blocks|key|912888|text|在使用remove或findAndModify时，不能设置限制。因此，如果要精确限制删除的文档数量，则需要分两步完成。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|912889|db.collectionName.find({},+{_id+:+1})
++++.limit(100)
++++.sort({timestamp:-1})
++++.toArray()
++++.map(function(doc)+{+return+doc._id;+});++//+Pull+out+just+the+_ids|code-block|syntax|javascript|912890|然后将返回的_ids传递给remove方法：|912891|db.collectionName.remove({_id:+{$in:+removeIdsArray}})|912892|FYI:不能从有上限的集合中删除文档。|912893|entityMap^0|3|6|A|D|0|0|6|3|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|11|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|12|8|@]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|13|8|@]|D|@]|E|$]]]|R|$]]

You can't set a limit when using <code>remove</code> or <code>findAndModify</code>. So, if you want to precisely limit the number of documents removed, you'll need to do it in two steps. 

<pre><code>db.collectionName.find({}, {_id : 1})
 .limit(100)
 .sort({timestamp:-1})
 .toArray()
 .map(function(doc) { return doc._id; }); // Pull out just the _ids
</code></pre>

Then pass the returned <code>_id</code>s to the remove method:

<pre><code>db.collectionName.remove({_id: {$in: removeIdsArray}})
</code></pre>

FYI: you cannot remove documents from a capped collection.

blocks|key|2257542|text|让N是要删除的记录数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2257543|++++db.collectionName.find().limit(N).forEach(doc+=>+
+++++{+
+++++++db.collectionName.remove({_id:doc._id})
+++++}
++++)|code-block|syntax|javascript|2257544|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Let N be number of records to delete.

<pre><code> db.collectionName.find().limit(N).forEach(doc =&gt; 
 { 
 db.collectionName.remove({_id:doc._id})
 }
 )
</code></pre>

blocks|key|912966|text|您可以将查找查询映射的输出管道到使用_id并执行基于$in查询的删除，例如：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|912967|db.collection.remove({_id:+
++++{+$in:+db.collection.find().sort({timestamp:-1}).limit(100).map(a+=>+a._id)+}
})|code-block|syntax|javascript|912968|entityMap^0|I|3|Q|3|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

You can pipeline the output of find query mapping to use <code>_id</code> and performing a remove based on <code>$in</code> query such as:
<pre><code>db.collection.remove({_id: 
 { $in: db.collection.find().sort({timestamp:-1}).limit(100).map(a =&gt; a._id) }
})
</code></pre>

blocks|key|912955|text|删除集合myCollection中的N个文档|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|912956|db.getCollection('myCollection').find({}).limit(N).forEach(function(doc){
++++db.getCollection('myCollection').remove({_id:+doc._id});
})|code-block|syntax|javascript|912957|entityMap^0|4|C|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

To remove N number of documents in your collection <code>myCollection</code>:

<pre><code>db.getCollection('myCollection').find({}).limit(N).forEach(function(doc){
 db.getCollection('myCollection').remove({_id: doc._id});
})
</code></pre>

blocks|key|913005|text|工作解决方案(受上面答案的启发)：|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|913006|(对于删除大量记录，$in操作符有一个限制，因此这是最终解决方案)，例如删除1,000,000条记录|CODE|913007|var+ITERATIONS=1000;
var+LIMIT=1000;
for(i=0;+i<ITERATIONS;+i%2B%2B)+{
++++arr+=+db.MyCollection.find({},+{_id+:+1}).limit(LIMIT).toArray().map(function(doc)+{+return+doc._id;+});
++++db.MyCollection.remove({_id:+{$in:+arr}});
++++//+sleep(1);++//+Optional+for+not+loading+mongo+in+case+of+huge+amount+of+deletions
}|code-block|syntax|javascript|913008|可以在mongo或Robomongo3T中运行|913009|entityMap^0|0|6|0|A|3|0|0|0|K|0^^$0|@$1|2|3|4|5|6|7|R|8|@$9|S|A|T|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|U|8|@$9|V|A|W|B|H]]|D|@]|E|$]]|$1|I|3|J|5|K|7|X|8|@]|D|@]|E|$L|M]]|$1|N|3|O|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|P|3|-4|5|6|7|11|8|@]|D|@]|E|$]]]|Q|$]]

