筛选出重复的数组，并在mongodb聚合中返回唯一的数组 [英] Filter Out duplicate arrays and return the unique array in mongodb aggregation
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问题描述
在构建以下mongodb数据收集方面，我已经走了很长一段路，但是我无法完成汇总阶段，
I have come a long way in structuring into the following mongodb data collection, but i couldn't finish the aggregation stage,
{
"test": [
{
"_id": "60014aee808bc5033b45c222",
"name": "a rogram",
"companyName": "company NAme",
"website": "https://www.example.comn",
"loginUrl": "https://www.example.comn",
"description": null,
"createdBy": "5fe5cbcdb9ac0f001dccfadf",
"createdAt": "2021-01-15T07:57:34.499Z",
"updatedAt": "2021-01-15T13:09:09.417Z",
"__v": 0,
"address": null,
"affiliatePlatform": "asdf",
"brands": 3,
"newList": [
{
"_id": "5feee63fd86250046d64d02e",
"name": "NEWBRAND",
"affiliateProgram": "60014aee808bc5033b45c222",
"operator": "scdscacscasc",
"description": null,
"established": "2021-01-20T12:39:25.000Z",
"createdAt": "2021-01-01T09:07:11.180Z",
"updatedAt": "2021-01-15T12:39:36.898Z",
"__v": 0,
"updatedBy": null,
"newPMBList": [
{
"_id": "5feee78ead36a5052cdaddc0",
"projectMarket": "5feb10c4906c880076ce5fa6",
"brand": "5feee63fd86250046d64d02e",
"createdAt": "2021-01-01T09:12:46.588Z",
"updatedAt": "2021-01-01T09:12:46.588Z",
"__v": 0,
"newPMList": [
{
"_id": "5feb10c4906c880076ce5fa6",
"project": "5feb10b6906c880076ce5fa5",
"market": "5feb0f44906c880076ce5f9d",
"url": "https://www.example.com/place",
"createdAt": "2020-12-29T11:19:32.945Z",
"updatedAt": "2020-12-29T11:19:32.945Z",
"__v": 0
}
]
}
]
},
{
"_id": "5ff07fce63da300174a014bb",
"name": "Nike",
"operator": "scdscacscasc",
"createdAt": "2021-01-02T14:14:38.607Z",
"updatedAt": "2021-01-15T08:05:09.475Z",
"__v": 0,
"description": "This is new afdkjnvlaf",
"established": "2021-01-27T08:02:38.000Z",
"updatedBy": null,
"affiliateProgram": "60014aee808bc5033b45c222",
"newPMBList": [
{
"_id": "5ff451891136a7006bc2b0eb",
"projectMarket": "5feb10c4906c880076ce5fa6",
"brand": "5ff07fce63da300174a014bb",
"createdAt": "2021-01-05T11:46:17.745Z",
"updatedAt": "2021-01-05T11:46:17.745Z",
"__v": 0,
"newPMList": [
{
"_id": "5feb10c4906c880076ce5fa6",
"project": "5feb10b6906c880076ce5fa5",
"market": "5feb0f44906c880076ce5f9d",
"url": "https://www.example.com/place",
"createdAt": "2020-12-29T11:19:32.945Z",
"updatedAt": "2020-12-29T11:19:32.945Z",
"__v": 0
}
]
}
]
},
{
"_id": "5fff0336d78339005812aaa4",
"name": "United Bank",
"affiliateProgram": "60014aee808bc5033b45c222",
"operator": "scdsc",
"description": "ascasccsac",
"established": null,
"createdAt": "2021-01-13T14:27:02.931Z",
"updatedAt": "2021-01-15T12:43:34.857Z",
"__v": 0,
"updatedBy": null,
"newPMBList": []
}
],
"projectsNumber": null
},
{
"_id": "6001923a1f1fb007437dc479",
"name": "Hotel Advertisement",
"companyName": "Atlas",
"website": "https://www.atlas.com",
"loginUrl": "https://www.atlas.com/luser/ogin",
"createdBy": "5fe5cbcdb9ac0f001dccfadf",
"createdAt": "2021-01-15T13:01:46.715Z",
"updatedAt": "2021-01-15T13:20:42.757Z",
"__v": 0,
"address": null,
