问题描述
我在任何地方都找不到它的记载.默认情况下，find()操作将从头开始获取记录.如何获取mongodb中的最后N条记录?
I can't find anywhere it has been documented this. By default, the find() operation will get the records from beginning. How can I get the last N records in mongodb?
我也希望返回的结果按从最近到最近的顺序排列，而不是相反.
also I want the returned result ordered from less recent to most recent, not the reverse.
推荐答案
如果我理解您的问题，则需要按升序排序.
If I understand your question, you need to sort in ascending order.
假设您有一个名为"x"的id或日期字段，那么您可以...
Assuming you have some id or date field called "x" you would do ...
db.foo.find().sort({x:1});
1 将按升序排序(从最旧到最新)， -1 将按降序排序(从最新到最旧).
The 1 will sort ascending (oldest to newest) and -1 will sort descending (newest to oldest.)
如果您使用自动创建的 _id 字段，则该字段中嵌入了一个日期...因此您可以使用该日期进行排序...
If you use the auto created _id field it has a date embedded in it ... so you can use that to order by ...
db.foo.find().sort({_id:1});
这将把您的所有文档从最早的到最新的返回.
That will return back all your documents sorted from oldest to newest.
您还可以使用上述自然顺序 ...
You can also use a Natural Order mentioned above ...
db.foo.find().sort({$natural:1});
再次使用 1 或 -1 ，具体取决于您想要的顺序.
Again, using 1 or -1 depending on the order you want.
最后，在进行这种广泛开放的查询时，最好添加一个限制，这样您就可以...
Lastly, it's good practice to add a limit when doing this sort of wide open query so you could do either ...
db.foo.find().sort({_id:1}).limit(50);
或
db.foo.find().sort({$natural:1}).limit(50);
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