s.f*_*frz 6 mongoose mongodb node.js
假设我有一个这样的产品系列:
{
"_id": "5a74784a8145fa1368905373",
"name": "This is my first product",
"description": "This is the description of my first product",
"category": "34/73/80",
"condition": "New",
"images": [
{
"length": 1000,
"width": 1000,
"src": "products/images/firstproduct_image1.jpg"
},
...
],
"attributes": [
{
"name": "Material",
"value": "Synthetic"
},
...
],
"variation": {
"attributes": [
{
"name": "Color",
"values": ["Black", "White"]
},
{
"name": "Size",
"values": ["S", "M", "L"]
}
]
}
}
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和这样的变体集合:
{
"_id": "5a748766f5eef50e10bc98a8",
"name": "color:black,size:s",
"productID": "5a74784a8145fa1368905373",
"condition": "New",
"price": 1000,
"sale": null,
"image": [
{
"length": 1000,
"width": 1000,
"src": "products/images/firstvariation_image1.jpg"
}
],
"attributes": [
{
"name": "Color",
"value": "Black"
},
{
"name": "Size",
"value": "S"
}
]
}
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我想将文档分开,为了便于浏览、搜索和分面搜索实现,我想在单个查询中获取所有数据,但我不想加入我的应用程序代码。我知道使用名为 summary 的第三个集合可以实现,它可能如下所示:
{
"_id": "5a74875fa1368905373",
"name": "This is my first product",
"category": "34/73/80",
"condition": "New",
"price": 1000,
"sale": null,
"description": "This is the description of my first product",
"images": [
{
"length": 1000,
"width": 1000,
"src": "products/images/firstproduct_image1.jpg"
},
...
],
"attributes": [
{
"name": "Material",
"value": "Synthetic"
},
...
],
"variations": [
{
"condition": "New",
"price": 1000,
"sale": null,
"image": [
{
"length": 1000,
"width": 1000,
"src": "products/images/firstvariation_image.jpg"
}
],
"attributes": [
"color=black",
"size=s"
]
},
...
]
}
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问题是,我不知道如何使汇总集合与产品和变体集合保持同步。我知道可以使用 mongo-connector 来完成,但我不确定如何实现它。请帮助我,我仍然是一个初学者程序员。
您实际上不需要维护摘要集合,将产品和变体摘要存储在另一个集合中是多余的
您可以使用聚合管道$lookup来使用 ProductID 外部连接产品和变体
骨料管道
db.products.aggregate(
[
{
$lookup : {
from : "variation",
localField : "_id",
foreignField : "productID",
as : "variations"
}
}
]
).pretty()
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