查询数组大小大于 1 的文档 [英] Query for documents where array size is greater than 1

问题描述

我有一个包含以下格式文档的 MongoDB 集合:

I have a MongoDB collection with documents in the following format:

{
  "_id" : ObjectId("4e8ae86d08101908e1000001"),
  "name" : ["Name"],
  "zipcode" : ["2223"]
}
{
  "_id" : ObjectId("4e8ae86d08101908e1000002"),
  "name" : ["Another ", "Name"],
  "zipcode" : ["2224"]
}

我目前可以获取匹配特定数组大小的文档:

I can currently get documents that match a specific array size:

db.accommodations.find({ name : { $size : 2 }})

这会正确返回 name 数组中包含 2 个元素的文档.但是，我无法执行 $gt 命令来返回 name 字段的数组大小大于 2 的所有文档:

This correctly returns the documents with 2 elements in the name array. However, I can't do a $gt command to return all documents where the name field has an array size of greater than 2:

db.accommodations.find({ name : { $size: { $gt : 1 } }})

如何选择name数组的大小大于1的所有文档(最好无需修改当前数据结构)?

How can I select all documents with a name array of a size greater than one (preferably without having to modify the current data structure)?

推荐答案

更新:

对于 mongodb 版本 2.2+，@JohnnyHK 在另一个 回答.

For mongodb versions 2.2+ more efficient way to do this described by @JohnnyHK in another answer.

1.使用$where

db.accommodations.find( { $where: "this.name.length > 1" } );

可是……

Javascript 的执行速度比上面列出的本机运算符慢这个页面，但是很灵活.查看服务器端处理页面了解更多信息.

Javascript executes more slowly than the native operators listed on this page, but is very flexible. See the server-side processing page for more information.

2.创建extra字段NamesArrayLength，用names数组长度更新它，然后在查询中使用:

2.Create extra field NamesArrayLength, update it with names array length and then use in queries:

db.accommodations.find({"NamesArrayLength": {$gt: 1} });

这将是更好的解决方案，并且运行速度会更快(您可以在其上创建索引).

It will be better solution, and will work much faster (you can create index on it).

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