我有以下数据集:
{
"_id" : ObjectId("59668a22734d1d48cf34de08"),
"name" : "Nobody Cares",
"menus" : [
{
"_id" : "menu_123",
"name" : "Weekend Menu",
"description" : "A menu for the weekend",
"groups" : [
{
"name" : "Spirits",
"has_mixers" : true,
"sizes" : [
"Single",
"Double"
],
"categories" : [
{
"name" : "Vodka",
"description" : "Maybe not necessary?",
"drinks" : [
{
"_id" : "drink_123",
"name" : "Absolut",
"description" : "Fancy ass vodka",
"sizes" : [
{
"_id" : "size_123",
"size" : "Single",
"price" : 300
}
]
}
]
}
]
}
],
"mixers" : [
{
"_id" : "mixer_1",
"name" : "Coca Cola",
"price" : 150
},
{
"_id" : "mixer_2",
"name" : "Lemonade",
"price" : 120
}
]
}
]
}
我尝试从该数据集中检索一杯饮料,我使用以下聚合查询:
db.getCollection('places').aggregate([
{ $match : {"menus.groups.categories.drinks._id" : "drink_123"} },
{ $unwind: "$menus" },
{ $project: { "_id": 1, "menus": { "groups": { "categories": { "drinks": { "name": 1 } } } } } }
])
但是,它会返回数据集的完整结构以及正确的数据。
所以而不是:
{
"_id": "drink_123",
"name": "Absolut"
}
我明白了:
{
"_id": ObjectId("59668a22734d1d48cf34de08"),
"menus": {
"groups": {
"categories": {
"drinks": { "name": "Absolut" }
}
}
}
}
例如。任何想法如何只检索子文档?
答案 0 :(得分:0)
如果您需要保留深层嵌套模型,则此调用将生成所需的输出:
db.getCollection('places').aggregate([
{ $match : {"menus.groups.categories.drinks._id" : "drink_123"} },
{ $project: {"_id": '$menus.groups.categories.drinks._id', name: '$menus.groups.categories.drinks.name'}},
{ $unwind: "$name" },
{ $unwind: "$name" },
{ $unwind: "$name" },
{ $unwind: "$name" },
{ $unwind: "$_id" },
{ $unwind: "$_id" },
{ $unwind: "$_id" },
{ $unwind: "$_id" }
])
许多unwinds是drinks
子文档深度嵌套的结果。
虽然,FWIW,这种查询可能暗示该模型不是“友好的”。