我正在处理包含音乐播放列表的文档。每个文档都有这样的结构：{ "user_id": "5858", "playlists": [ { "name": "My Playlist", "guild_ids": ["7575"], "items": [ { "title": "title", "url": "url", "duration": 200000 } ] } ]}我想提取同一公会的所有播放列表。但问题是我希望结果在单个文档中返回。包含播放列表列表的一个文档。guild_id=5656 的预期结果如下：{ "playlists": [ { "name": "My Playlist", "guild_ids": ["5656"], "items": [ { "title": "title", "url": "url", "duration": 200000 } ] }, // other playlists where guild_ids contains "5656" ]}我尝试使用聚合，但我总是得到与唯一 user_ids 数量相同的文档数量。我得到按 user_id 分组的播放列表。
1 回答
Smart猫小萌
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以下查询可以获得预期的输出：
db.collection.aggregate([
{
$unwind:"$playlists"
},
{
$match:{
"playlists.guild_ids":{
$in:["7575"]
}
}
},
{
$group:{
"_id":null,
"playlists":{
$push: "$playlists"
}
}
},
{
$project:{
"_id":0
}
}
]).pretty()
数据集：
{
"_id" : ObjectId("5d88225e38db7cf8d3f75cd6"),
"user_id" : "5858",
"playlists" : [
{
"name" : "My Playlist",
"guild_ids" : [
"7575"
],
"items" : [
{
"title" : "title",
"url" : "url",
"duration" : 200000
}
]
}
]
}
{
"_id" : ObjectId("5d88225e38db7cf8d3f75cd7"),
"user_id" : "5858",
"playlists" : [
{
"name" : "My Playlist 2",
"guild_ids" : [
"1234"
],
"items" : [
{
"title" : "title",
"url" : "url",
"duration" : 200000
}
]
}
]
}
{
"_id" : ObjectId("5d88225e38db7cf8d3f75cd8"),
"user_id" : "5858",
"playlists" : [
{
"name" : "My Playlist 3",
"guild_ids" : [
"7575"
],
"items" : [
{
"title" : "title",
"url" : "url",
"duration" : 200000
}
]
}
]
}
输出：
{
"playlists" : [
{
"name" : "My Playlist",
"guild_ids" : [
"7575"
],
"items" : [
{
"title" : "title",
"url" : "url",
"duration" : 200000
}
]
},
{
"name" : "My Playlist 3",
"guild_ids" : [
"7575"
],
"items" : [
{
"title" : "title",
"url" : "url",
"duration" : 200000
}
]
}
]
}
查询分析：我们正在展开playlists，仅过滤具有7575公会ID的那些，然后将它们重新分组。
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