我的檔案：
[{
"title": "lenovo x-100",
"brand": "lenovo",
"category": "laptops",
"variant": [{
"price": 30000,
"RAM": "4GB",
"storage": "256GB",
"screen": "full hd",
"chip": "i3"
}, {
"price": 35000,
"RAM": "8GB",
"storage": "512GB",
"screen": "full hd",
"chip": "i5"
}, {
"price": 40000,
"RAM": "12GB",
"storage": "2TB",
"screen": "uhd",
"chip": "i7"
}],
"salesCount": 32,
"buysCount": 35,
"viewsCount": 60
},
{
"title": "samsung12",
"brand": "lenovo",
"category": "mobile phones",
"variant": [{
"price": 11000,
"RAM": "4GB",
"ROM": "32GB"
}, {
"price": 16000,
"RAM": "6GB",
"ROM": "64GB"
}, {
"price": 21000,
"RAM": "8GB",
"ROM": "128GB"
}],
"salesCount": 48,
"buysCount": 39,
"viewsCount": 74
}
預期產出
{
_id:"lenovo",
minPrice:1100
}
我試過這種聚合方法
[{
$match: {
brand: 'lenovo'
}
}, {
$group: {
_id: '$brand',
prices: {
$min: '$variant.price'
}
}
}, {
$unwind: {
path: '$prices'
}
}, {
$group: {
_id: '$_id',
minPrice: {
$min: '$prices'
}
}
}]
我想根據品牌找到最低價格，這個查詢回傳預期的輸出，但是有沒有更好的方法來獲得預期的結果，因為使用 $unwind 運算子非常昂貴，因為它可能需要更長的執行時間，希望積極的回應。在此先感謝。
uj5u.com熱心網友回復：
您可以使用$reduce替換第二$group階段。
$match$group- 推variant.price入新陣列和結果嵌套陣列陣列。$project：3.1。
$reduce- 用于將陣列從結果2中的嵌套陣列展平$concat為一個。3.2. - 從結果3.1
$min中選擇最小值。
db.collection.aggregate([
{
$match: {
brand: "lenovo"
}
},
{
$group: {
_id: "$brand",
prices: {
$push: "$variant.price"
}
}
},
{
$project: {
_id: 1,
minPrice: {
$min: {
"$reduce": {
"input": "$prices",
"initialValue": [],
"in": {
"$concatArrays": [
"$$value",
"$$this"
]
}
}
}
}
}
}
])
示例 Mongo Playground
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