如何在Mongoid中强制执行唯一的嵌入式文档 [英] How to enforce unique embedded document in mongoid
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问题描述
我有以下型号
class Person
include Mongoid::Document
embeds_many :tasks
end
class Task
include Mongoid::Document
embedded_in :commit, :inverse_of => :tasks
field :name
end
如何确保以下内容?
person.tasks.create :name => "create facebook killer"
person.tasks.create :name => "create facebook killer"
person.tasks.count == 1
different_person.tasks.create :name => "create facebook killer"
person.tasks.count == 1
different_person.tasks.count == 1
即任务名称在特定人员中是唯一的
i.e. task names are unique within a particular person
检查了有关索引的文档后,我认为以下方法可能会起作用:
Having checked out the docs on indexes I thought the following might work:
class Person
include Mongoid::Document
embeds_many :tasks
index [
["tasks.name", Mongo::ASCENDING],
["_id", Mongo::ASCENDING]
], :unique => true
end
但是
person.tasks.create :name => "create facebook killer"
person.tasks.create :name => "create facebook killer"
仍然会产生重复.
上面Person中显示的索引配置将转换为mongodb
The index config shown above in Person would translate into for mongodb
db.things.ensureIndex({firstname : 1, 'tasks.name' : 1}, {unique : true})
推荐答案
默认情况下,索引不是唯一的.如果您查看 Mongo Docs ,那么唯一性就是一个额外的标志
Indexes are not unique by default. If you look at the Mongo Docs on this, uniqueness is an extra flag.
我不知道确切的Mongoid翻译,但是您正在寻找这样的东西:
I don't know the exact Mongoid translation, but you're looking for something like this:
db.things.ensureIndex({firstname : 1}, {unique : true, dropDups : true})
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