mongodb聚合框架-获取嵌套数组的第一个文档的字段 [英] mongodb aggregation framework - Fetch first document's field of the nested array

问题描述

考虑以下内容是我存储在集合Users

Consider the following is my document stored in collection Users

{
  _id : "User1",
  joined : ISODate("2011-03-02"),
  likes : {
           sublikes: [
                      {WebsiteID:'001': WebsiteName: 'ABCD'},
                      {WebsiteID:'002': WebsiteName: '2BC2'},
                      {WebsiteID:'003': WebsiteName: '3BC3'},
                      //........... 
                      //........... 
                      {WebsiteID:'999999': WebsiteName: 'SOME_NAME'}
                     ]
          }
}

现在使用mongodb聚合框架，我需要获取

Now using mongodb aggregation framework I need to fetch that

collection.aggregate([
                        { $project: {
                            _id: 1,
                            "UserID": "$_id", 
                            "WebsiteName": "$likes.sublikes[0]['WebsiteName']"
                        }
                        },
                         { $match: { _id: 'User1'} }
                      ], function (err, doc) {
 ///Not able to get the WebsiteName: 'ABCD'
});

如果我使用$unwind，则文档会变大(在上述情况下尤其如此)，因此，我不想为了仅获取数组中的第一项而展开它(与其他项无关)

If I use $unwind the document becomes bulk (specifically in the above case), So I don't want to unwind it for getting only first item in an array (irrespective of others)

谁能给我提示如何访问并重命名该字段?

---

更新:1:即使我尝试使用"WebsiteName": "$likes.sublikes.0.WebsiteName"进行投影.没用:-(

Can anyone give me hints on how to access is and rename that field?

---

Update 1: Even I tried to project with "WebsiteName": "$likes.sublikes.0.WebsiteName". Didn't work :-(

更新:2:未解决的问题- https://jira.mongodb.org/browse/SERVER-4589

Update 2: Open issue - https://jira.mongodb.org/browse/SERVER-4589

到目前为止，$at无法正常工作.引发错误:

As of now $at does not work. Throws an error:

{ [MongoError: exception: invalid operator '$at']
  name: 'MongoError',
  errmsg: 'exception: invalid operator \'$at\'',
  code: 15999,
  ok: 0 }

然后使用 $unwind

Till then using $unwind

// Array Remove - By John Resig (MIT Licensed)
Array.prototype.remove = function(from, to) {
  var rest = this.slice((to || from) + 1 || this.length);
  this.length = from < 0 ? this.length + from : from;
  return this.push.apply(this, rest);
};

推荐答案

获得结果的最简单方法是使用普通的find查询和$slice运算符:

The easiest way to achieve your result is using a normal find query and the $slice operator:

db.collection.find( {_id: "User1"}, {"likes.sublikes": {$slice: 1}} )

聚合框架(从MongoDB 2.4.1开始)不支持$slice或数组索引(投票/监视功能请求: SERVER-4589 )

The aggregation framework (as at MongoDB 2.4.1) does not support $slice or array indexes (vote/watch feature requests: SERVER-6074 and SERVER-4589).

您可以使用$unwind，$group和$first运算符在聚合框架中执行此操作，例如:

You could do this in aggregation framework using $unwind, $group and the $first operator, eg:

db.collection.aggregate([
    { $match: {
         _id : "User1"
    }},
    { $unwind: "$likes.sublikes" },
    { $group: {
        _id: "$_id",
        like: { $first: "$likes.sublikes" }
    }},
    { $project: {
        _id: 0,
        "UserID": "$_id",
        "WebsiteName": "$like.WebsiteName"
    }}
])

正常的$slice应该是性能最高的选项.

The normal $slice should be the most performant option.

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