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。

以上回答是正确的。

只是一个插件，任何人谁是异步等待而不是 承诺! !

const findAllFoo = async (req, resp, next) => {
    const pageSize = 10;
    const currentPage = 1;

    try {
        const foos = await FooModel.find() // find all documents
            .skip(pageSize * (currentPage - 1)) // we will not retrieve all records, but will skip first 'n' records
            .limit(pageSize); // will limit/restrict the number of records to display

        const numberOfFoos = await FooModel.countDocuments(); // count the number of records for that model

        resp.setHeader('max-records', numberOfFoos);
        resp.status(200).json(foos);

    } catch (err) {
        resp.status(500).json({
            message: err
        });
    }
};

MongoDB官方博客有一个关于分页的条目，在那里他们解释了为什么“跳过”可能会很慢，并提供了替代方案:https://www.mongodb.com/blog/post/paging-with-the-bucket-pattern--part-1

下面的代码是为我工作良好。 你也可以在countDocs查询中添加查找过滤器和user same来获得准确的结果。

export const yourController = async (req, res) => {
  const { body } = req;

  var perPage = body.limit,
  var page = Math.max(0, body.page);

  yourModel
    .find() // You Can Add Your Filters inside
    .limit(perPage)
    .skip(perPage * (page - 1))
    .exec(function (err, dbRes) {
      yourModel.count().exec(function (err, count) { // You Can Add Your Filters inside
        res.send(
          JSON.stringify({
            Articles: dbRes,
            page: page,
            pages: count / perPage,
          })
        );
      });
    });
};

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。

使用猫鼬，快递和翡翠的分页-这里有一个链接到我的博客与更多的细节

var perPage = 10
  , page = Math.max(0, req.params.page)

Event.find()
    .select('name')
    .limit(perPage)
    .skip(perPage * page)
    .sort({
        name: 'asc'
    })
    .exec(function(err, events) {
        Event.count().exec(function(err, count) {
            res.render('events', {
                events: events,
                page: page,
                pages: count / perPage
            })
        })
    })