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
使用这个简单的插件。
https://github.com/WebGangster/mongoose-paginate-v2
安装
NPM安装mongoose- pagate -v2 使用 添加插件到一个模式,然后使用模型paginate方法:
Const mongoose = require('mongoose'); const mongoosePaginate = require(' mongoosePaginate -v2'); const mySchema =新的猫鼬。模式({ /*你的模式定义*/ }); mySchema.plugin (mongoosePaginate); const myModel =猫鼬。模型(SampleModel, mySchema); myModel.paginate().then({}) //使用方法
其他回答
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) { ... });
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});
}
});
});
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);
});
};
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.
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,
})
);
});
});
};
推荐文章
- elasticsearch vs . MongoDB用于过滤应用程序
- ReferenceError: description没有定义NodeJs
- MongoDB记录所有查询
- 将一个二进制的NodeJS Buffer转换为JavaScript的ArrayBuffer
- 使用LIMIT/OFFSET运行查询,还可以获得总行数
- AngularJS只适用于单页应用程序吗?
- 如何在vue-cli项目中更改端口号
- MongoDB:如何找到安装的MongoDB的确切版本
- 同步和异步编程(在node.js中)的区别是什么?
- 如何编辑通过npm安装的节点模块?
- 如何使用mongoimport导入CSV文件?
- “node_modules”文件夹应该包含在git存储库中吗
- 使用package.json在全局和本地安装依赖项
- 在mongodb中存储日期/时间的最佳方法
- this.libOptions.parse不是一个函数