如何过滤产品和优化mongodb查询,
我们想得到流行的产品,根据一些条件,即产品是订单,view和likes。
db.products.aggregate([
{
"$lookup": {
"from": "orders",
"localField": "_id",
"foreignField": "product_id",
"as": "orders"
}
},
{
"$addFields": {
"orderCount": {
"$size": {
"$cond": [
{
"$isArray": "$orders"
},
"$orders",
[]
]
}
}
}
},
{
"$addFields": {
"likeCount": {
"$size": {
"$cond": [
{
"$isArray": "$likes"
},
"$likes",
[]
]
}
}
}
},
{
"$addFields": {
"sumCount": {
"$sum": [
"$orderCount",
"$likeCount",
"$view"
]
}
}
},
{
$sort: {
"sumCount": -1
}
}
])https://mongoplayground.net/p/fIG3-yHGuV6
必须使用多个$addFields,才能实现拥有最多orders、likes和views的产品。请指点
谢谢
发布于 2021-08-11 15:26:27
我建议进行两次修正,
$lookup阶段总是在数组中返回。orderCount阶段对likeCount和$addFields进行两种操作。你最后的疑问是,
db.products.aggregate([
{
"$lookup": {
"from": "orders",
"localField": "_id",
"foreignField": "product_id",
"as": "orders"
}
},
{
"$addFields": {
"orderCount": { "$size": "$orders" },
"likeCount": {
"$size": {
"$cond": [{ "$isArray": "$likes" }, "$likes", []]
}
}
}
},
{
"$addFields": {
"sumCount": {
"$sum": ["$orderCount", "$likeCount", "$view"]
}
}
},
{ "$sort": { "sumCount": -1 } }
])发布于 2021-08-11 13:26:19
您还可以使用投影将代码最小化。
{
"$project": {
"likes": 1,
"orderCount": {
"$size": {
"$cond": {
"if": {
"$isArray": [
"$orders"
]
},
"then": "$orders",
"else": []
}
}
},
"likeCount": {
"$size": {
"$cond": {
"if": {
"$isArray": [
"$likes"
]
},
"then": "$likes",
"else": []
}
}
},
"views": {
"$ifNull": [
"$view",
0
]
}
}
},去蒙古操场看看。
https://stackoverflow.com/questions/68742147
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