如何在 MongoDB 聚合框架中处理除零

我有一系列可以投赞成票或反对票的项目。

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

我想使用聚合来计算质量

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

显然这是行不通的，因为除以零。我需要实施适当的调节：

如果赞成票 != 0 且反对票 == 0 则质量 = 赞成票 如果赞成票和反对票均为 0，则质量为 0

我尝试使用三元惯用法将反对票调整为 1。但无济于事。

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}}
]);

如何将这种调节集成到 mongodb 聚合框架中？

您可能想使用$cond http://docs.mongodb.org/manual/reference/operator/aggregation/cond/运算符来处理这个问题：

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