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
});
}
};
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