"affiliatePlatform": "NetRefer",
"description": null,
"brands": 2,
"newList": [
{
"_id": "5ffff156f06d4700e255cde5",
"name": "Amazon",
"affiliateProgram": "6001923a1f1fb007437dc479",
"operator": "scdscacscasc",
"description": "vsdfvsfdv",
"established": null,
"createdAt": "2021-01-14T07:23:02.455Z",
"updatedAt": "2021-01-15T13:19:59.522Z",
"__v": 0,
"updatedBy": null,
"newPMBList": []
},
{
"_id": "60000d4c61316a01d2fbb1aa",
"name": "Toshiba",
"affiliateProgram": "6001923a1f1fb007437dc479",
"operator": "scdsc",
"description": "cvadfvfdvbfdav",
"established": null,
"createdAt": "2021-01-14T09:22:20.306Z",
"updatedAt": "2021-01-15T13:20:10.970Z",
"__v": 0,
"updatedBy": null,
"newPMBList": []
}
],
"projectsNumber": null
}
]
}
在上面的文档中，我只想返回 newPMList 的唯一列表以及除 newList 数组之外的其他数据，如下所示:
in the document above i would like to only return the distinct list of the newPMList along with the other data except the newList array like so:
"_id": "60014aee808bc5033b45c222",
"name": "21 Program",
"companyName": "Online Casino",
"website": "https://www.online.casinoe",
"loginUrl": "https://www.online.casino",
"description": null,
"createdBy": "5fe5cbcdb9ac0f001dccfadf",
"createdAt": "2021-01-15T07:57:34.499Z",
"updatedAt": "2021-01-15T13:09:09.417Z",
"__v": 0,
"address": null,
"affiliatePlatform": "Cellxpert",
"brands": 3,
"newPMList": [
{
"_id": "5feb10c4906c880076ce5fa6",
"project": "5feb10b6906c880076ce5fa5",
"market": "5feb0f44906c880076ce5f9d",
"url": "https://www.addissoftware.com/ethiopia",
"createdAt": "2020-12-29T11:19:32.945Z",
"updatedAt": "2020-12-29T11:19:32.945Z",
"__v": 0
}
]
每个文档，但我无法提供任何帮助.
per a document, but i couldn't any help is appreciated.
推荐答案
db.collection.aggregate([
{//Denormalize first level
"$unwind": "$newList"
},
{//Second nested level
"$unwind": "$newList.newPMBList"
},
{//Deep nested last level
"$unwind": "$newList.newPMBList.newPMList"
},
{
$group: {//Grouping back
"_id": null,
"newList": {
$push: "$newList.newPMBList.newPMList"
}
}
},
{
$project: {//Finding unique
newList: {
$setUnion: [
"$newList",
"$newList"
]
}
}
}
])
我建议您使用 first 累加器将其他字段包含在 group 中，并将其保存在 project 中.
I suggest you to include other fields using first accumulator in group and preserve them in project.
您可以进一步简化以下内容
You can simplify further as below
db.test.aggregate([
{
"$unwind": "$newList"
},
{
"$unwind": "$newList.newPMBList"
},
{
"$unwind": "$newList.newPMBList.newPMList"
},
{
$group: {
"_id": null,
"newList": {//addToSet keeps distinct
$addToSet: "$newList.newPMBList.newPMList"
}
}
}
])
此外，购买时可以跳过一个非规范化，但是它会返回数组数组.
Further, buy skipping one denormalisation, But it returns array of arrays.
db.test.aggregate([
{
"$unwind": "$newList"
},
{
"$unwind": "$newList.newPMBList"
},
{
$group: {
"_id": null,
"newList": {
$addToSet: "$newList.newPMBList.newPMList"
}
}
}
])
如果您跳过另一个级别，则会在结果中再添加一个嵌套级别.
If you skip another level, it will add one more nested level in the result.
这篇关于筛选出重复的数组，并在mongodb聚合中返回唯一的数组的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
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