如何用猫鼬对两个字段进行排序? [英] How to sort two fields with mongoose?

问题描述

我正试图允许用户查看热门帖子.总体思路是，按最新帖子(_id:-1)排序，然后按最新投票(upvotes_count:-1)排序，然后限制结果(.limit(3)).这有点简化，所以请忽略这种趋势文章"的实现.

I'm trying to allow users to see trending posts. The general idea is to sort by the most recent posts (_id: -1) and then sort those by most upvotes (upvotes_count: -1) and then limiting the results (.limit(3)). This is a bit simplified, so please ignore this implementation of "trending posts".

不幸的是，我无法以我想要的方式返回两种.因此，在收集了六个帖子之后，它会返回最近的三个帖子，但是并不会按照大多数投票对它们进行排序.例如:

Unfortunately, I'm not able to return two sorts in the way that I want. So with a collection of six posts, it returns the most recent three, but it doesn't then sort them by most upvotes. For instance:

第6个帖子(投票数:1) 第5个帖子(投票数:2) 第4个帖子(投票数:1)

Post 6 (upvotes: 1) Post 5 (upvotes: 2) Post 4 (upvotes: 1)

我希望对它们进行排序:

I want them to be sorted like so:

第5个帖子(投票数:2) 第6个帖子(投票数:1) 第4个帖子(投票数:1)

Post 5 (upvotes: 2) Post 6 (upvotes: 1) Post 4 (upvotes: 1)

我对平局中发生的事情不那么感兴趣，但是至少，我希望将具有较高投票权的帖子列出来高于具有较少投票权的帖子.

I'm not so interested in what happens with ties, but at a minimum, I want the posts that have more upvotes to be listed higher than those with less upvotes.

当然，我可以编写一种方法来对这些内容进行排序，但是肯定有一种方法可以对MongoDB进行处理.

Of course, I could write a method to sort these, but surely there is a way to do this with MongoDB.

下面是我尝试实现这种方式的一些方法.

Below are some of the ways I've tried to implement this sort.

// Use sort for date and then use it again for upvotes_count
Post.find()
    .sort({_id: -1})
    .sort({upvotes_count: -1})
    .limit(3)
    .exec( function(err, posts) {
        if (err) res.send(err);
        console.log(posts);
        res.json(posts);
     });

// Use sort for date, limit the results to three, and then
// use it again for upvotes_count
Post.find()
    .sort({_id: -1})
    .limit(3)
    .sort({upvotes_count: -1})
    .exec( function(err, posts) {
        if (err) res.send(err)
        console.log(posts);
        res.json(posts);
    });

// Use sort for date and upvotes_count in one step.
Post.find()
    .sort({_id: -1, upvotes_count: -1})
    .limit(3)
    .exec( function(err, posts) {
        if (err) res.send(err);
        console.log(posts);
        res.json(posts);
     });

没有一个工作.

推荐答案

请参阅 sort() 定义.

Refer to sort() definition.

sort({_id: -1, upvotes_count: -1})

表示首先对_id进行排序，然后仅对那些相同 _id帖子按降序对upvotes_count进行排序.不幸的是，_id是 ObjectId ，它是12个字节BSON类型，使用以下方式构造:

means sort the _id firstly, then sort upvotes_count by desc order only for those same _id posts. Unfortunately, the _id is ObjectId, which is 12-byte BSON type, constructed using:

• 一个4字节的值，表示自Unix时代以来的秒数，
• 3字节机器标识符，
• 2字节的进程ID，和
• 一个3字节计数器，以随机值开头.

很难获得相同的ObjectId.即，每个记录的_id在此文档中应该是唯一的.结果，测试代码的结果只是按_id desc排序.

It is hard to get the same ObjectId. Namely, the _id of every record should be unique in this document. As a result, the result of your test codes are just ordered by _id desc.

这里是一个例子，

+---------+---------------+
| _id     |  upvote_count |
+---------+---------------+
|  1      |      5        |
|  4      |      7        |
|  3      |      9        |
|  4      |      8        |

sort({_id: -1, upvotes_count: -1})的结果应为

+---------+---------------+
| _id     |  upvote_count |
+---------+---------------+
|  4      |      8        |
|  4      |      7        |
|  3      |      9        |
|  1      |      5        |

upvote_count将按相同的_id进行排序.

The upvote_count would be sorted for same _id.

但是，在这种情况下.在这种情况下，在同一_id上.

However, in this case. There is on same _id in this case.

+---------+---------------+
| _id     |  upvote_count |
+---------+---------------+
|  1      |      5        |
|  4      |      7        |
|  3      |      9        |
|  2      |      8        |

sort({_id: -1, upvotes_count: -1})的结果应为

+---------+---------------+
| _id     |  upvote_count |
+---------+---------------+
|  1      |      5        |
|  2      |      8        |
|  3      |      9        |
|  4      |      7        |

这篇关于如何用猫鼬对两个字段进行排序?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！