MongoDB的UpdateMany與$在和UPSERT

Question

命名persons1

蒙戈集合包含以下數據：

db.persons1.find().pretty(); 


{ "_id" : "Sims", "count" : 32 } 
{ "_id" : "Autumn", "count" : 35 } 
{ "_id" : "Becker", "count" : 35 } 
{ "_id" : "Cecile", "count" : 40 } 
{ "_id" : "Poole", "count" : 32 } 
{ "_id" : "Nanette", "count" : 31 } 

現在通過Java我寫的代碼遞增計數這是目前在列表中的用戶

MongoClient mongoclient = new MongoClient("localhost", 27017); 
MongoDatabase db = mongoclient.getDatabase("testdb1"); 
MongoCollection<Document> collection = db.getCollection("persons1"); 
List li = new ArrayList(); 
li.add("Sims"); 
li.add("Autumn"); 

collection.updateMany(
    in("_id",li), 
    new Document("$inc", new Document("count", 1)), 
    new UpdateOptions().upsert(true)); 

我運行上面的java程序後，我的輸出如下。

db.persons1.find().pretty(); 


{ "_id" : "Sims", "count" : 33 } 
{ "_id" : "Autumn", "count" : 36 } 
{ "_id" : "Becker", "count" : 35 } 
{ "_id" : "Cecile", "count" : 40 } 
{ "_id" : "Poole", "count" : 32 } 
{ "_id" : "Nanette", "count" : 31 } 

我的問題：是否可以插入，並從1開始計數，對於存在於數組列表和persons1集合不存在的項目嗎？

問題描述：

之前計劃數據庫包含詳情如下：

{ "_id" : "Sims", "count" : 33 } 
{ "_id" : "Autumn", "count" : 36 } 
{ "_id" : "Becker", "count" : 35 } 
{ "_id" : "Cecile", "count" : 40 } 
{ "_id" : "Poole", "count" : 32 } 
{ "_id" : "Nanette", "count" : 31 } 

樣品Java代碼：

MongoClient mongoclient = new MongoClient("localhost", 27017); 
MongoDatabase db = mongoclient.getDatabase("testdb1"); 
MongoCollection<Document> collection = db.getCollection("persons1"); 
List li = new ArrayList(); 

// Entry already Present so required to increment by 1 
li.add("Sims"); 

// Entry already Present so required to increment by 1 
li.add("Autumn"); 

// Entry is NOT Present, hence insert into persons data base with "_id" as User1 and count as 1 
li.add("User1"); 

// Entry is NOT Present, hence insert into persons data base with "_id" as User1 and count as 1 
li.add("User2"); 

// Code to be written 

應該是什麼代碼從數據庫放出來如下圖所示：

{ "_id" : "Sims", "count" : 34 } // Entry already Present, incremented by 1 
{ "_id" : "Autumn", "count" : 37 } // Entry already Present, incremented by 1 
{ "_id" : "Becker", "count" : 35 } 
{ "_id" : "Cecile", "count" : 40 } 
{ "_id" : "Poole", "count" : 32 } 
{ "_id" : "Nanette", "count" : 31 } 
{ "_id" : "User1", "count" : 1 } // Entry Not Present, start by 1 
{ "_id" : "User2", "count" : 1 } // Entry Not Present, start by 1 

Answer 1

這裏的「catch」是_id的$in參數不會被解釋爲_id字段在「多標記」更新中的有效「填充符」，這就是您正在執行的操作。所有的_id值將被默認填充ObjectId值，而不是「upsert」。

解決這個問題的方法是使用「批量」操作，以及與Java 3.x的驅動程序，您使用BulkWrite類和結構是這樣的：

MongoCollection<Document> collection = db.getCollection("persons1"); 

    List li = new ArrayList(); 

    li.add("Sims"); 
    li.add("User2"); 

    List<WriteModel<Document>> updates = new ArrayList<WriteModel<Document>>(); 

    ListIterator listIterator = li.listIterator(); 

    while (listIterator.hasNext()) { 
     updates.add(
      new UpdateOneModel<Document>(
       new Document("_id",listIterator.next()), 
       new Document("$inc",new Document("count",1)), 
       new UpdateOptions().upsert(true) 
      ) 
     ); 
    } 

    BulkWriteResult bulkWriteResult = collection.bulkWrite(updates); 

操縱你的基本List到UpdateOneModel對象並列出適合bulkWrite的列表，並且在一個請求中發送所有「個體」更新，即使它們是「技術上」多個更新語句。

這是通過更新操作一般有效設置多個_id密鑰或匹配通過$in的唯一方式。