MongoDB

Question

我有一个 Mongo table，其中包含年份、作者和 _id（比方说）。我需要做的就是过滤特定年份（比如 2006 年），对作者执行分组并获得每个作者的唯一 _id 列表。作者是一个数组。

我的table截图如下

{
  "_id": ObjectId("54898fbb306ba2153c3e637b"),
  "Author": [
    "Hullman, J.",
    "Drucker, S.",
    "Riche, N.H.",
    "Bongshin Lee",
    "Fisher, D.",
    "Adar, E."
  ],
  "Year": "2013"
}{
  "_id": ObjectId("54898fbb306ba2153c3e637c"),
  "Author": [
    "Schulz, H.",
    "Nocke, T.",
    "Heitzler, M.",
    "Schumann, H."
  ],
  "Year": "2013",
}{
  "_id": ObjectId("54898fbc306ba2153c3e6b58"),
  "Author": [
    "Boissonnat, J.-D.",
    "Geiger, B."
  ],
  "Year": "1993",
}

目前我的查询如下

db.list_input_file.aggregate([
  {'$match':{'Year':{$eq:'2006'}}},
  {'$unwind':'$Author'},
  {'$group':{'_id':'$Author', 'count': { '$sum': 1 }}}
])

我只需要输出格式为 [{"Author Name":[ids 列表]}, {"Author Name":[ids 列表]}]

Answer 1

我只需要使用 $addToSet 累加器运算符来获取 ID。我的最终查询如下

db.list_input_file.aggregate([
 {'$match':{'Year':{$eq:'2006'}}},
 {'$unwind':'$Author'},
 {'$group':{'_id':'$Author', 'count': { '$sum': 1 }, 'ids':{$addToSet:'$_id'}}}
])

上面查询的输出快照如下所示。

{ "_id" : "Hofacker, I.L.", "ids" : [ ObjectId("54898fbb306ba2153c3e66e4") ] }
{ "_id" : "Flamm, C.", "ids" : [ ObjectId("54898fbb306ba2153c3e66e4") ] }
{ "_id" : "Heine, C.", "ids" : [ ObjectId("54898fbb306ba2153c3e66e4") ] }
{ "_id" : "MacEachren, A.M.", "ids" : [ ObjectId("54898fbb306ba2153c3e66dd") ] }
{ "_id" : "Card, S.K.", "ids" : [ ObjectId("54898fbb306ba2153c3e66d4"), ObjectId("54898fbb306ba2153c3e66ca") ] }

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