获取数组中的第一个元素并使用Aggregate返回? [英] Get first element in array and return using Aggregate?

问题描述

如何使用Mongo聚合获取并返回数组中的第一个元素?

我尝试使用此代码:

db.my_collection.aggregate([
    { $project: {
        resp : { my_field: { $slice: 1 } }
    }}
])

但出现以下错误:

uncaught exception: aggregate failed: {
    "errmsg" : "exception: invalid operator '$slice'",
    "code" : 15999,
    "ok" : 0
}

请注意，'my_field'是一个由4个元素组成的数组，我只需要返回第一个元素.

Note that 'my_field' is an array of 4 elements, and I only need to return the first element.

推荐答案

当前，$slice运算符在聚合管道的$project操作中不可用. 因此，您可以做的是

Currently, the $slice operator is unavailable in the the $project operation, of the aggregation pipeline. So what you could do is,

首先是$unwind，是my_field数组，然后将它们分组在一起，并采用该组的$first元素.

First $unwind, the my_field array, and then group them together and take the $first element of the group.

db.my_collection.aggregate([
{$unwind:"$my_field"},
{$group:{"_id":"$_id","resp":{$first:"$my_field"}}},
{$project:{"_id":0,"resp":1}}
])

或使用find()命令，您可以在projection部分中使用$ slice运算符.

Or using the find() command, where you could make use of the $slice operator in the projection part.

db.my_collection.find({},{"my_field":{$slice:1}})

更新:根据您的评论，假设您只希望数组中的second项用于ID为id的记录.

Update: based on your comments, Say you want only the second item in an array, for the record with an id, id.

var field = 2;
var id = ObjectId("...");

然后，下面的聚合命令为您提供带有_id，id的记录的my_field数组中的第二项.

Then, the below aggregation command gives you the 2nd item in the my_field array of the record with the _id, id.

db.my_collection.aggregate([
{$match:{"_id":id}},
{$unwind:"$my_field"},
{$skip:field-1},
{$limit:1}
])

以上逻辑不能用于更多记录，因为它会在$unwind之后包含一个$group运算符. $group运算符为该特定组中的所有记录生成单个记录，从而使得在以后阶段应用的$limit或$skip运算符无效.

The above logic cannot be applied for more a record, since it would involve a $group, operator after $unwind. The $group operator produces a single record for all the records in that particular group making the $limit or $skip operators applied in the later stages to be ineffective.

在上面的find()查询中进行小的改动也会产生预期的结果.

A small variation on the find() query above would yield you the expected result as well.

db.my_collection.find({},{"my_field":{$slice:[field-1,1]}})

除了这些之外，总有一种方法可以在客户端进行，尽管如果记录数量很大，则可能会花费很多:

Apart from these, there is always a way to do it in the client side, though a bit costly if the number of records is very large:

var field = 2; 
db.my_collection.find().map(function(doc){
return doc.my_field[field-1];
})

从以上选项中进行选择取决于您的数据大小和应用程序设计.

Choosing from the above options depends upon your data size and app design.

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