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

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

其他回答

这是一个示例函数,用于获得具有分页和限制选项的技能模型的结果

 export function get_skills(req, res){
     console.log('get_skills');
     var page = req.body.page; // 1 or 2
     var size = req.body.size; // 5 or 10 per page
     var query = {};
     if(page < 0 || page === 0)
     {
        result = {'status': 401,'message':'invalid page number,should start with 1'};
        return res.json(result);
     }
     query.skip = size * (page - 1)
     query.limit = size
     Skills.count({},function(err1,tot_count){ //to get the total count of skills
      if(err1)
      {
         res.json({
            status: 401,
            message:'something went wrong!',
            err: err,
         })
      }
      else 
      {
         Skills.find({},{},query).sort({'name':1}).exec(function(err,skill_doc){
             if(!err)
             {
                 res.json({
                     status: 200,
                     message:'Skills list',
                     data: data,
                     tot_count: tot_count,
                 })
             }
             else
             {
                 res.json({
                      status: 401,
                      message: 'something went wrong',
                      err: err
                 })
             }
        }) //Skills.find end
    }
 });//Skills.count end

}

**//localhost:3000/asanas/?pageNo=1&size=3**

//requiring asanas model
const asanas = require("../models/asanas");


const fetchAllAsanasDao = () => {
    return new Promise((resolve, reject) => {

    var pageNo = parseInt(req.query.pageNo);
    var size = parseInt(req.query.size);
    var query = {};
        if (pageNo < 0 || pageNo === 0) {
            response = {
                "error": true,
                "message": "invalid page number, should start with 1"
            };
            return res.json(response);
        }
        query.skip = size * (pageNo - 1);
        query.limit = size;

  asanas
            .find(pageNo , size , query)
        .then((asanasResult) => {
                resolve(asanasResult);
            })
            .catch((error) => {
                reject(error);
            });

    });
}

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

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) { ... });

你可以像这样串起来:

var query = Model.find().sort('mykey', 1).skip(2).limit(5)

使用exec执行查询

query.exec(callback);