使用pymongo根据一个值查询两个字段的总和

Question

基本上，我有一个带有嵌套对象数组的对象数组。我想根据"userID"和"flashcardReversed.value" + "flashcardReversed.count"之和查询数据。

用户将提供所需的用户 ID 和（值 + 计数）值。

这是我的数据：

[
  {
    "_id": "608642db80a36336946620aa",
    "userID": "user1",
    "title": "title2",
    "flashcardReversed": [
      {
        "_id": "608d5b290e635ece6828141X",
        "front": "2front",
        "back": "2back",
        "value": 1,
        "count": 2
      },
      {
        "_id": "608t5b290e635ece6828141Y",
        "front": "2frontReversed",
        "back": "2backReversed",
        "value": 2,
        "count": 3
      },
      {
        "_id": "608a5b31a3f9806de253726X",
        "front": "2front2",
        "back": "2back2",
        "value": 3,
        "count": 4
      },
      {
        "_id": "608a5b31a3f9806de253726Y",
        "front": "2frontReversed2",
        "back": "2backReversed2",
        "value": 4,
        "count": 5
      }
    ]
  },
  {
    "_id": "608642db80a36336946620aa",
    "userID": "user1",
    "title": "title3",
    "flashcardReversed": [
      {
        "_id": "608d5b290e635ece6828142X",
        "front": "2front",
        "back": "2back",
        "value": 12,
        "count": 6
      },
      {
        "_id": "608t5b290e635ece6828143Y",
        "front": "2frontReversed",
        "back": "2backReversed",
        "value": 21,
        "count": 7
      },
      {
        "_id": "608a5b31a3f9806de253727X",
        "front": "2front2",
        "back": "2back2",
        "value": 34,
        "count": 8
      },
      {
        "_id": "608a5b31a3f9806de253729Y",
        "front": "2frontReversed2",
        "back": "2backReversed2",
        "value": 42,
        "count": 9
      }
    ]
  },
  {
    "_id": "608642db80a36336946620aa",
    "userID": "user2",
    "title": "title4",
    "flashcardReversed": [
      {
        "_id": "608d5b290e635ece6828131X",
        "front": "2front",
        "back": "2back",
        "value": 41,
        "count": 10
      },
      {
        "_id": "608t5b290e635ece6828161Y",
        "front": "2frontReversed",
        "back": "2backReversed",
        "value": 54,
        "count": 11
      },
      {
        "_id": "608a5b31a3f9806de253526X",
        "front": "2front2",
        "back": "2back2",
        "value": 63,
        "count": 12
      },
      {
        "_id": "608a5b31a3f9806de253326Y",
        "front": "2frontReversed2",
        "back": "2backReversed2",
        "value": 29,
        "count": 13
      }
    ]
  }
]

例如查询应该是这样的：

{
  "userID": {"$eq": provided_user_id}
  {"flashcardReversed.value" + "flashcardReversed.count"}: {"$eq": provided_value_plus_count}
    }

是否可以使用 pymongo 来实现这一点？

Answer 1

假设这里的唯一性是userID+title，并且我们希望在flashcardReversed 字段和多个title 之间进行聚合，即为用户获取一个数字，那么这是一个解决方案。

var targetUser = "user1";
var targetAmt = 163;

c = db.foo.aggregate([
    {$match: {"userID": targetUser}} // will fetch one OR MORE records...                                       
    ,{$addFields: {
        X: {$reduce: {
            input: "$flashcardReversed",
            initialValue: 0,
                // $$value is the running result of the reduce operation ($add)
                // $$this is the current object from the array; it just so
                // happens that a field in your object is named "value" so
                // do not confuse the two.  The expression below translates
                // to:  n = n + oneitem.value + oneitem.count
                in:{$add: [ "$$value", "$$this.value", "$$this.count" ] }
        }}
    }}

    // Now, add up X over multiple docs.  We can group by null because we                                       
    // know ONLY docs where userID = targetUser are coming through:                                             
    ,{$group: {_id: null, n: {$sum: "$X"}}}

    // ...and match against input:                                                                              
    ,{$match: {n: targetAmt}}                                                                                 
]);

上述解决方案演示了单独的 $reduce 和 $group 行为，但这些行为可以组合使用。另外，让我们探讨一下要为目标数量获取多个用户需要做什么：

var targetUser = ["user1","user2"];
var targetAmt = 163;

c = db.foo.aggregate([
    {$match: {"userID": {$in: targetUser}}}                                 

    // Combine the $group and $reduce together.  Saves a stage!  Also, since                                    
    // more than one userID can come through, we must change _id to group                                       
    // by $userID instead of null:                                                                              
    ,{$group: {_id: "$userID", n: {$sum: {$reduce: {
            input: "$flashcardReversed",
            initialValue: 0,
                in:{$add: [ "$$value", "$$this.value", "$$this.count" ] }
    }} }  }}

    // Probably want to use >= or <= instead of exact match sometimes                                           
    // so here is an example of that:                                                                           
    ,{$match: {$expr: {$gte: ["$n", targetAmt]}} }
]);

{ "_id" : "user1", "n" : 163 }
{ "_id" : "user2", "n" : 233 }

Answer 2

查询

• 匹配用户 ID
• 过滤数组并只保留(= (+ value count) 3)的成员
• 您可以将 3 替换为变量或任何数字
• 如果过滤后数组为空，则拒绝文档
• 在最终结果中，我们只在数组中保留通过的那些，并且只保留至少有 1 个通过的文档

*不确定你是否想要这个

Test code here

aggregate(
[{"$match": {"userID": {"$eq": "user1"}}},
 {"$set":  
   {"flashcardReversed": 
     {"$filter": 
       {"input": "$flashcardReversed",
        "cond": 
        {"$eq": [{"$add": ["$$this.value", "$$this.count"]}, 3]}}}}},
 {"$match": {"$expr": {"$ne": ["$flashcardReversed", []]}}}])