英文:
MongoDb Aggregate Total Count Before Grouping
问题
我有一个聚合管道，用于分组对象并在分组对象的特定字段上保存计数。您可以在此处重现问题：https://mongoplayground.net/p/2DGaiQDYDBP。
模式如下：
[
{
"_id": {
"$oid": "63ce93ffb6e06322db59fdc0"
},
"fruit": "apple",
"source": "tree",
"is_fruit_important": "true"
},
{
"_id": {
"$oid": "63ce93ffb6e06322db59fdc1"
},
"fruit": "orange",
"source": "tree",
"is_fruit_important": "false"
}
]
当前查询按源分组水果，并保存每个组的重要水果计数。应用聚合后，查询结果类似于以下内容：
[
{
"count": {
"number_of_important_fruits": 1
},
"objects": [
{
"fruit": "apple",
"id": "63ce93ffb6e06322db59fdc0",
"is_fruit_important": "true",
"source": "tree"
},
{
"fruit": "orange",
"id": "63ce93ffb6e06322db59fdc1",
"is_fruit_important": "false",
"source": "tree"
}
],
"source": {
"source-of": "tree"
}
}
]
是否有一种方法可以将数据库中的所有水果数量添加到响应对象中？例如，像这样：
{
"total-count": 2,
"result": [
{
"count": {
"number_of_important_fruits": 1
},
"objects": [
{
"fruit": "apple",
"id": "63ce93ffb6e06322db59fdc0",
"is_fruit_important": "true",
"source": "tree"
},
{
"fruit": "orange",
"id": "63ce93ffb6e06322db59fdc1",
"is_fruit_important": "false",
"source": "tree"
}
],
"source": {
"source-of": "tree"
}
}
]
}
它们可以在单独的聚合管道中处理，但我不想实现这种方式。任何帮助将不胜感激。
英文:
I have an aggregation pipeline that groups objects and holds count for some specific field for grouped objects. You can reproduce the problem here: https://mongoplayground.net/p/2DGaiQDYDBP .
The schema is like this;
[
{
"_id": {
"$oid": "63ce93ffb6e06322db59fdc0"
},
"fruit": "apple",
"source": "tree",
"is_fruit_important": "true"
},
{
"_id": {
"$oid": "63ce93ffb6e06322db59fdc1"
},
"fruit": "orange",
"source": "tree",
"is_fruit_important": "false"
},
]
and the current query groups fruits by the source, and holds the count of important fruits for every group. After applying aggregation I get something like this after query:
[
{
"count": {
"number_of_important_fruits": 1
},
"objects": [
{
"fruit": "apple",
"id": "63ce93ffb6e06322db59fdc0",
"is_fruit_important": "true",
"source": "tree"
},
{
"fruit": "orange",
"id": "63ce93ffb6e06322db59fdc1",
"is_fruit_important": "false",
"source": "tree"
}
],
"source": {
"source-of": "tree"
}
}
]
Is there a way to put the number of all fruits in the database to the response object. For example like this:
{
"total-count": 2,
"result": [
{
"count": {
"number_of_important_fruits": 1
},
"objects": [
{
"fruit": "apple",
"id": "63ce93ffb6e06322db59fdc0",
"is_fruit_important": "true",
"source": "tree"
},
{
"fruit": "orange",
"id": "63ce93ffb6e06322db59fdc1",
"is_fruit_important": "false",
"source": "tree"
}
],
"source": {
"source-of": "tree"
}
}
]
}
They can be handled in separate aggregation pipelines but that's what I would not like to implement. Any help would be highly appreciated.
答案1
得分: 2
{
"$group": {
"_id": null,
"result": { "$push": "$$ROOT" },
"count_total": { "$sum": { "$size": "$objects" } },
"count_important": { "$sum": "$count.number_of_important_fruits" }
}
}
<details>
<summary>英文:</summary>
Add one additional group stage just before the final $project, using `$sum` with `$size` for a total count, or add up the important counts for a total important count.
{$group: {
_id: null,
result: {$push: "$$ROOT"},
"count_total": {$sum: {$size: "$objects"}},
"count_important": {$sum: "$count.number_of_important_fruits"}
}},
[Playground](https://mongoplayground.net/p/mocYDXQjHCF)
</details>
# 答案2
**得分**: 1
你只需在聚合管道中添加一个`$facet`阶段，将所有结果推送到`result`中。然后对`result`执行`$size`操作以获取`total-count`。
```js
db.collection.aggregate([
...,
{
"$facet": {
"result": [],
"total-important-count": [
{
$group: {
_id: null,
cnt: {
$sum: "$count.number_of_important_fruits"
}
}
}
]
}
},
{
"$addFields": {
"total-count": {
$size: "$result"
},
"total-important-count": {
$first: "$total-important-count.cnt"
}
}
}
])
英文:
You can simply add a $facet stage to push all your results into result. Then perform a $size on result to get total-count.
db.collection.aggregate([
...,
{
"$facet": {
"result": [],
"total-important-count": [
{
$group: {
_id: null,
cnt: {
$sum: "$count.number_of_important_fruits"
}
}
}
]
}
},
{
"$addFields": {
"total-count": {
$size: "$result"
},
"total-important-count": {
$first: "$total-important-count.cnt"
}
}
}
])
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