mongodb Mongo 按数组长度排序
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Mongo order by length of array
提问by Evgenius
Lets say I have mongo documents like this:
假设我有这样的 mongo 文件：
Question 1
问题 1
{
answers:[
{content: 'answer1'},
{content: '2nd answer'}
]
}
Question 2
问题2
{
answers:[
{content: 'answer1'},
{content: '2nd answer'}
{content: 'The third answer'}
]
}
Is there a way to order the collection by size of answers?
有没有办法按答案的大小对集合进行排序？
After a little research I saw suggestions of adding another field, that would contain number of answers and use it as a reference but may be there is native way to do it?
经过一些研究，我看到了添加另一个领域的建议，该领域将包含许多答案并将其用作参考，但可能有本地方法可以做到这一点？
回答by Eve Freeman
I thought you might be able to use $size, but that's only to find arrays of a certain size, not ordering.
我以为您可以使用$size，但这只是为了找到特定大小的数组，而不是排序。
From the mongo documentation: http://www.mongodb.org/display/DOCS/Advanced+Queries#AdvancedQueries-%24size
来自 mongo 文档：http: //www.mongodb.org/display/DOCS/Advanced+Queries#AdvancedQueries-%24size
You cannot use $size to find a range of sizes (for example: arrays with more than 1 element). If you need to query for a range, create an extra size field that you increment when you add elements. Indexes cannot be used for the $size portion of a query, although if other query expressions are included indexes may be used to search for matches on that portion of the query expression.
您不能使用 $size 来查找大小范围（例如：具有超过 1 个元素的数组）。如果您需要查询范围，请创建一个额外的大小字段，在添加元素时增加该字段。索引不能用于查询的 $size 部分，但如果包含其他查询表达式，则索引可用于搜索查询表达式该部分的匹配项。
Looks like you can probably fairly easily do this with the new aggregation framework, edit:which isn't out yet. http://www.mongodb.org/display/DOCS/Aggregation+Framework
看起来您可以使用新的聚合框架很容易地做到这一点，编辑：尚未发布。 http://www.mongodb.org/display/DOCS/Aggregation+Framework
UpdateNow the Aggregation Framework is out...
现在更新聚合框架已经出来......
> db.test.aggregate([
{$unwind: "$answers"},
{$group: {_id:"$_id", answers: {$push:"$answers"}, size: {$sum:1}}},
{$sort:{size:1}}]);
{
"result" : [
{
"_id" : ObjectId("5053b4547d820880c3469365"),
"answers" : [
{
"content" : "answer1"
},
{
"content" : "2nd answer"
}
],
"size" : 2
},
{
"_id" : ObjectId("5053b46d7d820880c3469366"),
"answers" : [
{
"content" : "answer1"
},
{
"content" : "2nd answer"
},
{
"content" : "The third answer"
}
],
"size" : 3
}
],
"ok" : 1
}
回答by Oleg
I use $projectfor this:
我$project为此使用：
db.test.aggregate([
{
$project : { answers_count: {$size: { "$ifNull": [ "$answers", [] ] } } }
},
{
$sort: {"answers_count":1}
}
])
It also allows to include documents with empty answers.
But also has a disadvantage (or sometimes advantage): you should manually add all needed fields in $projectstep.
它还允许包含答案为空的文档。但也有一个缺点（或有时是优点）：您应该$project逐步手动添加所有需要的字段。
回答by bhavin jalodara
you can use mongodb aggregation stage $addFieldswhich will add extra field to store count and then followed by $sortstage.
您可以使用 mongodb 聚合阶段$addFields，它将添加额外的字段来存储计数，然后是$sort阶段。
db.test.aggregate([
{
$addFields: { answers_count: {$size: { "$ifNull": [ "$answers", [] ] } } }
},
{
$sort: {"answers_count":1}
}
])
