我没有找到排序修饰符的doc。唯一的洞见在单元测试中: spec.lib.query.js # L12

writer.limit(5).sort(['test', 1]).group('name')

但这对我不起作用:

Post.find().sort(['updatedAt', 1]);

当前回答

// Ascending with updatedAt field
Post.find().sort('updatedAt').exec((err, post) => {...});

// Descending with updatedAt field
Post.find().sort('-updatedAt').exec((err, post) => {...});

参考网址:https://mongoosejs.com/docs/queries.html

其他回答

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
})
app.get('/getting',function(req,res){
    Blog.find({}).limit(4).skip(2).sort({age:-1}).then((resu)=>{
        res.send(resu);
        console.log(resu)
        // console.log(result)
    })
})

输出

[ { _id: 5c2eec3b8d6e5c20ed2f040e, name: 'e', age: 5, __v: 0 },
  { _id: 5c2eec0c8d6e5c20ed2f040d, name: 'd', age: 4, __v: 0 },
  { _id: 5c2eec048d6e5c20ed2f040c, name: 'c', age: 3, __v: 0 },
  { _id: 5c2eebf48d6e5c20ed2f040b, name: 'b', age: 2, __v: 0 } ]

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。

其他人为我工作,但这个做到了:

  Tag.find().sort('name', 1).run(onComplete);