还可以使用aggregate()进行排序

 const sortBy = req.params.sort;
  const limitNum = req.params.limit;
  const posts = await Post.aggregate([
    { $unset: ['field-1', 'field-2', 'field-3', 'field-4'] },
    { $match: { field-1: value} },
    { $sort: { [sortBy]: -1 } },  //-------------------> sort the result
    { $limit: Number(limitNum) },
  ]);

从Mongoose 3.8.x开始:

model.find({ ... }).sort({ field : criteria}).exec(function(err, model){ ... });

地点:

条件包括asc、desc、升序、降序、1、-1

注意:使用引号或双引号

使用“asc”，“desc”，“ascending”，“descent”，1或-1

在Mongoose 4中使用查询生成器接口进行链接。

// Build up a query using chaining syntax. Since no callback is passed this will create an instance of Query.
var query = Person.
    find({ occupation: /host/ }).
    where('name.last').equals('Ghost'). // find each Person with a last name matching 'Ghost'
    where('age').gt(17).lt(66).
    where('likes').in(['vaporizing', 'talking']).
    limit(10).
    sort('-occupation'). // sort by occupation in decreasing order
    select('name occupation'); // selecting the `name` and `occupation` fields


// Excute the query at a later time.
query.exec(function (err, person) {
    if (err) return handleError(err);
    console.log('%s %s is a %s.', person.name.first, person.name.last, person.occupation) // Space Ghost is a talk show host
})

有关查询的更多信息，请参阅文档。

更新:

Post.find().sort({'updatedAt': -1}).all((posts) => {
  // do something with the array of posts
});

Try:

Post.find().sort([['updatedAt', 'descending']]).all((posts) => {
  // do something with the array of posts
});

这是我如何得到排序工作在猫鼬2.3.0:)

// Find First 10 News Items
News.find({
    deal_id:deal._id // Search Filters
},
['type','date_added'], // Columns to Return
{
    skip:0, // Starting Row
    limit:10, // Ending Row
    sort:{
        date_added: -1 //Sort by Date Added DESC
    }
},
function(err,allNews){
    socket.emit('news-load', allNews); // Do something with the array of 10 objects
})

我就是这么做的，效果很好。

User.find({name:'Thava'}, null, {sort: { name : 1 }})