如何检索Mongo DB中数组中存在的所有匹配元素? [英] How to retrieve all matching elements present inside array in Mongo DB?
问题描述
我的文档如下所示:
{
name: "testing",
place:"London",
documents: [
{
x:1,
y:2,
},
{
x:1,
y:3,
},
{
x:4,
y:3,
}
]
}
我想检索所有匹配的文档,即我想以以下格式进行o/p:
I want to retrieve all matching documents i.e. I want o/p in below format:
{
name: "testing",
place:"London",
documents: [
{
x:1,
y:2,
},
{
x:1,
y:3,
}
]
}
我尝试过的是:
db.test.find({"documents.x": 1},{_id: 0, documents: {$elemMatch: {x: 1}}});
但是,它只给出第一个条目.
But, it gives first entry only.
推荐答案
As JohnnyHK said, the answer in MongoDB: select matched elements of subcollection explains it well.
在您的情况下,汇总看起来像这样:
In your case, the aggregate would look like this:
(注意:第一次匹配不是严格必要的,但是它在性能(可以使用索引)和内存使用($ unwind在有限的集合上)方面有帮助
(note: the first match is not strictly necessary, but it helps in regards of performance (can use index) and memory usage ($unwind on a limited set)
> db.xx.aggregate([
... // find the relevant documents in the collection
... // uses index, if defined on documents.x
... { $match: { documents: { $elemMatch: { "x": 1 } } } },
... // flatten array documennts
... { $unwind : "$documents" },
... // match for elements, "documents" is no longer an array
... { $match: { "documents.x" : 1 } },
... // re-create documents array
... { $group : { _id : "$_id", documents : { $addToSet : "$documents" } }}
... ]);
{
"result" : [
{
"_id" : ObjectId("515e2e6657a0887a97cc8d1a"),
"documents" : [
{
"x" : 1,
"y" : 3
},
{
"x" : 1,
"y" : 2
}
]
}
],
"ok" : 1
}
有关Aggregate()的更多信息,请参见> http://docs.mongodb.org /manual/applications/aggregation/
For more information about aggregate(), see http://docs.mongodb.org/manual/applications/aggregation/
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