如何在 mongoDB 中分组并返回结果中的所有字段 [英] how to group in mongoDB and return all fields in result

问题描述

我在 mongoDB 中使用聚合方法进行分组，但是当我使用 $group 时，它返回我用来分组的唯一字段.我试过 $project 但它也不起作用.我也尝试了 $first 并且它有效，但结果数据现在采用不同的格式.

I am using aggregate method in mongoDB to group but when I use $group it returns the only field which I used to group. I have tried $project but it is not working either. I also tried $first and it worked but the result data is now in different format.

我需要的响应格式如下:

The response format I need looks like:

{
    "_id" : ObjectId("5b814b2852d47e00514d6a09"),
    "tags" : [],
    "name" : "name here",
    "rating" : "123456789"
}

在我的 query.response 中添加 $group 后，_id 的值发生了变化.(并且 $group 只使用 _id，如果我尝试任何其他关键字，它会引发累加器错误.请解释一下.)

and after adding $group in my query.response is like this, the value of _id changes. (and the $group is taking only _id, if i try any other keyword it throws an error of accumulator something. please explain this also.)

{
    "_id" :"name here" //the value of _id changed to the name field which i used in $group condition
}

我必须删除名称字段中的重复项，而不更改任何结构和字段.我也在使用 nodeJS 和猫鼬，所以请提供适用的解决方案.

I have to remove the duplicates in name field, without changing any structure and fields. also I am using nodeJS with mongoose, so please provide the solution that works with it.

推荐答案

你可以使用下面的聚合查询.

You can use below aggregation query.

$$ROOT 保留每个名称的整个文档，后跟 $replaceRoot 将文档提升到顶部.

$$ROOT to keep the whole document per each name followed by $replaceRoot to promote the document to the top.

db.col.aggregate([
  {"$group":{"_id":"$name","doc":{"$first":"$$ROOT"}}},
  {"$replaceRoot":{"newRoot":"$doc"}}
])

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