使用ts-mongoose-pagination

    const trainers = await Trainer.paginate(
        { user: req.userId },
        {
            perPage: 3,
            page: 1,
            select: '-password, -createdAt -updatedAt -__v',
            sort: { createdAt: -1 },
        }
    )

    return res.status(200).json(trainers)

您可以使用skip()和limit()，但效率非常低。更好的解决方案是对索引字段加上limit()进行排序。 我们在Wunderflats发布了一个小库:https://github.com/wunderflats/goosepage 它用了第一种方法。

const ITEMS_PER_PAGE = 2;

exports.getProducts = (req, res, next) => {
  // + will turn the string to a number
  const page = +req.query.page || 1;
  let totalItems;
  //Product model
  Product.find()
    .countDocuments()
    .then((numProducts) => {
      totalItems = numProducts;
      return Product.find()
         //If query param is 3, since ITEMS_PER_PAGE = 2, we skip 2*2 items   
         // we show only 5th and 6th item
        .skip((page - 1) * ITEMS_PER_PAGE)
        .limit(ITEMS_PER_PAGE);
    })
    .then((products) => {
      res.render("shop/products", {
        // maybe sending the products object to templating engine
      });
    })
    .catch((err) => {
      const error = new Error(err);
      error.httpStatusCode = 500;
      // if you are set express error handler, use this
      // when we call next() with an argument passed in, we let express know, we skip all other middlewares, we move to error handling middleware

      return next(error);
    });
};

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
            })
        })
    })

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
        });
    }
};

在用Rodolphe提供的信息仔细研究了Mongoose API后，我想出了这个解决方案:

MyModel.find(query, fields, { skip: 10, limit: 5 }, function(err, results) { ... });