在 mongoDB 中使用 Monk 限制查找

Question

I have a large collection of documents. I want to get the first 100 of these. From the Monk Docs, this is the find method I am using

var documents = [];
users.find({}, function (err, docs){
  for(i=0;i<100;i++)
     documents.push(docs[i]);
});

This is highly wasteful since, the entire documents are anyway retrieved. I want something like this (from the mongodb docs)

docs =  db.users.find().limit( 100 );

I tried in monk,

users.find({}, function (err, docs){
  for(i=0;i<docs.length;i++)
     documents.push(docs[i]);
}).limit(100);

But it gives an error saying that there is no function limit in the "promise" object that is returned before it.

Is there such an option in Monk to limit the number of documents?

Answer 1

Yes, you can pass it as an option in the second parameter:

users.find({}, { limit : 100 }, function (err, docs){
  for(i=0;i<docs.length;i++)
     documents.push(docs[i]);
});

This comes from the native node mongodb driver, which monk wraps via mongoskin:

http://mongodb.github.io/node-mongodb-native/markdown-docs/queries.html#query-options

Answer 2

您可以将options对象作为第二个参数传递给.find() ：

users.find({}, {limit: 100}, next);

Answer 3

If you want add pagenumber and sorting, here is how to do it with monk:

users.find({}, {limit: 100, skip: pagenumber, sort: {'username': 1}})  

where limit is for page size, skip is page number, sorting result by username ascending.