我正在使用MongoDB的聚合管道,以我想要的形式获取我的文档。作为聚合的最后一步,我使用$project
将文档放入最终形式。
但是我在投影和子文档数组时遇到了麻烦。这是我目前从aggrgation得到的:
{
"_id": "581c8c3df1325f68ffd23386",
"count": 14,
"authors": [
{
"author": {
"author": "57f246b9e01e6c6f08e1d99a",
"post": "581c8c3df1325f68ffd23386"
},
"count": 13
},
{
"author": {
"author": "5824382511f16d0f3fd5aaf2",
"post": "581c8c3df1325f68ffd23386"
},
"count": 1
}
]
}
我想$project
authors
数组,以便返回:
{
"_id": "581c8c3df1325f68ffd23386",
"count": 14,
"authors": [
{
"_id": "57f246b9e01e6c6f08e1d99a",
"count": 13
},
{
"_id": "5824382511f16d0f3fd5aaf2",
"count": 1
}
]
}
我将如何实现这一目标?
答案 0 :(得分:7)
您可以在投射后将阵列展开并再次缠绕它。 像这样:
db.collectionName.aggregate([
{$unwind:'$authors'},
{$project:{_id:1,count:1,'author.id':'$authors.author.author','author.count':'$authors.count'}},
{$group:{_id:{_id:'$_id',count:'$count'},author:{$push:{id:'$author.id',count:'$author.count'}}}},
{$project:{_id:0,_id:'$_id._id',count:'$_id.count',author:1}}
])
上面的输出将是:
{
"_id" : "581c8c3df1325f68ffd23386",
"author" : [
{
"id" : "57f246b9e01e6c6f08e1d99a",
"count" : 13.0
},
{
"id" : "5824382511f16d0f3fd5aaf2",
"count" : 1.0
}
],
"count" : 14.0
}
答案 1 :(得分:3)
我遇到了同样的问题,只是找到了一个简单优雅的解决方案,这一点在任何地方都没有提及,所以我想在这里分享一下:
您可以使用$ map迭代数组并投影每个作者。使用给定的结构,聚合应该看起来像这样
db.collectionName.aggregate([
$project: {
_id: 1,
count:1,
authors: {
$map: {
input: "$authors",
as: "author",
in: {
id: "$$author.author.author",
count: $$author.author.count
}
}
}
}
])
希望这对正在寻找的人(例如我)有帮助:)
答案 2 :(得分:0)
问题:
"customFields" : [
{
"index" : "1",
"value" : "true",
"label" : "isOffline",
"dataType" : "check_box",
"placeholder" : "cf_isoffline",
"valueFormatted" : "true"
},
{
"index" : "2",
"value" : "false",
"label" : "tenure_extended",
"dataType" : "check_box",
"placeholder" : "cf_tenure_extended",
"valueFormatted" : "false"
}
],
答案:
db.subscription.aggregate([
{$match:{"autoCollect" : false,"remainingBillingCycles" : -1,"customFields.value":"false", "customFields.label" : "isOffline"}},
{$project: {first: { $arrayElemAt: [ "$customFields", 1 ] }}}
])