mongodb 查询数组大小大于 1 的文档
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Query for documents where array size is greater than 1
提问by emson
I have a MongoDB collection with documents in the following format:
我有一个包含以下格式文档的 MongoDB 集合：
{
"_id" : ObjectId("4e8ae86d08101908e1000001"),
"name" : ["Name"],
"zipcode" : ["2223"]
}
{
"_id" : ObjectId("4e8ae86d08101908e1000002"),
"name" : ["Another ", "Name"],
"zipcode" : ["2224"]
}
I can currently get documents that match a specific array size:
我目前可以获得匹配特定数组大小的文档：
db.accommodations.find({ name : { $size : 2 }})
This correctly returns the documents with 2 elements in the namearray. However, I can't do a $gtcommand to return all documents where the namefield has an array size of greater than 2:
这将正确返回name数组中包含 2 个元素的文档。但是，我无法执行$gt命令来返回该name字段的数组大小大于 2 的所有文档：
db.accommodations.find({ name : { $size: { $gt : 1 } }})
How can I select all documents with a namearray of a size greater than one (preferably without having to modify the current data structure)?
如何选择name大小大于 1的数组的所有文档（最好无需修改当前数据结构）？
采纳答案by Andrew Orsich
Update:
更新：
For mongodb versions 2.2+more efficient way to do this described by @JohnnyHKin another answer.
MongoDB的版本2.2+更有效的方式来做到这一点的描述@JohnnyHK在另一个答案。
1.Using $where
1.使用$where
db.accommodations.find( { $where: "this.name.length > 1" } );
But...
但...
Javascript executes more slowly than the native operators listed on this page, but is very flexible. See the server-side processing page for more information.
Javascript 的执行速度比此页面上列出的本机运算符慢，但非常灵活。有关更多信息，请参阅服务器端处理页面。
2.Create extrafield NamesArrayLength, update it with names array length and then use in queries:
2.创建额外的字段NamesArrayLength，用名称数组长度更新它，然后在查询中使用：
db.accommodations.find({"NamesArrayLength": {$gt: 1} });
It will be better solution, and will work much faster (you can create index on it).
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）。
回答by JohnnyHK
There's a more efficient way to do this in MongoDB 2.2+ now that you can use numeric array indexes in query object keys.
现在您可以在查询对象键中使用数字数组索引，在 MongoDB 2.2+ 中有一种更有效的方法来执行此操作。
// Find all docs that have at least two name array elements.
db.accommodations.find({'name.1': {$exists: true}})
You can support this query with an index that uses a partial filter expression (requires 3.2+):
您可以使用使用部分过滤器表达式的索引支持此查询（需要 3.2+）：
// index for at least two name array elements
db.accommodations.createIndex(
{'name.1': 1},
{partialFilterExpression: {'name.1': {$exists: true}}}
);
回答by Tobia
I believe this is the fastest query that answers your question, because it doesn't use an interpreted $whereclause:
我相信这是回答您问题的最快查询，因为它不使用解释$where子句：
{$nor: [
{name: {$exists: false}},
{name: {$size: 0}},
{name: {$size: 1}}
]}
It means "all documents except those without a name (either non existant or empty array) or with just one name."
它的意思是“除了没有名称（不存在或空数组）或只有一个名称的文档之外的所有文档”。
Test:
测试：
> db.test.save({})
> db.test.save({name: []})
> db.test.save({name: ['George']})
> db.test.save({name: ['George', 'Raymond']})
> db.test.save({name: ['George', 'Raymond', 'Richard']})
> db.test.save({name: ['George', 'Raymond', 'Richard', 'Martin']})
> db.test.find({$nor: [{name: {$exists: false}}, {name: {$size: 0}}, {name: {$size: 1}}]})
{ "_id" : ObjectId("511907e3fb13145a3d2e225b"), "name" : [ "George", "Raymond" ] }
{ "_id" : ObjectId("511907e3fb13145a3d2e225c"), "name" : [ "George", "Raymond", "Richard" ] }
{ "_id" : ObjectId("511907e3fb13145a3d2e225d"), "name" : [ "George", "Raymond", "Richard", "Martin" ] }
>
回答by one_cent_thought
You can use aggregate, too:
您也可以使用聚合：
db.accommodations.aggregate(
[
{$project: {_id:1, name:1, zipcode:1,
size_of_name: {$size: "$name"}
}
},
{$match: {"size_of_name": {$gt: 1}}}
])
// you add "size_of_name" to transit document and use it to filter the size of the name
// 您将“size_of_name”添加到传输文档并使用它来过滤名称的大小
回答by Aman Goel
Try to do something like this:
尝试做这样的事情：
db.getCollection('collectionName').find({'ArrayName.1': {$exists: true}})
1 is number, if you want to fetch record greater than 50 then do ArrayName.50 Thanks.
1 是数字，如果你想获取大于 50 的记录，那么做 ArrayName.50 谢谢。
回答by lesolorzanov
None of the above worked for me. This one did so I'm sharing it:
以上都不适合我。这个是这样做的，所以我要分享它：
db.collection.find( {arrayName : {$exists:true}, $where:'this.arrayName.length>1'} )
回答by Sagar Veeram
You can use $expr( 3.6 mongo version operator ) to use aggregation functions in regular query.
您可以使用$expr( 3.6 mongo version operator ) 在常规查询中使用聚合函数。
Compare query operatorsvs aggregation comparison operators.
比较query operatorsvs aggregation comparison operators。
db.accommodations.find({$expr:{$gt:[{$size:"$name"}, 1]}})
回答by Yadvendar
db.accommodations.find({"name":{"$exists":true, "$ne":[], "$not":{"$size":1}}})
回答by Daniele Tassone
MongoDB 3.6 include $expr https://docs.mongodb.com/manual/reference/operator/query/expr/
MongoDB 3.6 包括 $expr https://docs.mongodb.com/manual/reference/operator/query/expr/
You can use $expr in order to evaluate an expression inside a $match, or find.
您可以使用 $expr 来计算 $match 或 find 中的表达式。
{ $match: {
$expr: {$gt: [{$size: "$yourArrayField"}, 0]}
}
}
or find
或找到
collection.find({$expr: {$gte: [{$size: "$yourArrayField"}, 0]}});
回答by Barrard
I found this solution, to find items with an array field greater than certain length
我找到了这个解决方案，以查找数组字段大于特定长度的项目
db.allusers.aggregate([
{$match:{username:{$exists:true}}},
{$project: { count: { $size:"$locations.lat" }}},
{$match:{count:{$gt:20}}}
])
The first $match aggregate uses an argument thats true for all the documents. If blank, i would get
第一个 $match 聚合使用一个对所有文档都为真的参数。如果空白，我会得到
"errmsg" : "exception: The argument to $size must be an Array, but was of type: EOO"
