英文:
MongoDB filter nested array
问题
[
{
"_id": 1,
"title": "dummy title",
"settings": [
{
"type": "light",
"status": "enabled"
},
{
"type": "toolbar",
"status": "enabled"
}
]
}
]
我想按ID获取它，但只想获取“enabled”设置，不确定聚合管道应该是什么样子的。
我应该使用Java / Kotlin中的mongoTemplate创建查询，但仅有Mongo查询即可。
英文:
I have a collection that looks something like:
[
{
"_id": 1,
"title": "dummy title",
"settings": [
{
"type": "light",
"status": "enabled"
},
{
"type": "flare",
"status": "disabled"
},
{
"type": "toolbar",
"status": "enabled"
}
]
}
]
I wanna fetch it by Id but only with the "enabled" settings and not sure how the aggregation pipeline should look like.
I supposed to create the query using mongoTemplate in Java / Kotlin but even just the mongo query would be enough.
答案1
得分: 1
// MongoDB shell/CLI版本4.26的工作代码（在Windows 10机器/操作系统上）
> db.titles.find().pretty();
{
"_id" : 1,
"title" : "虚拟标题",
"settings" : [
{
"type" : "light",
"status" : "已启用"
},
{
"type" : "flare",
"status" : "已禁用"
},
{
"type" : "toolbar",
"status" : "已启用"
}
]
}
> db.titles.aggregate([
... {$unwind:"$settings"},
... {$match:{"settings.status":"已启用"}}
... ]);
{ "_id" : 1, "title" : "虚拟标题", "settings" : { "type" : "light", "status" : "已启用" } }
{ "_id" : 1, "title" : "虚拟标题", "settings" : { "type" : "toolbar", "status" : "已启用" } }
> print("MongoDB: ",db.version());
MongoDB: 4.2.6
>
英文:
//working code from MongoDB shell/CLI version 4.26(on windows 10 machine/OS)
> db.titles.find().pretty();
{
"_id" : 1,
"title" : "dummy title",
"settings" : [
{
"type" : "light",
"status" : "enabled"
},
{
"type" : "flare",
"status" : "disabled"
},
{
"type" : "toolbar",
"status" : "enabled"
}
]
}
> db.titles.aggregate([
... {$unwind:"$settings"},
... {$match:{"settings.status":"enabled"}}
... ]);
{ "_id" : 1, "title" : "dummy title", "settings" : { "type" : "light", "status" : "enabled" } }
{ "_id" : 1, "title" : "dummy title", "settings" : { "type" : "toolbar", "status" : "enabled" } }
> print("MongoDB: ",db.version());
MongoDB: 4.2.6
>
答案2
得分: 1
你可以使用$filter来实现。
db.collection.aggregate([
{
$project: {
_id: 1,
title: 1,
settings: {
$filter: {
input: "$settings",
as: "item",
cond: {
$eq: [
"$$item.status",
"enabled"
]
}
}
}
}
}
])
在此链接中尝试。
英文:
You can do it with a $filter
db.collection.aggregate([
{
$project: {
_id: 1,
title: 1,
settings: {
$filter: {
input: "$settings",
as: "item",
cond: {
$eq: [
"$$item.status",
"enabled"
]
}
}
}
}
}
])
try it here
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