Question

这是我的预约系列：

{ _id: ObjectId("518ee0bc9be1909012000002"), date: ISODate("2013-05-13T22:00:00Z"), patient:ObjectId("518ee0bc9be1909012000002") }

{ _id: ObjectId("518ee0bc9be1909012000002"), date: ISODate("2013-05-13T22:00:00Z"), patient:ObjectId("518ee0bc9be1909012000002") }

{ _id: ObjectId("518ee0bc9be1909012000002"), date: ISODate("2013-05-13T22:00:00Z"), patient:ObjectId("518ee0bc9be1909012000002") }

我使用聚合来获得以下结果

{date: ISODate("2013-05-13T22:00:00Z"),
patients:[ObjectId("518ee0bc9be1909012000002"),ObjectId("518ee0bc9be1909012000002"),ObjectId("518ee0bc9be1909012000002")] }

像这样：

Appointments.aggregate([
{$group: {_id: '$date', patients: {$push: '$patient'}}},
{$project: {date: '$_id', patients: 1, _id: 0}}
], ...)

如何填写患者文档我纠结了这个，但它不起作用...... Appointments.find({}).populate("patient").aggregate ....

换句话说，我可以在同一个语句中使用populate和aggregate

任何帮助，请

Answer 1

使用最新版本的mongoose（mongoose＆gt; = 3.6），可以，但它需要第二个查询，并使用不同的填充。聚合后，请执行以下操作：

Patients.populate(result, {path: "patient"}, callback);

点击Mongoose API和Mongoose docs了解详情。

Answer 2

您可以使用与填充类似的$ lookup。

在一个不相关的例子中，我使用$ match来查询记录，使用$ lookup来填充外部模型作为这些记录的子属性：

  Invite.aggregate(
      { $match: {interview: req.params.interview}},
      { $lookup: {from: 'users', localField: 'email', foreignField: 'email', as: 'user'} }
    ).exec( function (err, invites) {
      if (err) {
        next(err);
      }

      res.json(invites);
    }
  );

Answer 3

简答： 你不能。

答案很长： 在聚合框架中，返回的字段由您构建，您可以“重命名”文档属性。

这意味着Mongoose无法确定您的参考文件是否可用于最终结果。

在这种情况下，您可以做的最好的事情是在返回查询后填充所需的字段。是的，这将导致两次数据库调用，但这是MongoDB允许我们做的事情。

有点像这样：

Appointments.aggregate([ ... ], function( e, result ) {
  if ( e ) return;

  // You would probably have to do some loop here, as probably 'result' is array
  Patients.findOneById( result.patient, function( e, patient ) {
    if ( e ) return;

    result.patient = patient;
  });
});

Answer 4

使用$ lookup执行加入

收集订单包含以下文件：

{ "_id" : 1, "item" : "abc", "price" : 12, "quantity" : 2 }
{ "_id" : 2, "item" : "jkl", "price" : 20, "quantity" : 1 }
{ "_id" : 3  }

另一个集合清单包含以下文档：

{ "_id" : 1, "sku" : "abc", description: "product 1", "instock" : 120 }
{ "_id" : 2, "sku" : "def", description: "product 2", "instock" : 80 }
{ "_id" : 3, "sku" : "ijk", description: "product 3", "instock" : 60 }
{ "_id" : 4, "sku" : "jkl", description: "product 4", "instock" : 70 }
{ "_id" : 5, "sku": null, description: "Incomplete" }
{ "_id" : 6 }

订单集合的以下聚合操作使用orders集合中的字段项和库存集合中的sku字段将订单中的文档与库存集合中的文档相连接：

db.orders.aggregate([
    {
      $lookup:
        {
          from: "inventory",
          localField: "item",
          foreignField: "sku",
          as: "inventory_docs"
        }
   }
])

