如何在MongoDB彙總框架中處理零除

Question

我有一個項目的集合可以upvoted或downvoted。

{"_id" : 1, "name": "foo", "upvotes" : 30, "downvotes" : 10} 
{"_id" : 2, "name": "bar", "upvotes" : 20, "downvotes" : 0} 
{"_id" : 3, "name": "baz", "upvotes" : 0, "downvotes" : 0} 

我想用零來使用聚合計算質量

db.items.aggregate([ 
    {"$project": 
     { 
      "name": "$name", 
      "upvotes": "$upvotes" 
      "downvotes": "$downvotes", 
      "quality": {"$divide":["$upvotes", "$downvotes"]} 
     } 
    }, 
    {"$sort": {"quality":-1}} 
]); 

顯然，這是不行的，因爲分工。我需要實現適當的調節：

如果upvotes = 0和downvotes == 0則質量= upvotes 如果upvotes和downvotes均爲0，那麼質量是0

我試圖調整downvotes 1使用！三元成語。但無濟於事。

db.items.aggregate([ 
    {"$project": 
     { 
      "name": "$name", 
      "upvotes": "$upvotes", 
      "downvotes": "$downvotes" ? "$downvotes": 1 
     } 
    }, 
    {"$project": 
     { 
      "name": "$name", 
      "upvotes": "$upvotes" 
      "downvotes": "$downvotes", 
      "quality": {"$divide":["$upvotes", "$downvotes"]} 
     } 
    }, 
    {"$sort": {"quality":-1}} 
]); 

如何在mongodb聚合框架中集成這種調節？

Answer 1

您可能需要使用$cond操作者處理這個問題：

db.items.aggregate([ 
    {"$project": 
     { 
      "name": "$name", 
      "upvotes": "$upvotes", 
      "downvotes": "$downvotes", 
      "quality": { $cond: [ { $eq: [ "$downvotes", 0 ] }, "N/A", {"$divide":["$upvotes", "$downvotes"]} ] } 
     } 
    }, 
    {"$sort": {"quality":-1}} 
]); 

Answer 2

你想要$ cond操作符。 http://docs.mongodb.org/manual/reference/operator/aggregation/cond/

喜歡的東西：

{"$project": 
    { 
    "name": "$name", 
    "upvotes": "$upvotes", 
    "downvotes": { $cond: [ { $eq: ["$downvotes", 0] }, 1, "$downvotes"] } 
    } 
},