Mongodb 按 _id 的聚合查找无效

Question

MongoDb 当 foreignField 为 _id 时聚合查找未生成结果。我有两个集合说 users 和 discussions

样本users数据：

[{
    _id: 5f9c50dcfac1f091400225e3,
    email: 'Peter.Parker@gmail.com',
    details: { fname: 'Peter Test', lname: 'Fulton' },
},
{
    _id: 5fa432bfb91fab7db60c70eb,
    email: 'Spidy@xxx.com',
    details: { fname: 'Frodo', lname: 'Baggins' },
},
{
    _id: 5fa8ec7d3ce22610e5d15190,
    email: 'tommy@xxx.com',
    details: { fname: 'Tommy', lname: 'test' },
},
{
    _id: 5fc38bb0b3683651be970180,
    email: 'jerry@xxx.io',
},
{
    _id: 5fd2340cc443d155ab38383b,
    email: 'Dexter@xxx.io',
    details: { fname: 'Dexter', lname: 'Lab' },
}]

示例 discussions 数据：

 {_id: ObjectId("5fb2abd6b14fa5683979df58"),
  tags: [ 'javascritp', 'css', 'html' ],
  title: 'Why is this inline-block element pushed downward?',
  post: 'Test Post',
  learnerId: ObjectId("5f9c50dcfac1f091400225e3"),
 }

此处 users 的“_id”与“discussions”的“learnerId”链接。

我的聚合查询如下所示。

db.users.aggregate([
            { $project: { "details.fname": 1, "details.lname":1,email:1, _id:1}},
            {$lookup: {
                from: "discussions",
                localField: "learnerId",
                foreignField: "_id",
                as: "discussions"
            }}
        ])

此处 'Peter Test' 的 _id 5f9c50dcfac1f091400225e3 与 discussions LeanerId 链接。但我希望讨论会出现在我的结果中。我在所有用户集合中看到空的讨论数组。

 [{
    _id: 5f9c50dcfac1f091400225e3,
    email: 'Peter.Parker@gmail.com',
    details: { fname: 'Peter Test', lname: 'Fulton' },
    discussions: []
},
{
    _id: 5fa432bfb91fab7db60c70eb,
    email: 'Spidy@xxx.com',
    details: { fname: 'Frodo', lname: 'Baggins' },
    discussions: []
},
{
    _id: 5fa8ec7d3ce22610e5d15190,
    email: 'tommy@xxx.com',
    details: { fname: 'Tommy', lname: 'test' },
    discussions: []
},
{
    _id: 5fc38bb0b3683651be970180,
    email: 'jerry@xxx.io',
    discussions: []
},
{
    _id: 5fd2340cc443d155ab38383b,
    email: 'Dexter@xxx.io',
    details: { fname: 'Dexter', lname: 'Lab' },
    discussions: []
}]

你能指出我的聚合查询有什么问题吗？

Answer 1

您的 localField 和 foreignField 不匹配

db.users.aggregate([
  {
    $project: {
      "details.fname": 1,
      "details.lname": 1,
      email: 1,
      _id: 1
    }
  },
  {
    $lookup: {
      from: "discussions",
      localField: "_id",
      foreignField: "learnerId",
      as: "discussions"
    }
  }
])

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