该操作返回以下文档：

{
  "_id" : 1,
   "item" : "abc",
  "price" : 12,
  "quantity" : 2,
  "inventory_docs" : [
    { "_id" : 1, "sku" : "abc", description: "product 1", "instock" : 120       }
  ]
 }
{
  "_id" : 2,
  "item" : "jkl",
  "price" : 20,
  "quantity" : 1,
  "inventory_docs" : [
    { "_id" : 4, "sku" : "jkl", "description" : "product 4", "instock" : 70 }
  ]
}
{
  "_id" : 3,
  "inventory_docs" : [
    { "_id" : 5, "sku" : null, "description" : "Incomplete" },
    { "_id" : 6 }
  ]
}

Reference $lookup

Answer 5

您可以在一个查询中执行此操作：

Appointments.aggregate([{
        $group: {
            _id: '$date',
            patients: {
                $push: '$patient'
            }
        }
    },
    {
        $project: {
            date: '$_id',
            patients: 1,
            _id: 0
        }
    },
    {
        $lookup: {
            from: "patients",
            localField: "patient",
            foreignField: "_id",
            as: "patient_doc"
        }
    }
])

populate基本上使用引擎盖下的$ lookup。在这种情况下，不需要第二个查询。有关详情，请查看MongoDB aggregation lookup

Answer 6

domain.Farm.aggregate({
    $match: {
        "_id": mongoose.Types.ObjectId(farmId)
    }
}, {
    $unwind: "$SelfAssessment"
}, {
    $match: {
        "SelfAssessment.questionCategoryID": QuesCategoryId,
        "SelfAssessment.questionID": quesId
    }
},function(err, docs) {
      var options = {
          path: 'SelfAssessment.actions',
          model: 'FarmAction'
     };
     domain.Farm.populate(docs, options, function (err, projects) {
       callback(err,projects);

      });

});

结果我得到了行动模型填充

{   "error": false,   "object": [
    {
      "_id": "57750cf6197f0b5137d259a0",
      "createdAt": "2016-06-30T12:13:42.299Z",
      "updatedAt": "2016-06-30T12:13:42.299Z",
      "farmName": "abb",
      "userId": "57750ce2197f0b5137d2599e",
      "SelfAssessment": {
        "questionName": "Aquatic biodiversity",
        "questionID": "3kGTBsESPeYQoA8ae2Ocoy",
        "questionCategoryID": "5aBe7kuYWIEoyqWCWcAEe0",
        "question": "Waterways protected from nutrient runoff and stock access through fencing, buffer strips and off stream watering points",
        "questionImage": "http://images.contentful.com/vkfoa0gk73be/4pGLv16BziYYSe2ageCK04/6a04041ab3344ec18fb2ecaba3bb26d5/thumb1_home.png",
        "_id": "57750cf6197f0b5137d259a1",
        "actions": [
          {
            "_id": "577512c6af3a87543932e675",
            "createdAt": "2016-06-30T12:38:30.314Z",
            "updatedAt": "2016-06-30T12:38:30.314Z",
            "__v": 0,
            "Evidence": [],
            "setReminder": "",
            "description": "sdsdsd",
            "priority": "High",
            "created": "2016-06-30T12:38:30.312Z",
            "actionTitle": "sdsd"
          }
        ],
        "answer": "Relevant"
      },
      "locations": []
    }   ],   "message": "",   "extendedMessage": "",   "timeStamp": 1467351827979 }

Answer 7

您必须分两次而不是一次声明。

在 异步等待 方案中，确保等待直到填充。

const appointments = await Appointments.aggregate([...]);

await Patients.populate(appointments, {path: "patient"});

return appointments;

或（如果您想限制）

await Patients.populate(appointments, {path: "patient",  select:  {_id: 1, fullname: 1}});

Answer 8

我改用了查找，而且效果很好。请参阅下面截取的代码。

Post.aggregate([
        {
            $group: {
                // Each `_id` must be unique, so if there are multiple
                // posts with the same category, MongoDB will increment `count`.
                _id: '$category',
                count: { $sum: 1 }
            }
        },
        //from: is collection name in MongoDB, localField are primary and foreign keys in Model.
        {$lookup: {from: 'categories', localField: '_id', foreignField:'_id', as: 'category'}}
    ]).then(categoryCount => {
        console.log(categoryCount);
        let json = [];
        categoryCount.forEach(cat => {
            console.log(json);
        });

如何在同一语句中使用populate和aggregate？

8 个答案: