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。

其他回答

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

你可以使用mongoose- pagate -v2。欲了解更多信息,请点击这里

const mongoose         = require('mongoose');
const mongoosePaginate = require('mongoose-paginate-v2');

const mySchema = new mongoose.Schema({
    // your schema code
}); 
mySchema.plugin(mongoosePaginate); 
const myModel = mongoose.model('SampleModel',  mySchema);

myModel.paginate().then({}) // Usage

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) { ... });

如果你正在使用mongoose作为一个restful api的源,请看看 ' retify -mongoose'和它的查询。它内置了这个功能。

集合上的任何查询都提供了在这里有用的标头

test-01:~$ curl -s -D - localhost:3330/data?sort=-created -o /dev/null
HTTP/1.1 200 OK
link: </data?sort=-created&p=0>; rel="first", </data?sort=-created&p=1>; rel="next", </data?sort=-created&p=134715>; rel="last"
.....
Response-Time: 37

所以基本上你得到了一个通用服务器,它对集合的查询具有相对线性的加载时间。这是非常棒的,如果你想要进行自己的实现,可以参考一下。

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。