MongoDB$facet聚合
我的$facet聚合有问题。我还想知道它是否是副产品,我将知道该产品是畅销书,新添加的,以及最受欢迎的MongoDB$facet聚合,mongodb,mongodb-query,aggregation-framework,Mongodb,Mongodb Query,Aggregation Framework,我的$facet聚合有问题。我还想知道它是否是副产品,我将知道该产品是畅销书,新添加的,以及最受欢迎的 $facet: { "byBestSeller": [ { $sort: { "sold":-1 }}, { $limit: 1 } ], "byMostReviewed": [ { $addFields: { "noOfReview
$facet: {
"byBestSeller": [
{ $sort: { "sold":-1 }},
{ $limit: 1 }
],
"byMostReviewed": [
{ $addFields: { "noOfReviews": {$size:"$reviews.message"}}},
{ $sort: { "noOfReviews":-1 }},
{ $limit: 1 }
],
"byNewlyAdded": [
{ $sort: { "createdAt":-1 }},
{ $limit: 1 }
],
"byAllProducts": [
{ $limit: 100 }
],
}
现在,我的结果是这样的
"byBestSeller": [
{
"sold": 150,
"name": "Big Mac",
"shop": [
{
"name": "McDonald's"
}
],
"createdAt": {
"2020-05-11T01:55:17.623Z"
},
"reviews": [
{
"message": "Yummylicious"
},
{
"message": "Yummy burger"
}
]
}
],
"byMostReviewed": [
{
"sold": 100,
"name": "Spaghetti",
"shop": [
{
"name": "McDonald's"
}
],
"createdAt": {
"2020-06-11T01:55:17.623Z"
},
"reviews": [
{
"message": "Yummylicious"
},
{
"message": "Yummy Spaghetti"
},
{
"message": "Yummy super rich spag."
}
],
"noOfReviews": 3
}
],
"byNewlyArrived": [
{
"sold": 0,
"name": "Fries",
"shop": [
{
"name": "McDonald's"
}
],
"createdAt": {
"2020-7-11T01:55:17.623Z"
},
"reviews": [
{
"message": "Yummylicious"
},
]
}
],
"byAllProducts": [
{
"sold": 150,
"name": "Big Mac",
"shop": [
{
"name": "McDonald's"
}
],
"createdAt": {
"2020-05-11T01:55:17.623Z"
},
"reviews": [
{
"message": "Yummylicious"
},
{
"message": "Yummy burger"
}
]
}
{
"sold": 100,
"name": "Fries",
"shop": [
{
"name": "McDonald's"
}
],
"createdAt": {
"2020-06-11T01:55:17.623Z"
},
"reviews": [
{
"message": "Yummylicious"
},
{
"message": "Yummy Spaghetti"
},
{
"message": "Yummy super rich spag."
}
]
},
{
"sold": 0,
"name": "Fries",
"shop": [
{
"name": "McDonald's"
}
],
"createdAt": {
"2020-07-11T01:55:17.623Z"
},
"reviews": [
{
"message": "Yummylicious"
},
]
}
],
有没有一种方法,即使它包含在所有产品中,我也能知道它是最畅销、最受欢迎和最新推出的