MongoDB的UpdateMany与$在和UPSERT

Question

命名persons1

蒙戈集合包含以下数据：

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 } 

现在通过Java我写的代码递增计数这是目前在列表中的用户

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)); 

我运行上面的java程序后，我的输出如下。

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 } 

我的问题：是否可以插入，并从1开始计数，对于存在于数组列表和persons1集合不存在的项目吗？

问题描述：

之前计划数据库包含详情如下：

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

样品Java代码：

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 

应该是什么代码从数据库放出来如下图所示：

{ "_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

这里的“catch”是_id的$in参数不会被解释为_id字段在“多标记”更新中的有效“填充符”，这就是您正在执行的操作。所有的_id值将被默认填充ObjectId值，而不是“upsert”。

解决这个问题的方法是使用“批量”操作，以及与Java 3.x的驱动程序，您使用BulkWrite类和结构是这样的：

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); 

操纵你的基本List到UpdateOneModel对象并列出适合bulkWrite的列表，并且在一个请求中发送所有“个体”更新，即使它们是“技术上”多个更新语句。

这是通过更新操作一般有效设置多个_id密钥或匹配通过$in的唯一方式。