mongodb 如何删除mongodb中的N个文档
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How to delete N numbers of documents in mongodb
提问by Yogesh
In my collections, documents contains key like status and timestamp. When I want to find latest ten documents then I write following query
在我的收藏中，文档包含状态和时间戳等键。当我想查找最新的十个文档时，我会编写以下查询
db.collectionsname.find().sort({"timestamp"-1}).limit(10)
This query gives me results which I want but when I want to delete latest ten documents then I was writing the following query
此查询为我提供了我想要的结果，但是当我想删除最新的十个文档时，我正在编写以下查询
db.collectionsname.remove({"status":0},10).sort({"timestamp":-1})
but it shows following error
TypeError: Cannot call method 'sort' of undefinedand again I wrote the same query as below
db.collectionsname.remove({"status":0},10)It deletes only one document. So how can I write a query which deletes ten latest documents and sorts on timestamp?
但它显示以下错误
TypeError: Cannot call method 'sort' of undefined，我再次编写了与下面相同的查询
db.collectionsname.remove({"status":0},10)它只删除了一个文档。那么如何编写一个删除十个最新文档并按时间戳排序的查询呢？
回答by WiredPrairie
You can't set a limit when using removeor findAndModify. So, if you want to precisely limit the number of documents removed, you'll need to do it in two steps.
使用remove或时不能设置限制findAndModify。因此，如果您想精确限制删除的文档数量，则需要分两步完成。
db.collectionName.find({}, {_id : 1})
.limit(100)
.sort({timestamp:-1})
.toArray()
.map(function(doc) { return doc._id; }); // Pull out just the _ids
Then pass the returned _ids to the remove method:
然后将返回的_ids传递给 remove 方法：
db.collectionName.remove({_id: {$in: removeIdsArray}})
FYI: you cannot remove documents from a capped collection.
仅供参考：您不能从有上限的集合中删除文档。
回答by deusxmachine
Let N be number of records to delete.
设 N 为要删除的记录数。
db.collectionName.find().limit(N).forEach(doc =>
{
db.collectionName.remove({_id:doc._id})
}
)
回答by Manoj Pandey
To remove N number of documents in your collection myCollection:
要删除集合中的 N 个文档myCollection：
db.getCollection('myCollection').find({}).limit(N).forEach(function(doc){
db.getCollection('myCollection').remove({_id: doc._id});
})
回答by Anthony Awuley
db.collection.remove({_id:
{ $in: db.collection.find().sort({timestamp:-1}).limit(100).map(a => a._id) }
})
回答by jellyDean
Another way is to write a python script.
另一种方法是编写python脚本。
from pymongo import MongoClient
def main():
local_client = MongoClient()
collection = local_client.database.collection
cursor = collection.find()
total_number_of_records = 10000
for document in cursor:
id = document.get("_id")
if total_number_of_records == 100:
break
delete_query = {"_id": id}
collection.delete_one(delete_query)
total_number_of_records -= 1
if __name__ == "__main__":
# execute only if run as a script
main()
回答by yitai wei
query sql is
查询sql是
db.order.find({"业务员姓名" : "吊炸天"},{"业务员编号":0}).sort({ "订单时间" : -1 })
the result is
结果是
{
"_id" : ObjectId("5c9c875fdadfd961b4d847e7"),
"推送ID" : "248437",
"订单时间" : ISODate("2019-03-28T08:35:52Z"),
"订单状态" : "1",
"订单编号" : "20190328163552306694",
"业务员姓名" : "吊炸天"
}
{
"_id" : ObjectId("5c9c875fdadfd961b4d847e8"),
"推送ID" : "248438",
"订单时间" : ISODate("2019-03-28T08:35:52Z"),
"订单状态" : "1",
"订单编号" : "20190328163552178132",
"业务员姓名" : "吊炸天"
}
{
"_id" : ObjectId("5c9c875fdadfd961b4d847e5"),
"推送ID" : "248435",
"订单时间" : ISODate("2019-03-28T08:35:51Z"),
"订单状态" : "1",
"订单编号" : "20190328163551711074",
"业务员姓名" : "吊炸天"
}
{
"_id" : ObjectId("5c9c875fdadfd961b4d847e6"),
"推送ID" : "248436",
"订单时间" : ISODate("2019-03-28T08:35:51Z"),
"订单状态" : "1",
"订单编号" : "20190328163551758179",
"业务员姓名" : "吊炸天"
}
now delete the 3 and 4 data
现在删除3和4数据
var name = ["吊炸天"]
var idArray = db.order.find({"业务员姓名" : {$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}})
return result is
返回结果是
{
"acknowledged" : true,
"deletedCount" : 2
}
回答by Pratiksha Aggarwal
Below query will find and delete the latest 10 documents from collection:-
以下查询将从集合中查找并删除最新的 10 个文档：-
db.collectionsname.findAndModify({
query: { 'status':0 },
sort: { 'timestamp': -1 },
limit: 10,
remove: true
});
