我正在嘗試用陣列聚合一個物件。非常感謝幫助。
這是我的示例物件。
[
{
"tipo": "A",
"prices": [
{
"min_pax": 1,
"max_pax": 3,
"type": "One Way",
"price": 35
},
{
"min_pax": 1,
"max_pax": 3,
"type": "Round Trip",
"price": 63
},
{
"min_pax": 4,
"max_pax": 6,
"type": "One Way",
"price": 40
},
{
"min_pax": 4,
"max_pax": 6,
"type": "Round Trip",
"price": 65
},
{
"min_pax": 7,
"max_pax": 10,
"type": "One Way",
"price": 50
},
{
"min_pax": 7,
"max_pax": 10,
"type": "Round Trip",
"price": 80
}
],
}
]
我想合并具有相同最小和最大人數的物件，例如：
1 至 3 名乘客單程費用為 35，往返費用為 63。
我認為看起來更好的是我可以合并兩個物件。
我想要這樣的東西：
[
{
"tipo": "A",
"prices": [
{
"min_pax": 1,
"max_pax": 3,
"one_way": 35,
"round_trip": 63
},
{
"min_pax": 4,
"max_pax": 6,
"one_way": 40,
"round_trip":65
},
{
"min_pax": 7,
"max_pax": 10,
"one_way": 50,
"round_trip": 80
},
],
}
]
我將衷心感謝您的幫助
uj5u.com熱心網友回復：
一種選擇是$unwind根據：$grouppax
db.collection.aggregate([
{$unwind: "$prices"},
{$group: {
_id: {max_pax: "$prices.max_pax", min_pax: "$prices.min_pax"},
tipo: {$first: "$tipo"},
data: {$push: {k: "$prices.type", v: "$prices.price"}}
}
},
{$project: {
_id: 0,
tipo: 1,
max_pax: "$_id.max_pax",
min_pax: "$_id.min_pax",
data: {$arrayToObject: "$data"}
}
},
{$set: {"data.max_pax": "$max_pax", "data.min_pax": "$min_pax"}},
{$group: {_id: "$tipo", prices: {$push: "$data"}}},
{$project: {_id: 0, tipo: "$_id", prices: 1}}
])
看看它在操場上的例子是如何作業的
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