您也可以使用下面的代码行

per_page = parseInt(req.query.per_page) || 10
page_no = parseInt(req.query.page_no) || 1
var pagination = {
  limit: per_page ,
  skip:per_page * (page_no - 1)
}
users = await User.find({<CONDITION>}).limit(pagination.limit).skip(pagination.skip).exec()

这段代码将在最新版本的mongo中工作

我对这个问题的公认答案感到非常失望。这是无法缩放的。如果你读了游标上的小字。跳过():

cursor.skip()方法通常开销很大，因为它需要服务器从集合或索引的开始处遍历以获得偏移或跳过位置，然后才开始返回结果。随着偏移量(例如上面的pageNumber)的增加，cursor.skip()将变得更慢，更消耗CPU。对于较大的集合，cursor.skip()可能成为IO绑定。

要以可伸缩的方式将limit()与至少一个筛选条件结合起来实现分页，createdOn日期适合多种用途。

MyModel.find( { createdOn: { $lte: request.createdOnBefore } } )
.limit( 10 )
.sort( '-createdOn' )

实现这一点的可靠方法是使用查询字符串从前端传递值。假设我们想要获得第2页，并将输出限制为25个结果。 page=2&limit=25 //这将被添加到您的URL: http:localhost:5000?= 2限制= 25页

让我们看看代码:

// We would receive the values with req.query.<<valueName>>  => e.g. req.query.page
// Since it would be a String we need to convert it to a Number in order to do our
// necessary calculations. Let's do it using the parseInt() method and let's also provide some default values:

  const page = parseInt(req.query.page, 10) || 1; // getting the 'page' value
  const limit = parseInt(req.query.limit, 10) || 25; // getting the 'limit' value
  const startIndex = (page - 1) * limit; // this is how we would calculate the start index aka the SKIP value
  const endIndex = page * limit; // this is how we would calculate the end index

// We also need the 'total' and we can get it easily using the Mongoose built-in **countDocuments** method
  const total = await <<modelName>>.countDocuments();

// skip() will return a certain number of results after a certain number of documents.
// limit() is used to specify the maximum number of results to be returned.

// Let's assume that both are set (if that's not the case, the default value will be used for)

  query = query.skip(startIndex).limit(limit);

  // Executing the query
  const results = await query;

  // Pagination result 
 // Let's now prepare an object for the frontend
  const pagination = {};

// If the endIndex is smaller than the total number of documents, we have a next page
  if (endIndex < total) {
    pagination.next = {
      page: page + 1,
      limit
    };
  }

// If the startIndex is greater than 0, we have a previous page
  if (startIndex > 0) {
    pagination.prev = {
      page: page - 1,
      limit
    };
  }

 // Implementing some final touches and making a successful response (Express.js)

const advancedResults = {
    success: true,
    count: results.length,
    pagination,
    data: results
 }
// That's it. All we have to do now is send the `results` to the frontend.
 res.status(200).json(advancedResults);

我建议将这个逻辑实现到中间件中，这样你就可以将它用于各种路由/控制器。

尝试使用mongoose函数进行分页。限制是每页的记录数量和页的数量。

var limit = parseInt(body.limit);
var skip = (parseInt(body.page)-1) * parseInt(limit);

 db.Rankings.find({})
            .sort('-id')
            .limit(limit)
            .skip(skip)
            .exec(function(err,wins){
 });

有一些很好的答案给出了使用skip()和limit()的解决方案，但是，在某些情况下，我们还需要文档计数来生成分页。以下是我们在项目中所做的:

const PaginatePlugin = (schema, options) => {
  options = options || {}
  schema.query.paginate = async function(params) {
    const pagination = {
      limit: options.limit || 10,
      page: 1,
      count: 0
    }
    pagination.limit = parseInt(params.limit) || pagination.limit
    const page = parseInt(params.page)
    pagination.page = page > 0 ? page : pagination.page
    const offset = (pagination.page - 1) * pagination.limit

    const [data, count] = await Promise.all([
      this.limit(pagination.limit).skip(offset),
      this.model.countDocuments(this.getQuery())
    ]);
    pagination.count = count;
    return { data, pagination }
  }
}

mySchema.plugin(PaginatePlugin, { limit: DEFAULT_LIMIT })

// using async/await
const { data, pagination } = await MyModel.find(...)
  .populate(...)
  .sort(...)
  .paginate({ page: 1, limit: 10 })

// or using Promise
MyModel.find(...).paginate(req.query)
  .then(({ data, pagination }) => {

  })
  .catch(err => {

  })

简单而强大的分页解决方案

async getNextDocs(no_of_docs_required: number = 5, last_doc_id?: string) {
    let docs

    if (!last_doc_id) {
        // get first 5 docs
        docs = await MySchema.find().sort({ _id: -1 }).limit(no_of_docs_required)
    }
    else {
        // get next 5 docs according to that last document id
        docs = await MySchema.find({_id: {$lt: last_doc_id}})
                                    .sort({ _id: -1 }).limit(no_of_docs_required)
    }
    return docs
}

Last_doc_id:您获得的最后一个文档id

No_of_docs_required:你想要获取的文档数量，例如5、10、50等。

如果你不提供last_doc_id给方法，你会得到5个最新的文档 如果你提供了last_doc_id，那么你会得到下一个，即5个文档。