从 mongoDB 数组中获取特定元素
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Get particular element from mongoDB array
提问by Swapnil Sonawane
I have mongo collection like below
我有如下 mongo 集合
{
"auther" : "xyz" ,
"location" : "zzz" ,
"books" :
[
{"book1" : "b1" , "date" : 2-3-00} ,
{"book1" : "b2" , "date" : 4-9-00}
]
}
{
"auther" : "pqr",
"location" : "zzz" ,
"books" :
[
{"book1" : "b1" , "date" : 2-4-00}
]
}
I want to get the only the date of book b1 and author xyz .
我想获得书 b1 和作者 xyz 的唯一日期。
i have make query like below
我有如下查询
db.coll.find({"auther" : "xyz" , "books.book1" : "b1"} , {"books.date" : 1})
but it's gives output as follows
但它给出的输出如下
"books" : {"date" : 2-4-00} , "books" : {"date" : 4-9-00}
I want to get the only the date of book b1 and other xyz .means only "books" : {"date" : 2-4-00}
我想获得书籍 b1 和其他 xyz 的唯一日期。 "books" : {"date" : 2-4-00}
is it possible in mongo or am I doing something wrong?
在mongo中可能还是我做错了什么？
采纳答案by Gates VP
The MongoDB query language is designed to return all matching Documents.
MongoDB 查询语言旨在返回所有匹配的文档。
There is no support for returning only sub-documents.
不支持仅返回子文档。
This issue has an outstanding ticketin MongoDB's ticket tracker.
这个问题在 MongoDB 的票务跟踪器中有一张未清票。
UPDATE:it looks like the ticket has been marked as fixed.
更新：看起来票已被标记为固定。
See herefor an example of how to use this.
有关如何使用它的示例，请参见此处。
回答by bm_i
It can be done using map/reduce, just emit the sub element into a temporary inline collection. Its a Hack and it works however I'd advise against it as map/reduce is single threaded, and has a large overhead for what you want to achieve, it is much easier to just extract the sub-element in your application.
可以使用 map/reduce 来完成，只需将子元素发送到临时内联集合中即可。它是一个 Hack 并且它可以工作，但是我建议不要使用它，因为 map/reduce 是单线程的，并且对于您想要实现的目标有很大的开销，在您的应用程序中提取子元素要容易得多。
Something like this...
像这样的东西...
map:
地图：
m = function() {
this.books.forEach(function(book){
if(book1 == 'b1'){
emit("books", {date: book.date,});
}
});
}
reduce:
降低：
r = function(key, values) {
return this;
}
query:
询问：
db.coll.mapReduce(m, r, {query : {"auther" : "xyz" , "books.book1" : "b1"}, out: { inline : 1}})
回答by Swapnil Sonawane
if you want to select only matching element you can query like this.
如果您只想选择匹配的元素，您可以这样查询。
b.coll.find({"auther" : "xyz" , "books.book1" : "b1"},{"books.$.date" : 1})
b.coll.find({"auther" : "xyz" , "books.book1" : "b1"},{"books.$.date" : 1})
回答by Jason Sebring
With a little imagination (pre mongo v 2.6) ...
有点想象力（pre mongo v 2.6）......
You can do this with aggregate or map reduce. Aggregate is newer, easier and more optimized. Here's a sample of returning a sub document with aggregate assuming your collection is named "Authors". I took the liberty of spelling things correctly.
您可以使用聚合或映射缩减来做到这一点。聚合更新、更简单、更优化。这是返回带有聚合的子文档的示例，假设您的集合名为“作者”。我冒昧地拼写正确。
Authors.aggregate([
{ $match: { author: 'xyz' } },
{ $unwind: '$books' },
{
$project: {
_id: '$books.book1',
date: '$books.date'
}
},
{ $match: { '$_id' : 'b1' } }
]);
You'll get back an array with a single entry like so:
您将返回一个包含单个条目的数组，如下所示：
[{ _id: 'b1', date: '2-4-00' }]
Otherwise, if mongo 2.6+you can do the really easy way:
否则，如果mongo 2.6+你可以做一个非常简单的方法：
Authors.find({
author: 'xyz',
books: { $elemMatch: { book1: 'b1' } }
},'books')
Where you will get back the books collection if found and only a single record within:
如果找到并且只有一个记录，您将在哪里取回书籍收藏：
{ _id: 'xyz', books: [ { book1: 'b1', date: '2-4-00' } ] }