Working solution (Inspired by answers above):
(For deleting Huge amount of records the <code>$in</code> operator has a limitation so this is the ultimate solution)
Example for deleting 1,000,000 records
<pre><code>var ITERATIONS=1000;
var LIMIT=1000;
for(i=0; i&lt;ITERATIONS; i++) {
 arr = db.MyCollection.find({}, {_id : 1}).limit(LIMIT).toArray().map(function(doc) { return doc._id; });
 db.MyCollection.remove({_id: {$in: arr}});
 // sleep(1); // Optional for not loading mongo in case of huge amount of deletions
}
</code></pre>
You can run this in mongo shell or in Robomongo 3T

blocks|key|3398575|text|以下是另一种方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3398576|步骤1:临时集合中的排序和限制以及$out文档|3398577|db.my_coll.aggregate([
++{+$sort:+{+_id:+-+1}+},
++{+$limit:+10000+},
++{+$out:+'my_coll_temp'+}
])|code-block|syntax|javascript|3398578|步骤2:删除原始集合中的所有文档(不删除，因为这样可以保留索引)。同样，这一步骤可能会使这种方法变得不那么通用，因为删除可能并不总是可能的。|3398579|db.my_coll.remove({})|3398580|步骤3:将文档从临时集合移回原始集合|3398581|db.my_coll_temp.aggregate([+{+$out:+'my_coll'+}+])|3398582|步骤4:删除临时集合|3398583|db.my_coll_temp.drop()|3398584|entityMap^0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|X|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|Z|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|10|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|11|8|@]|9|@]|A|$]]|$1|O|3|P|5|F|7|12|8|@]|9|@]|A|$G|H]]|$1|Q|3|R|5|6|7|13|8|@]|9|@]|A|$]]|$1|S|3|T|5|F|7|14|8|@]|9|@]|A|$G|H]]|$1|U|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|V|$]]

Here is another approach:

Step 1: Sort and limit and $out documents in temporary collection

<pre><code>db.my_coll.aggregate([
 { $sort: { _id: - 1} },
 { $limit: 10000 },
 { $out: 'my_coll_temp' }
])
</code></pre>

Step 2: Remove (Not dropping since, this preserves indexes) all the documents from the original collection. Also this step is what might make this approach less versatile since, removing might not always be possible.

<pre><code>db.my_coll.remove({})
</code></pre>

Step 3: Move back the documents from temporary collection to original collection

<pre><code>db.my_coll_temp.aggregate([ { $out: 'my_coll' } ])
</code></pre>

Step 4: Drop the temporary collection

<pre><code>db.my_coll_temp.drop()
</code></pre>

blocks|key|2257803|text|我发现的方法是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2257804|++++++++let+uid+=+"your_user";

++++++++let+a+=+await+db
++++++++++.collection(`notifications`)
++++++++++.find({
++++++++++++owner:+uid,
++++++++++})
++++++++++.sort({+_id:+-1+})
++++++++++.skip(1000)
++++++++++.limit(1)
++++++++++.toArray();

++++++++if+(a.length+>+0)+{
++++++++++let+_id+=+new+ObjectId(a[0]._id);
++++++++++db.collection(
++++++++++++`notifications`
++++++++++).deleteMany({
++++++++++++owner:+uid,
++++++++++++_id:+{+$lt:+_id+},
++++++++++});
++++++++}|code-block|syntax|javascript|2257805|说明：“a”变量获取要删除旧项的最后一个条目的_id。您可以使用'skip‘指定什么n值。例子:你有5000条条目，但你想保留400条新条目，只需将其设置为“400”即可。在本例中，我们将按_id进行排序，但您可以使用您想要的。您可以使用find筛选结果，使其与所需的完全匹配。|2257806|算法的下半部分进行删除。使用相同的过滤器(owner:uid)来匹配第一个查找，并删除许多小于"_id“的内容。|2257807|您还可以在此代码块之前使用“count”if语句来检查是否有必要，以避免服务器海岸。|2257808|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

