英文:
MongoDB, return selected fields from document with aggregation operations
问题
根据您提供的信息，您想要重新编写查询，以便在每个返回的评论中包含productId。以下是重写的查询：
db.comments.aggregate([
{
$match: {
"comments._id": {
$in: [
ObjectId('630b7868f51e10876223b4aa'),
ObjectId('630bd277f919a9e9c0e7a559')
]
}
}
},
{
$unwind: "$comments"
},
{
$match: {
"comments._id": {
$in: [
ObjectId('630b7868f51e10876223b4aa'),
ObjectId('630bd277f919a9e9c0e7a559')
]
}
}
},
{
$project: {
_id: 1,
"comment.userId": "$comments.userId",
"comment.userName": "$comments.userName",
"comment.date": "$comments.date",
"comment.confirmed": "$comments.confirmed",
"comment.likes": "$comments.likes",
"comment.content": "$comments.content",
"comment._id": "$comments._id",
"comment.usersWhoLiked": "$comments.usersWhoLiked",
"comment.productId": "$productId" // Add productId to each comment
}
}
])
这个查询首先使用$match阶段筛选包含指定评论ID的文档，然后使用$unwind阶段展开comments数组，接着再次使用$match来匹配指定的评论ID。最后，使用$project阶段将所需字段包括productId添加到每个评论中。
请注意，这只是一种方式，您可以根据自己的需求进行进一步的调整和优化。
英文:
With given query I also want to return productId.
I have collection comments that contains documents with data about productId and comments for given product
Example document in this collection:
{
"_id": {
"$oid": "635ee64f55460d1796447662"
},
"productId": "63413800d36ed477adc763d0",
"__v": 0,
"comments": [
{
"userId": "",
"userName": "test",
"date": "2022.12.18.21.51.36",
"confirmed": false,
"likes": {
"up": 0,
"down": 0
},
"content": {
"rating": 6,
"description": "testtesttest"
},
"image": {
"added": false,
"images": []
},
"_id": {
"$oid": "639f7d58b6206a863c4a7aba"
},
"usersWhoLiked": []
},
{
"userId": "",
"userName": "test",
"date": "2022.12.18.21.52.19",
"confirmed": false,
"likes": {
"up": 0,
"down": 0
},
"content": {
"rating": 6,
"description": "testtesttest"
},
"image": {
"added": true,
"images": [
"comments/63413800d36ed477adc763d0/639f7d83b6206a863c4a7ad6/dell.jpg"
]
},
"_id": {
"$oid": "639f7d83b6206a863c4a7ad6"
},
"usersWhoLiked": []
}
]
}
My exmaple query:
db.comments.aggregate([{$match: {"comments._id": {$in: [ObjectId('630b7868f51e10876223b4aa'), ObjectId('630bd277f919a9e9c0e7a559')]}}},
{$project: {comment: {$filter: {input: "$comments", as: "comment", cond: {$in: ["$$comment._id", [ObjectId("630b7868f51e10876223b4aa"), ObjectId("630bd277f919a9e9c0e7a559")]]}}}}}])
With this query I get the result :
{ _id: ObjectId("630b7868f51e10876223b4a6"),
comment:
[ { userId: '62f29c2c324f4778dff443f6',
userName: 'User',
date: '2022.08.19',
confirmed: false,
likes: { up: 3, down: 0 },
content: { rating: 4, description: 'Super laptop <3' },
_id: ObjectId("630b7868f51e10876223b4aa"),
usersWhoLiked:
[ { userId: '62f29c2c324f4778dff443f6',
likeUp: true,
_id: ObjectId("630d2b0494370efb37107983") },
{ userId: '6322434f2b5bbac87f0e7aba',
likeUp: true,
_id: ObjectId("632243702b5bbac87f0e7afa") },
{ userId: '62f2991e324f4778dff443d4',
likeUp: true,
_id: ObjectId("63af4d77c8991b74d6986995") } ] } ] }
{ _id: ObjectId("630bd277f919a9e9c0e7a555"),
comment:
[ { userId: '62f29c2c324f4778dff443f6',
userName: 'User',
date: '2022.08.28',
confirmed: false,
likes: { up: 1, down: 1 },
content:
{ rating: 6,
description: 'Laptop posiada przyjemna klawiature, nie grzeje się. Do codziennego grania wystarczy.' },
_id: ObjectId("630bd277f919a9e9c0e7a559"),
usersWhoLiked:
[ { userId: '62f29c2c324f4778dff443f6',
likeUp: true,
_id: ObjectId("630d2dfc94370efb37107991") },
{ userId: '62fa2549f029348f75bc9c81',
likeUp: false,
_id: ObjectId("631241fe755c641525dc9cfa") } ] } ] }
As you see in the result, the productId is missing.
I was trying to rebuild query with #group operator but still with out effect...
So my questsion is:
How shall I rewrite that query to get the same result but with productId in it for each returned comment
答案1
得分: 1
这是$project的工作原理，如果未指定字段，则不会输出。
所以只需在$project阶段添加productId: 1，它就会显示出来。
查看此示例。
英文:
This is how $project works, if a field is not specified will not be output.
So just add productId: 1 into the $project stage and it will be shown.
Check this example.
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