计算字段在 collection 中出现的次数，同时保持所有数据不重复

Question

假设我有这样一个 collection：

{
  _id : 544e97123c9ef694fc68e21b,
  title: "First Title",
  notebook: {
    title: "Misc",
    slug: "misc"
  }
}
{
  _id: 54ab035a849788d0921d8eb2,
  title: "Second Title",
  notebook: {
    title: "Personal",
    slug: "personal"
  }
}
{
  _id: 544e97123c9ef694fc68e21b,
  title: "Third Title",
  notebook: {
    title: "Misc",
    slug: "misc"
  }
}

在我看来，我希望能够显示笔记本标题与 link 一起用于特定笔记本的 slug 的次数，没有特定的顺序。例如：

<a href="/notebooks/misc">Misc(2)</a>
<a href="/notebooks/personal">Personal(1)</a>

我通过遍历每个文档实现了这一点，但问题是它有重复项，因为它正在遍历每个文档。所以在我看来它看起来像这样：

<a href="/notebooks/misc">Misc(2)</a>
<a href="/notebooks/personal">Personal(1)</a>
<a href="/notebooks/misc">Misc(2)</a>

如何抓取notebook.title、notebook.slug，数一数，不重复？

这是我目前执行此操作的怪异方式（这会导致重复）：

function countNotebooks(notes) {
  var table = Object.create(null);
  for (var i = 0; i < notes.length; i++) {
    if (typeof table[notes[i].notebook.slug] === 'undefined') {
      table[notes[i].notebook.slug] = 1;
    } else {
      table[notes[i].notebook.slug] += 1;
    }
  }
  return table;
}

app.get('/notebooks', function(req, res) {
  Note.find(function(err, notes) {
    if (err) {
      throw err;
    }
    res.render('notebooks/index.html', {
      title: 'All Notebooks',
      jumbotron: 'Notebooks',
      notes: notes,
      notesTable: countNotebooks(notes)
    });
  });
});

notebooks/index.html:

{% for note in notes %}
  <article class="note">
    <h3 class="note-title">
      <a href="/notebooks/{{ note.notebook.slug }}">{{ note.notebook.title }}</a> <span class="count">({{ notesTable[note.notebook.slug] }})</span>
    </h3>
  </article>
{% endfor %}

Answer 1

您可以通过基于公用密钥对数据进行实质性分组来实现此目的。 MongoDB 的 aggregation framework 就是为这种聚合和操作而设计的。

首先，我可以更正你的数据样本吗，因为你有一个重复的 _id 值，这是不允许的。

{
    "_id" : ObjectId("544e97123c9ef694fc68e21b"),
    "title" : "First Title",
    "notebook" : {
            "title" : "Misc",
            "slug" : "misc"
    }
},
{
    "_id" : ObjectId("54ab035a849788d0921d8eb2"),
    "title" : "Second Title",
    "notebook" : {
            "title" : "Personal",
            "slug" : "personal"
    }
},
{
    "_id" : ObjectId("54ac074fa8a621d3fd49ac91"),
    "title" : "Third Title",
    "notebook" : {
            "title" : "Misc",
            "slug" : "misc"
    }
}

要为数据中出现的 "slug" 添加 "count" 值，您可以形成这样的管道：

Note.aggregate([

    // Group on the slug values and put other fields in an array
    { "$group": {
         "_id": "$notebook.slug",
         "count": { "$sum": 1 },
         "docs": { 
             "$push": {
                 "_id": "$_id",
                 "title": "$title",
                 "notebook": "$notebook"
             }
         }
    }},

    // Unwind the created array elements
    { "$unwind": "$docs" },

    // Re-structure back to original form
    { "$project": {
        "_id": "$docs._id",
        "title": "$docs.title",
        "count": "$count",
        "notebook": "$docs.notebook"
    }},

    // Sort in original order (or as desired)
    { "$sort": { "_id": 1 } }

],function(err,result) {

});

这会给你带来这样的结果：

{
    "_id" : ObjectId("544e97123c9ef694fc68e21b"),
    "count" : 2,
    "title" : "First Title",
    "notebook" : {
            "title" : "Misc",
            "slug" : "misc"
    }
},
{
    "_id" : ObjectId("54ab035a849788d0921d8eb2"),
    "count" : 1,
    "title" : "Second Title",
    "notebook" : {
            "title" : "Personal",
            "slug" : "personal"
    }
},
{
    "_id" : ObjectId("54ac074fa8a621d3fd49ac91"),
    "count" : 2,
    "title" : "Third Title",
    "notebook" : {
            "title" : "Misc",
            "slug" : "misc"
    }
}

那是如果你想要 "retain the documents"，但如果你只想要像 "facet count" 这样的唯一 "slugs"，那么只需使用第一个 $group 与 $first 在笔记本标题上而不是 $push 与其他内容：

Note.aggregate([

    // Group on the slug values and put other fields in an array
    { "$group": {
         "_id": "$notebook.slug",
         "count": { "$sum": 1 },
         "title": { "$first": "$notebook.title" }
    }},
],function(err,result) {

});

它应该是不言自明的，但只是总结一下。初始$group is done using the value of "slug" to count the occurrences using the $sum operator. In order to keep the rest of the document data it is placed in an array under the "slug" using $push.

数组分组后 de-normalized 使用 $unwind 生成文档，然后使用 $project. The final $sort operation provides the original order or whatever you want, since the ordering was changed during the $group re-structured 返回原始格式]管道阶段。

这不仅可以获得结果，而且还允许您使用 $limit and $skip 运算符对数据进行 "page" 计算，甚至可以根据这些计数值对数据进行排序如果需要的话。

查看完整的 aggregation pipeline operator reference 以获得完整的描述和其他可以在这里完成的事情。

Answer 2

您需要：

Group 通过 notebook.slug 和 notebook.title 字段，到消除重复项。
由于显示顺序并不重要，您可以跳过此处的 sort 阶段。
Project 他们

代码：

Model.aggregate([
{$group:{"_id":{"title":"$notebook.title",
                "slug":"$notebook.slug"},
                "count":{$sum:1}}},
{$project:{"title":"$_id.title",
           "slug":"$_id.slug",
           "count":1,"_id":0}}
],function(err,data){
      // handle response and store it in the `notes` variable to be
      // used to display.
  })

o/p:

{ "count" : 1, "title" : "Personal", "slug" : "personal" }
{ "count" : 2, "title" : "Misc", "slug" : "misc" }

并显示为，

{% for note in notes %}
  <article class="note">
    <h3 class="note-title">
      <a href="/notebooks/{{ note.slug }}">{{ note.title }}</a> 
        <span class="count">({{note.count}})</span>
    </h3>
  </article>
{% endfor %}

计算字段在 collection 中出现的次数，同时保持所有数据不重复

Count how many times a field shows up in a collection while keeping all data with no duplicates

node.js

mongodb

mongoose

mongodb-query

aggregation-framework