[英]How to sort Arrays of Object using aggregation in mongoDB
[英]Aggregation using two sub object arrays MongoDB
在不同的 collections object 中有兩個 arrays。 我想在一個 object 指定和常見的 uuid 中從兩個 arrays 獲取所有字段。
這是來自兩個不同對象的兩個 arrays:
{
"_id" : ObjectId("5a1ea4a4a1a13eaecf571267"),
"storage" : "events",
"list" : [
{
"uuid" : "5a03c1e0e31a11e7b8c2e398bcd9f882",
"type" : "TYPE",
"desc" : "DESC",
"when" : 1513934100000,
"loc" : "LOC",
"schedule" : [ ]
},
{
"uuid" : "1b45a340e70911e7b8c2e398bcd9f882",
"type" : "TYPE1",
"desc" : "DESC1",
"when" : 1514624400000,
"loc" : "LOC1",
"schedule" : []
}
}
}
{
"_id" : ObjectId("5a297001840b2aba87a5b1eb"),
"storage" : "control",
"list" : [
{
"uuid" : "5a03c1e0e31a11e7b8c2e398bcd9f882",
"missing" : [
{
"user" : "user",
"instrument" : "inst"
},
{
"user" : "user",
"instrument" : "inst"
}
],
"when" : 1513934100000
},
{
"uuid" : "1b45a340e70911e7b8c2e398bcd9f882",
"missing" : [
{
"user" : "user",
"instrument" : "inst"
},
{
"user" : "user",
"instrument" : "inst"
}
],
"when" : 1514624400000
},
}
}
期望的結果是:(因為給定的 uuid 是 5a03c1e0e31a11e7b8c2e398bcd9f882)
{
"list" : [
{
"uuid" : "5a03c1e0e31a11e7b8c2e398bcd9f882",
"type" : "TYPE",
"desc" : "DESC",
"when" : 1513934100000,
"loc" : "LOC",
"schedule" : [ ],
"missing" : [
{
"user" : "user",
"instrument" : "inst"
},
{
"user" : "user",
"instrument" : "inst"
}
]
}
}
問題是如果沒有兩個匹配的 uuid,則結果必須是 null。 我有多個 $list,所以展開時必須有兩個具有相同 uuid 的對象,如果不是,結果必須是 null。
以下查詢可以得到我們預期的 output:
db.collection.aggregate([
{
$unwind:"$list"
},
{
$group:{
"_id":"$list.uuid",
"list":{
$push:"$list"
}
}
},
{
$project:{
"merged":{
$reduce:{
"input":"$list",
"initialValue":{},
"in":{
$mergeObjects:["$$this","$$value"]
}
}
}
}
},
{
$group:{
"_id":null,
"list":{
$push: "$merged"
}
}
},
{
$project:{
"_id":0,
"storage" : "result",
"list":1
}
}
]).pretty()
數據集:
{
"_id" : ObjectId("5a1ea4a4a1a13eaecf571267"),
"storage" : "events",
"list" : [
{
"uuid" : "5a03c1e0e31a11e7b8c2e398bcd9f882",
"type" : "TYPE",
"desc" : "DESC",
"when" : 1513934100000,
"loc" : "LOC",
"schedule" : [ ]
},
{
"uuid" : "1b45a340e70911e7b8c2e398bcd9f882",
"type" : "TYPE1",
"desc" : "DESC1",
"when" : 1514624400000,
"loc" : "LOC1",
"schedule" : [ ]
}
]
}
{
"_id" : ObjectId("5a297001840b2aba87a5b1eb"),
"storage" : "control",
"list" : [
{
"uuid" : "5a03c1e0e31a11e7b8c2e398bcd9f882",
"missing" : [
{
"user" : "user",
"instrument" : "inst"
},
{
"user" : "user",
"instrument" : "inst"
}
],
"when" : 1513934100000
},
{
"uuid" : "1b45a340e70911e7b8c2e398bcd9f882",
"missing" : [
{
"user" : "user",
"instrument" : "inst"
},
{
"user" : "user",
"instrument" : "inst"
}
],
"when" : 1514624400000
}
]
}
Output:
{
"list" : [
{
"uuid" : "1b45a340e70911e7b8c2e398bcd9f882",
"missing" : [
{
"user" : "user",
"instrument" : "inst"
},
{
"user" : "user",
"instrument" : "inst"
}
],
"when" : 1514624400000,
"type" : "TYPE1",
"desc" : "DESC1",
"loc" : "LOC1",
"schedule" : [ ]
},
{
"uuid" : "5a03c1e0e31a11e7b8c2e398bcd9f882",
"missing" : [
{
"user" : "user",
"instrument" : "inst"
},
{
"user" : "user",
"instrument" : "inst"
}
],
"when" : 1513934100000,
"type" : "TYPE",
"desc" : "DESC",
"loc" : "LOC",
"schedule" : [ ]
}
],
"storage" : "result"
}
查詢分析:我們首先展開list
數組,然后根據list.uuid
進行分組。 每個組包含具有相同 UUID 的所有列表元素的集合。 稍后,列表元素合並為一個。
`db.a9e34aa77f3474fbab5e9827942fbdd.aggregate([
{
$project: {
'list': {
$filter: {
input: '$list',
as: 'item',
cond: {
$eq: ['$$item.uuid', '15d055a0d5d711e8bd45bff08b1fb980']
}
}
}
}
},
{
$unwind:"$list"
},
{
$group:{
"_id":"$list.uuid",
"list":{
$push:"$list"
}
}
},
{
$project:{
"merged":{
$reduce:{
"input":"$list",
"initialValue":{},
"in":{
$mergeObjects:["$$this","$$value"]
}
}
}
}
},
{
$group:{
"_id":null,
"list":{
$push: "$merged"
}
}
},
{
$project:{
"_id":0,
"list":1
}
}
]).pretty()`
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