問題描述
有沒有辦法在 MongoDB 的一個語句中添加遞增的 id？ (Is there a way to add an incrementing id in one statement in MongoDB?)
所以我有一個小型數據庫，它不會增長太多，我正試圖從我用 python 實現的 API 中的數據庫中獲取一個文檔，以便使用給定的文檔 ID 我在D b。但是，我發現讓用戶從數據庫中寫入一個隨機數有點困難。我所需要的只是一個通過設置 id 字段和自動增量來修改每個文檔的函數。正如我所說，它不會增長那麼多，性能也不是問題。
到目前為止，我能做的是：
var i = 0
db.MyCollection.update({},
{$set : {"new_field":1}},
{upsert:false,
multi:true}
i ++;),
我實現了設置一個 id 字段，但它為每個文檔設置了相同的數字（每個文檔的計數）所以假設數據庫有 10 個文檔，它會將 ID 設置為 10。
參考解法
方法 1:
Find‑and‑modify operation returns the document updated (before or after the update depending on returnDocument setting). You can use this with $inc to implement a counter. Ruby example where c is a collection:
irb(main):005:0> c['foo'].insert_one(counter:true,count:1)
=> #<Mongo::Operation::Insert::Result:0x8040 documents=[{"n"=>1, "opTime"=>{"ts"=>#<BSON::Timestamp:0x00005609f260b7e0 @seconds=1594961771, @increment=2>, "t"=>1}, "electionId"=>BSON::ObjectId('7fffffff0000000000000001'), "ok"=>1.0, "$clusterTime"=>{"clusterTime"=>#<BSON::Timestamp:0x00005609f260b538 @seconds=1594961771, @increment=2>, "signature"=>{"hash"=><BSON::Binary:0x8060 type=generic data=0x0000000000000000...>, "keyId"=>0}}, "operationTime"=>#<BSON::Timestamp:0x00005609f260b290 @seconds=1594961771, @increment=2>}]>
irb(main):011:0> c['foo'].find_one_and_update({counter:true},{'$inc':{count:1}})
=> {"_id"=>BSON::ObjectId('5f112f6b2c97a6281f63f575'), "counter"=>true, "count"=>1}
irb(main):012:0> c['foo'].find_one_and_update({counter:true},{'$inc':{count:1}})
=> {"_id"=>BSON::ObjectId('5f112f6b2c97a6281f63f575'), "counter"=>true, "count"=>2}
irb(main):013:0> c['foo'].find_one_and_update({counter:true},{'$inc':{count:1}})
=> {"_id"=>BSON::ObjectId('5f112f6b2c97a6281f63f575'), "counter"=>true, "count"=>3}
irb(main):014:0> c['foo'].find_one_and_update({counter:true},{'$inc':{count:1}})
=> {"_id"=>BSON::ObjectId('5f112f6b2c97a6281f63f575'), "counter"=>true, "count"=>4}
方法 2:
Why not just use this logic? Instead of updating all via one query, just launch multiple queries one by one? Mongo will do it pretty fast, even if you have >1M docs in database (according to your phrase: I got a small database) because pre‑builded index on _id field.
this is a javasript code, but I guess, you'll understand the logic of it
let all_documents = db.MyCollection.find({});
for (let i = 0; i < all_documents.length; i++) {
db.MyCollection.update({_id: all_documents[i]._id }, {$set : {"new_field": i}}, {upsert:false})
}
(by Diego Duarte、D. SM、AlexZeDim)