The way I find is:
<pre><code> let uid = &quot;your_user&quot;;

 let a = await db
 .collection(`notifications`)
 .find({
 owner: uid,
 })
 .sort({ _id: -1 })
 .skip(1000)
 .limit(1)
 .toArray();

 if (a.length &gt; 0) {
 let _id = new ObjectId(a[0]._id);
 db.collection(
 `notifications`
 ).deleteMany({
 owner: uid,
 _id: { $lt: _id },
 });
 }
</code></pre>
Explanation: The 'a' variable get the _id of the last entry you want to delete olders. You can specify what n value by using 'skip'. Example: You have 5000 entries but you wanna stay with 400 new ones, just set skip to '400'. In this case we are sorting by _id, but you can use what you want. You can use find to filter results to match exactly what you need.
The second half of algorithm do the delete. Use the same filter (owner:uid) to match the first find, and delete many less than &quot;_id&quot;.
You also use a 'count' if statement before this code block to check if it is necessary, to avoid server coasts.

blocks|key|912915|text|另一种方法是编写python脚本。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|912916|from+pymongo+import+MongoClient

def+main():
++++local_client+=+MongoClient()
++++collection+=+local_client.database.collection
++++cursor+=+collection.find()
++++total_number_of_records+=+10000

++++for+document+in+cursor:
++++++++id+=+document.get("_id")

++++++++if+total_number_of_records+==+100:
++++++++++++break

++++++++delete_query+=+{"_id":+id}
++++++++collection.delete_one(delete_query)

++++++++total_number_of_records+-=+1

if+__name__+==+"__main__":
++++#+execute+only+if+run+as+a+script
++++main()|code-block|syntax|javascript|912917|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Another way is to write a python script.

<pre><code>from pymongo import MongoClient

def main():
 local_client = MongoClient()
 collection = local_client.database.collection
 cursor = collection.find()
 total_number_of_records = 10000

 for document in cursor:
 id = document.get("_id")

 if total_number_of_records == 100:
 break

 delete_query = {"_id": id}
 collection.delete_one(delete_query)

 total_number_of_records -= 1

if __name__ == "__main__":
 # execute only if run as a script
 main()
</code></pre>

blocks|key|912903|text|以下查询将从收藏中找到并删除最新的10份文件：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|912904|db.collectionsname.findAndModify({
++++query:+{+'status':0+},
++++sort:+{+'timestamp':+-1+},
++++limit:+10,
++++remove:+true
});|code-block|syntax|javascript|912905|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Below query will find and delete the latest 10 documents from collection:-

<pre><code>db.collectionsname.findAndModify({
 query: { 'status':0 },
 sort: { 'timestamp': -1 },
 limit: 10,
 remove: true
});
</code></pre>

blocks|key|912942|text|查询sql是|type|unstyled|depth|inlineStyleRanges|entityRanges|data|912943|db.order.find({"业务员姓名"+:+"吊炸天"},{"业务员编号":0}).sort({+"订单时间"+:+-1+})|code-block|syntax|javascript|912944|结果是|912945|{
"_id"+:+ObjectId("5c9c875fdadfd961b4d847e7"),
"推送ID"+:+"248437",
"订单时间"+:+ISODate("2019-03-28T08:35:52Z"),
"订单状态"+:+"1",
"订单编号"+:+"20190328163552306694",
"业务员姓名"+:+"吊炸天"
}
{
"_id"+:+ObjectId("5c9c875fdadfd961b4d847e8"),
"推送ID"+:+"248438",
"订单时间"+:+ISODate("2019-03-28T08:35:52Z"),
"订单状态"+:+"1",
"订单编号"+:+"20190328163552178132",
"业务员姓名"+:+"吊炸天"
}
{
"_id"+:+ObjectId("5c9c875fdadfd961b4d847e5"),
"推送ID"+:+"248435",
"订单时间"+:+ISODate("2019-03-28T08:35:51Z"),
"订单状态"+:+"1",
"订单编号"+:+"20190328163551711074",
"业务员姓名"+:+"吊炸天"
}
{
"_id"+:+ObjectId("5c9c875fdadfd961b4d847e6"),
"推送ID"+:+"248436",
"订单时间"+:+ISODate("2019-03-28T08:35:51Z"),
"订单状态"+:+"1",
"订单编号"+:+"20190328163551758179",
"业务员姓名"+:+"吊炸天"
}|912946|现在删除3和4数据|912947|var+name+=+["吊炸天"]
var+idArray+=+db.order.find({"业务员姓名"+:+{$in:name}},{"订单编号":1,})
++++++++++++++++.sort({+"订单时间"+:+-1+})
++++++++++++++++.skip(2)
++++++++++++++++.map(function(doc){return+doc.订单编号})

