如何仅显示MongoDB中的唯一记录，而忽略重复记录？
> db.demo613.insertOne({"Name":"Chris"});{
"acknowledged" : true, "insertedId" : ObjectId("5e988bd4f6b89257f5584d88")
}
> db.demo613.insertOne({"Name":"Bob"});{
"acknowledged" : true, "insertedId" : ObjectId("5e988bdbf6b89257f5584d89")
}
> db.demo613.insertOne({"Name":"Bob"});{
"acknowledged" : true, "insertedId" : ObjectId("5e988bddf6b89257f5584d8a")
}
> db.demo613.insertOne({"Name":"David"});{
"acknowledged" : true, "insertedId" : ObjectId("5e988be0f6b89257f5584d8b")
}
> db.demo613.insertOne({"Name":"Chris"});{
"acknowledged" : true, "insertedId" : ObjectId("5e988be2f6b89257f5584d8c")
}
> db.demo613.insertOne({"Name":"Sam"});{
"acknowledged" : true, "insertedId" : ObjectId("5e988be8f6b89257f5584d8d")
}在find()方法的帮助下显示集合中的所有文档-
> db.demo613.find();
这将产生以下输出-
{ "_id" : ObjectId("5e988bd4f6b89257f5584d88"), "Name" : "Chris" }
{ "_id" : ObjectId("5e988bdbf6b89257f5584d89"), "Name" : "Bob" }
{ "_id" : ObjectId("5e988bddf6b89257f5584d8a"), "Name" : "Bob" }
{ "_id" : ObjectId("5e988be0f6b89257f5584d8b"), "Name" : "David" }
{ "_id" : ObjectId("5e988be2f6b89257f5584d8c"), "Name" : "Chris" }
{ "_id" : ObjectId("5e988be8f6b89257f5584d8d"), "Name" : "Sam" }以下是获取唯一记录的查询-
> db.demo613.distinct("Name");这将产生以下输出-
[ "Chris", "Bob", "David", "Sam" ]