    "name":"Office",
                  }
   },
   {
    "_id": 3,
    "name": "Blue Monday",
    "cat_id": 2,
    "status": "Active",
    "catdetails": {
                    "_id": 2,
                    "name":"Restaurant",
                  }
   },
]

如何将查询更改为此类结果？

Answer 1

请尝试这个:

var langId = 15;
db.Property.aggregate([
    {
        $lookup:
        {
            from: 'category',
            localField: 'cat_id',
            foreignField: '_id',
            as: 'catdetails',
        },
    }, { $addFields: { name: `$name.${langId}`, catdetails: { $arrayElemAt: ["$catdetails", 0] } } },
    { $addFields: { 'catdetails.name': `$catdetails.name.${langId}` } }])

一些更改包括删除 _id :1，因为默认情况下它将保留在不需要包含的 $project 中，此外，如果您有更多字段要保留，则可以使用 $addFields而不是笨拙的$project。

引用: $arrayElemAt , $addFields