下面是样本收集文档记录,我希望将相同的收集与不同的子数组元素连接起来。
样品采集记录:
{
"_id": "052dc2aa-043b-4cd7-a3f2-f3fe6540ae52",
"Details": [
{
"Id": "104b0bb1-d4a5-469b-b1fd-b4822e96dcb0",
"Number": "12345",
"Percentages": [
{
"Code": "55555",
"Percentage": "45"
},
{
"Code": "55333",
"Percentage": "50"
}
]
},
{
"Id": "104b0bb1-d4a5-469b-b1fd-b4822e96dcb0",
"Number": "55555",
"Percentages": [
{
"Code": "55555",
"Percentage": "45"
}
]
}
],
"Payments": [
{
"Id": "61ee1a6f-3334-4f33-ab6c-51c646b75c41",
"Number": "12345"
}
]
}
我想获取具有匹配条件的 Percentages 数组的mongo Pipeline查询,其 Details.Number 和 Payment.Number 应该为相同
结果:
"Percentages": [
{
"Code": "55555",
"Percentage": "45"
},
{
"Code": "55333",
"Percentage": "50"
}]
如何通过使用聚合将相同集合的子元素合并来带来结果?
答案 0 :(得分:1)
以下查询可满足您的需求:
db.collection.aggregate([
{$unwind : "$Details"},
{$unwind : "$Details.Percentages"},
{$unwind : "$Payments"}, // $unwind all your arrays
{
$addFields : { //This include new `isMatch` field, which is gonna be true, only if Details.Number = Payment.Number
"isMatch" : {$cond: { if: { $eq: [ "$Details.Number", "$Payments.Number" ] }, then: true, else: false }}
}
},
{
$match : { // This ignores all others, for which Details.Number != Payment.Number
"isMatch" : true
}
},
{
$group : { // This will return only the Percentage objects
_id : null,
"Percentages" : {$push : "$Details.Percentages"}
}
},
{
$project : { // To ignore "_id" field
_id : 0,
"Percentages" : 1
}
}
])
结果:
{
"Percentages" : [
{
"Code" : "55555",
"Percentage" : "45"
},
{
"Code" : "55333",
"Percentage" : "50"
}
]
}
希望这会有所帮助!