我们有三个嵌套数组:
-
principalCredits，带2个对象 -
credits个，每个对象2个 -
awardNominations.edges，总计从0到3不等
任务是基于eventsCollection的查找向第三个对象数组awardNominations.edges添加字段.
以下是我拥有的数据(经过简化，可以复制并粘贴到MongoDB Compass中):
[{
"principalCredits": [
{
"category": {
"id": "director",
"text": "Directors"
},
"totalCredits": 2,
"credits": [
{
"name": {
"id": "nm11813828",
"nameText": {
"text": "Pippa Ehrlich"
},
"awardNominations": {
"total": 2,
"edges": [
{
"node": {
"id": "an1393007",
"isWinner": true,
"award": {
"id": "an1393007",
"year": 2020,
"text": "Green Warsaw Award",
"event": {
"id": "ev0003786",
"text": "Millennium Docs Against Gravity"
},
"category": {
"text": null
}
}
}
},
{
"node": {
"id": "an1428940",
"isWinner": false,
"award": {
"id": "an1428940",
"year": 2021,
"text": "IDA Award",
"event": {
"id": "ev0000351",
"text": "International Documentary Association"
},
"category": {
"text": "Best Writing"
}
}
}
},
]
}
},
"category": {
"id": "director",
"text": "Director"
}
},
{
"name": {
"id": "nm1624755",
"nameText": {
"text": "James Reed"
},
"awardNominations": {
"total": 3,
"edges": [
{
"node": {
"id": "an0694012",
"isWinner": true,
"award": {
"id": "an0694012",
"year": 2015,
"text": "Best of Festival",
"event": {
"id": "ev0001486",
"text": "Jackson Wild Media Awards"
},
"category": {
"text": "Best of Festival"
}
}
}
},
{
"node": {
"id": "an0975779",
"isWinner": true,
"award": {
"id": "an0975779",
"year": 2017,
"text": "RTS West Television Award",
"event": {
"id": "ev0000571",
"text": "Royal Television Society, UK"
},
"category": {
"text": "Documentary"
}
}
}
},
{
"node": {
"id": "an0975781",
"isWinner": true,
"award": {
"id": "an0975781",
"year": 2015,
"text": "Grand Teton Prize",
"event": {
"id": "ev0001356",
"text": "Jackson Hole Film Festival"
},
"category": {
"text": "Best in Festival"
}
}
}
}
]
}
},
"category": {
"id": "director",
"text": "Director"
}
}
]
},
{
"category": {
"id": "writer",
"text": "Writers"
},
"totalCredits": 2,
"credits": [
{
"name": {
"id": "nm11813828",
"nameText": {
"text": "Pippa Ehrlich"
},
"awardNominations": {
"total": 2,
"edges": [
{
"node": {
"id": "an1393007",
"isWinner": true,
"award": {
"id": "an1393007",
"year": 2020,
"text": "Green Warsaw Award",
"event": {
"id": "ev0003786",
"text": "Millennium Docs Against Gravity"
},
"category": {
"text": null
}
}
}
},
{
"node": {
"id": "an1428940",
"isWinner": false,
"award": {
"id": "an1428940",
"year": 2021,
"text": "IDA Award",
"event": {
"id": "ev0000351",
"text": "International Documentary Association"
},
"category": {
"text": "Best Writing"
}
}
}
}
]
}
},
"category": {
"id": "writer",
"text": "Writer"
},
},
{
"name": {
"id": "nm1624755",
"nameText": {
"text": "James Reed"
},
"awardNominations": {
"total": 0,
"edges": []
}
},
"category": {
"id": "writer",
"text": "Writer"
},
}
]
}
]
}]
获奖示例如下:
{
"id": "an0975781",
"isWinner": true,
"award": { ... },
"score": 1.5
}
一旦完成所有操作，数据需要与最初的形状完全相同，并且没有空值.所以在最后一个数组awardsNominations.edges的情况下，它应该是[]，而不是{ node: { score: null }}或其他任何东西.
为了实现这一点，我创建了一个聚合管道:
[
{
'$unwind': {
'path': '$principalCredits',
'preserveNullAndEmptyArrays': true
}
}, {
'$unwind': {
'path': '$principalCredits.credits',
'preserveNullAndEmptyArrays': true
}
}, {
'$unwind': {
'path': '$principalCredits.credits.name.awardNominations.edges',
'preserveNullAndEmptyArrays': true
}
}, {
'$lookup': {
'from': 'eventsCollection',
'localField': 'principalCredits.credits.name.awardNominations.edges.node.award.event.id',
'foreignField': 'id',
'as': 'matchingEvent'
}
}, {
'$unwind': {
'path': '$matchingEvent',
'preserveNullAndEmptyArrays': true
}
}, {
'$addFields': {
'principalCredits.credits.name.awardNominations.edges.node.score': {
'$multiply': [
'$matchingEvent.importance', {
'$cond': {
'if': '$principalCredits.credits.name.awardNominations.edges.node.isWinner',
'then': 1.5,
'else': 1.2
}
}
]
}
}
}
]
上述管道将score分配给每个奖励.然而，空值仍然存在，我完全不知道如何将其恢复到一起.我曾try 与以下群体合作:
{
'$group': {
'_id': '$id',
'titleDoc': {
'$first': '$$ROOT'
},
'allPrincipalCredits': {
'$push': '$principalCredits'
}
}
}
保留根，然后以某种方式将所有记录排序回形状，但无法返回到原始对象 struct .
任何有助于把这一切放在一起的人都将不胜感激！
我很擅长简单的聚合，但这对我来说似乎太多了，我很想学习如何正确地返回$group个东西.
我试着把迄今为止从不同来源和类似答案获得的所有知识整合在一起，但似乎无法实现.
查找集合eventsCollection包含以下对象:
{
"_id": { "$oid": "62c57125d6943d92f83f6fff" },
"id": "ev0030197",
"text": "#AmLatino Film Festival",
"importance": 1
}