db.order.deleteMany({"订单编号":{$in:idArray}})|912948|返回结果是|912949|{
"acknowledged"+:+true,
"deletedCount"+:+2
}|912950|entityMap^0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|Y|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|10|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|S|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|T|$]]

query sql is

<pre><code>db.order.find({"业务员姓名" : "吊炸天"},{"业务员编号":0}).sort({ "订单时间" : -1 })
</code></pre>

the result is 

<pre><code>{
"_id" : ObjectId("5c9c875fdadfd961b4d847e7"),
"推送ID" : "248437",
"订单时间" : ISODate("2019-03-28T08:35:52Z"),
"订单状态" : "1",
"订单编号" : "20190328163552306694",
"业务员姓名" : "吊炸天"
}
{
"_id" : ObjectId("5c9c875fdadfd961b4d847e8"),
"推送ID" : "248438",
"订单时间" : ISODate("2019-03-28T08:35:52Z"),
"订单状态" : "1",
"订单编号" : "20190328163552178132",
"业务员姓名" : "吊炸天"
}
{
"_id" : ObjectId("5c9c875fdadfd961b4d847e5"),
"推送ID" : "248435",
"订单时间" : ISODate("2019-03-28T08:35:51Z"),
"订单状态" : "1",
"订单编号" : "20190328163551711074",
"业务员姓名" : "吊炸天"
}
{
"_id" : ObjectId("5c9c875fdadfd961b4d847e6"),
"推送ID" : "248436",
"订单时间" : ISODate("2019-03-28T08:35:51Z"),
"订单状态" : "1",
"订单编号" : "20190328163551758179",
"业务员姓名" : "吊炸天"
}
</code></pre>

now delete the 3 and 4 data

<pre><code>var name = ["吊炸天"]
var idArray = db.order.find({"业务员姓名" : {$in:name}},{"订单编号":1,})
 .sort({ "订单时间" : -1 })
 .skip(2)
 .map(function(doc){return doc.订单编号})

db.order.deleteMany({"订单编号":{$in:idArray}})
</code></pre>

return result is

<pre><code>{
"acknowledged" : true,
"deletedCount" : 2
}
</code></pre>

In my collections, documents contains key like status and timestamp. When I want to find latest ten documents then I write following query

<pre><code>db.collectionsname.find().sort({"timestamp"-1}).limit(10)
</code></pre>

This query gives me results which I want but when I want to delete latest ten documents then I was writing the following query 

<pre><code>db.collectionsname.remove({"status":0},10).sort({"timestamp":-1})
</code></pre>

but it shows following error 
 <code>TypeError: Cannot call method 'sort' of undefined</code>
and again I wrote the same query as below 
 <code>db.collectionsname.remove({"status":0},10)</code>
It deletes only one document. So how can I write a query which deletes ten latest documents and sorts on timestamp?

How to delete N numbers of documents in mongodb

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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在我的集合中，文档包含状态和时间戳之类的键。当我想找到最新的10个文档时，我会编写以下查询db.collectionsname.find().sort({"timestamp"-1}).limit(10)这个查询给出了我想要的结果，但是当我想删除最新的10个文档时，我编写了下面的查询db.collectionsname.remove({"status":0},10).sort({"timestam

问如何删除mongodb中的N个文档号
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