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
实现这一点的可靠方法是使用查询字符串从前端传递值。假设我们想要获得第2页,并将输出限制为25个结果。 page=2&limit=25 //这将被添加到您的URL: http:localhost:5000?= 2限制= 25页
让我们看看代码:
// We would receive the values with req.query.<<valueName>> => e.g. req.query.page
// Since it would be a String we need to convert it to a Number in order to do our
// necessary calculations. Let's do it using the parseInt() method and let's also provide some default values:
const page = parseInt(req.query.page, 10) || 1; // getting the 'page' value
const limit = parseInt(req.query.limit, 10) || 25; // getting the 'limit' value
const startIndex = (page - 1) * limit; // this is how we would calculate the start index aka the SKIP value
const endIndex = page * limit; // this is how we would calculate the end index
// We also need the 'total' and we can get it easily using the Mongoose built-in **countDocuments** method
const total = await <<modelName>>.countDocuments();
// skip() will return a certain number of results after a certain number of documents.
// limit() is used to specify the maximum number of results to be returned.
// Let's assume that both are set (if that's not the case, the default value will be used for)
query = query.skip(startIndex).limit(limit);
// Executing the query
const results = await query;
// Pagination result
// Let's now prepare an object for the frontend
const pagination = {};
// If the endIndex is smaller than the total number of documents, we have a next page
if (endIndex < total) {
pagination.next = {
page: page + 1,
limit
};
}
// If the startIndex is greater than 0, we have a previous page
if (startIndex > 0) {
pagination.prev = {
page: page - 1,
limit
};
}
// Implementing some final touches and making a successful response (Express.js)
const advancedResults = {
success: true,
count: results.length,
pagination,
data: results
}
// That's it. All we have to do now is send the `results` to the frontend.
res.status(200).json(advancedResults);
我建议将这个逻辑实现到中间件中,这样你就可以将它用于各种路由/控制器。
其他回答
使用猫鼬,快递和翡翠的分页-这里有一个链接到我的博客与更多的细节
var perPage = 10
, page = Math.max(0, req.params.page)
Event.find()
.select('name')
.limit(perPage)
.skip(perPage * page)
.sort({
name: 'asc'
})
.exec(function(err, events) {
Event.count().exec(function(err, count) {
res.render('events', {
events: events,
page: page,
pages: count / perPage
})
})
})
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
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});
}
});
});
你可以使用一个叫Mongoose Paginate的小包,让它更容易。
$ npm install mongoose-paginate
在你的路由或控制器后,只需添加:
/**
* querying for `all` {} items in `MyModel`
* paginating by second page, 10 items per page (10 results, page 2)
**/
MyModel.paginate({}, 2, 10, function(error, pageCount, paginatedResults) {
if (error) {
console.error(error);
} else {
console.log('Pages:', pageCount);
console.log(paginatedResults);
}
}
您可以使用skip()和limit(),但效率非常低。更好的解决方案是对索引字段加上limit()进行排序。 我们在Wunderflats发布了一个小库:https://github.com/wunderflats/goosepage 它用了第一种方法。
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