Question

我需要从数据库中检索整个单个对象层次结构作为JSON。我尝试聚合几个小时，无法解决如何使用我的数据。所以我有三个系列：

形式

{ "_id" : "1", "name" : "My first form" }
{ "_id" : "2", "name" : "Second one" }
{ "_id" : "3", "name" : "Another" }

问题

{ "_id" : "q1", "form" : "1", "title": "What's your country?"}
{ "_id" : "q2", "form" : "1", "title": "What your favorite color?"}
{ "_id" : "q3", "form" : "1", "title": "Where do you live?"}
{ "_id" : "q4", "form" : "2", "title": "Where to go?"}

选项

{ "_id" : "o1", "question" : "q1", "text" : "Brazil" }
{ "_id" : "o2", "question" : "q1", "text" : "EUA" }
{ "_id" : "o3", "question" : "q1", "text" : "China" }
{ "_id" : "o4", "question" : "q2", "text" : "Red" }
{ "_id" : "o5", "question" : "q2", "text" : "Blue" }
{ "_id" : "o6", "question" : "q2", "text" : "Green" }

我需要使用所有相应的问题检索每个表单，并在每个问题中检索它的选项。像这样：

[
   {
      _id:"q1",
      name: "My first form",
      questions: [
          { "_id" : "q1",
            "form" : "1", 
            "title": "What's your country?",
            "options": [
                  { "_id" : "o1", "question" : "q1", "text" : "Brazil" }
                  { "_id" : "o2", "question" : "q1", "text" : "EUA" },
                  { "_id" : "o3", "question" : "q1", "text" : "China" }
            ]
          },
          { "_id" : "q2",
            "form" : "1", 
            "title": "What your favorite color",
            "options": [
                  { "_id" : "o4", "question" : "q2", "text" : "Red" }
                  { "_id" : "o5", "question" : "q2", "text" : "Blue" },
                  { "_id" : "o6", "question" : "q2", "text" : "Green" }
            ]
          },
          { "_id" : "q3", 
            "form" : "1", 
            "title": "Where do you live?",
            "options": []
          }
      ]
   },
   ...
]

我尝试了很多$ lookup，$ unwind，另一个$ lookup和$ project，但没有给我带来结果（里面有问题的表单，里面有选项的问题）。

请帮帮我！：）

Answer 1

我认为周围的人正在查询question集合，查找他们的question并按form分组，最后按顺序查找表单和项目。

这应该这样做。请注意，此聚合的输出中的_id是_id形式。

db.question.aggregate([
    {$match: {}},
    {$lookup: {
        from: 'option',
        localField: '_id',
        foreignField: 'question',
        as: 'options'
    }},
    {$group: {
        _id: "$form",
        questions: {$push: {
            title: "$title",
            options: "$options",
            form: "$form"
        }}
    }},
    {$lookup: {
        from: 'form',
        localField: "_id",
        foreignField: "_id",
        as: 'form'
    }},
    {$project: {
        name: {$arrayElemAt: ["$form.name", 0]},
        questions: true
    }}
]);

实际上......这似乎是一个更好的选择。它也将返回form s而没有question。

db.form.aggregate([
    {$match: {}},
    {$lookup: {
        from: 'question',
        localField: '_id',
        foreignField: 'form',
        as: 'questions'
    }},
    {$unwind: {
        path: "$questions",
        preserveNullAndEmptyArrays: true
    }},
    {$lookup: {
        from: 'option',
        localField: 'questions._id',
        foreignField: 'question',
        as: 'options'
    }},
    {$group: {
        _id: "$_id",
        name: {$first: "$name"},
        question: {$push: {
            title: "$questions.title",
            form: "$questions.form",
            options: "$options"
        }}
    }}
])

MongoDB嵌套查找具有3个子级别

1 个答案: