如果子文档值不存在，Mongo DB将插入到子文档中 [英] Mongo DB insert into a sub document if sub document value doesn't exist

问题描述

我对mongodb很新，我有点失落。

I'm very new to mongodb and i'm a little lost.

我有mongo db集合，如下所示：

I have mongo db collection that looks like this :

  ({
    _id:id ,
    createdAt: new Date(),
    name:name,
    friends : [{name:1,children:[{name:sarah,age:12}]}],
    dogs : [{}]
  });

如果名称不存在，我希望能够在friends数组中插入新元素新数组的子元素。

I would like to be able to insert a new element in the friends array if the name doesnt exist and a children element to that new array.

伪代码

如果目录.find id.friends.name = true

更新并添加一个新的子文档{name：sarah ，年龄：5}

else

使用新的子子文档创建一个新的朋友元素

经过大量研究后，似乎人们推荐使用$存在，但我找不到在子文档中使用它的方法'

After alot of research it seems that people are recommending the use of $exist but i can't find a way to use it in a sub document'

我已经想出了这个，但它不是真的很有效，我想知道我怎么能用它作为我的if else stament：

I've come up with this but it's not really working and i'm wondering how could i use that for my if else stament:

db.Directory.find({_id:id},{friends.name:"foo", {"$exists": True}})

对于实际查询

db.Directory.update(
    { _id: id, 'friends.name': "foo" },
    {$push: {'friends.$.children':  {name:"x",age:"yy"}}}
)

如果它不存在：

db.Directory.insert(
    { _id: id, 'friends.name': "foo" },
    {$push: {'friends.$.children':  {name:"x",age:"yy"}}}
)

但它并没有真正起作用，我不确定我的公式是否有什么问题，因为这是研究中的大量试验和错误。
我也不知道是否正确的方法，因为我找不到任何关于子文档上存在的值的东西，并且正计划尝试进行Find（）搜索存储该值，然后在JS的结果中测试true或false以进行更新/插入调用。

But it's not really working , i'm not sure if there is something wrong in my formulas as it's a lot of trials and errors from research. I also don't really know if that the proper way to do it as i couldn't find anything on a value existing on sub documents and was planning on trying to do a Find() search store the value and then test for true or false on the result in JS to make either of the update/insert calls.

希望有人可以提供帮助

编辑：
假设我想将{name：john，年龄：15}添加到该朋友姓名：1

Edit : Let's say i want to add {name:john,age:15} to that friend name:1

friends : [{name:1,children:[{name:sarah,age:12}]}]

我希望输出为

friends : [{name:1,children:[{name:sarah,age:12},{name:john,age:15}]}]

或者如果名称1不存在

friends : []

使它输出

friends : [{name:1,children:[{name:john,age:15}]}]

更新示例：

案例一：

friends : []

我添加了新朋友名字josh和一个孩子莎拉，12岁我得到：

i add a new friend name "josh" with one child sarah , age 12 i get:

friends : [{name:"josh",children:[{name:sarah,age:12}]}]

这很完美。

Now i add a new friend name 2 with children tom 15.

没有任何事情发生。

之后，我想给josh添加一个新孩子，时间15岁

after , I want to add a new child to josh , timmy age 15

我得到：

friends : [{name:"josh",children:[{name:timmy,age:12}]}]

莎拉消失了。

推荐答案

请尝试通过 批量操作 ，通过 $ addToSet operator and update fri通过 $ set <结束/ a>运算符，在您更清楚地提出问题后更新它。

Please try to do it through Bulk operation, add one new friend through $addToSet operator and update friend through $set operator, update it after you making your question more clearly.

var bulk = db.Directory.initializeOrderedBulkOp();

// if `friends` is `[]`, push the empty children firstly through addToSet
bulk.find({_id: id, 'friends.name': {$exists: false}}).updateOne(
     {$addToSet: {friends: {name: 1, children: []}});

// if we find the match friend, update this one through `$set`
bulk.find({_id: id, 'friends.children.name': 'john'}).updateOne(
    {$set: {'friends.$.children': {name: 'john', age: 22}}});

// if we cannot find match friend, insert the new one through `$addToSet`
bulk.find({_id: id, 'friends.children.name': {$ne: "john"}}).updateOne( 
    {$addToSet: {'friends.0.children': {name: 'john', age: 12}}});

bulk.execute();

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