我正在用Node.js和mongoose写一个web应用程序。如何对我从.find()调用得到的结果进行分页?我想要一个功能可比的“限制50,100”在SQL。


当前回答

最简单和更快速的方法是,用objectId进行分页 例子;

初始加载条件

condition = {limit:12, type:""};

从响应数据中获取第一个和最后一个ObjectId

下一页条件

condition = {limit:12, type:"next", firstId:"57762a4c875adce3c38c662d", lastId:"57762a4c875adce3c38c6615"};

下一页条件

condition = {limit:12, type:"next", firstId:"57762a4c875adce3c38c6645", lastId:"57762a4c875adce3c38c6675"};

在猫鼬

var condition = {};
    var sort = { _id: 1 };
    if (req.body.type == "next") {
        condition._id = { $gt: req.body.lastId };
    } else if (req.body.type == "prev") {
        sort = { _id: -1 };
        condition._id = { $lt: req.body.firstId };
    }

var query = Model.find(condition, {}, { sort: sort }).limit(req.body.limit);

query.exec(function(err, properties) {
        return res.json({ "result": result);
});

其他回答

最简单和更快速的方法是,用objectId进行分页 例子;

初始加载条件

condition = {limit:12, type:""};

从响应数据中获取第一个和最后一个ObjectId

下一页条件

condition = {limit:12, type:"next", firstId:"57762a4c875adce3c38c662d", lastId:"57762a4c875adce3c38c6615"};

下一页条件

condition = {limit:12, type:"next", firstId:"57762a4c875adce3c38c6645", lastId:"57762a4c875adce3c38c6675"};

在猫鼬

var condition = {};
    var sort = { _id: 1 };
    if (req.body.type == "next") {
        condition._id = { $gt: req.body.lastId };
    } else if (req.body.type == "prev") {
        sort = { _id: -1 };
        condition._id = { $lt: req.body.firstId };
    }

var query = Model.find(condition, {}, { sort: sort }).limit(req.body.limit);

query.exec(function(err, properties) {
        return res.json({ "result": result);
});

这是一个版本,我附加到我所有的模型。为了方便,它依赖于下划线,为了性能,它依赖于异步。opts允许使用mongoose语法进行字段选择和排序。

var _ = require('underscore');
var async = require('async');

function findPaginated(filter, opts, cb) {
  var defaults = {skip : 0, limit : 10};
  opts = _.extend({}, defaults, opts);

  filter = _.extend({}, filter);

  var cntQry = this.find(filter);
  var qry = this.find(filter);

  if (opts.sort) {
    qry = qry.sort(opts.sort);
  }
  if (opts.fields) {
    qry = qry.select(opts.fields);
  }

  qry = qry.limit(opts.limit).skip(opts.skip);

  async.parallel(
    [
      function (cb) {
        cntQry.count(cb);
      },
      function (cb) {
        qry.exec(cb);
      }
    ],
    function (err, results) {
      if (err) return cb(err);
      var count = 0, ret = [];

      _.each(results, function (r) {
        if (typeof(r) == 'number') {
          count = r;
        } else if (typeof(r) != 'number') {
          ret = r;
        }
      });

      cb(null, {totalCount : count, results : ret});
    }
  );

  return qry;
}

将它附加到您的模型模式。

MySchema.statics.findPaginated = findPaginated;

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

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);
    });
};

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
        });
    }
};