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

也可以用async/await实现结果。

下面的代码示例使用hapi v17和mongoose v5的异步处理程序

{
            method: 'GET',
            path: '/api/v1/paintings',
            config: {
                description: 'Get all the paintings',
                tags: ['api', 'v1', 'all paintings']
            },
            handler: async (request, reply) => {
                /*
                 * Grab the querystring parameters
                 * page and limit to handle our pagination
                */
                var pageOptions = {
                    page: parseInt(request.query.page) - 1 || 0, 
                    limit: parseInt(request.query.limit) || 10
                }
                /*
                 * Apply our sort and limit
                */
               try {
                    return await Painting.find()
                        .sort({dateCreated: 1, dateModified: -1})
                        .skip(pageOptions.page * pageOptions.limit)
                        .limit(pageOptions.limit)
                        .exec();
               } catch(err) {
                   return err;
               }

            }
        }

其他回答

这是一个你可以尝试的例子,

var _pageNumber = 2,
  _pageSize = 50;

Student.count({},function(err,count){
  Student.find({}, null, {
    sort: {
      Name: 1
    }
  }).skip(_pageNumber > 0 ? ((_pageNumber - 1) * _pageSize) : 0).limit(_pageSize).exec(function(err, docs) {
    if (err)
      res.json(err);
    else
      res.json({
        "TotalCount": count,
        "_Array": docs
      });
  });
 });
app.get("/:page",(req,res)=>{
        post.find({}).then((data)=>{
            let per_page = 5;
            let num_page = Number(req.params.page);
            let max_pages = Math.ceil(data.length/per_page);
            if(num_page == 0 || num_page > max_pages){
                res.render('404');
            }else{
                let starting = per_page*(num_page-1)
                let ending = per_page+starting
                res.render('posts', {posts:data.slice(starting,ending), pages: max_pages, current_page: num_page});
            }
        });
});

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);
let page,limit,skip,lastPage, query;
 page = req.params.page *1 || 1;  //This is the page,fetch from the server
 limit = req.params.limit * 1 || 1; //  This is the limit ,it also fetch from the server
 skip = (page - 1) * limit;   // Number of skip document
 lastPage = page * limit;   //last index 
 counts = await userModel.countDocuments() //Number of document in the collection

query = query.skip(skip).limit(limit) //current page

const paginate = {}

//For previous page
if(skip > 0) {
   paginate.prev = {
       page: page - 1,
       limit: limit
} 
//For next page
 if(lastPage < counts) {
  paginate.next = {
     page: page + 1,
     limit: limit
}
results = await query //Here is the final results of the query.

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
        });
    }
};