MongoDB:如何在查询中获得 N 个小数精度 [英] MongoDB: How to get N decimals precision in a query

问题描述

给定 MongoDB 集合中的以下文档...

Given the following documents in a MongoDB collection...

{ "_id": 1, "amount": { "value": 1.123456789999, "rate": 1.2 }}
{ "_id": 2, "amount": { "value": 2.9844, "rate": 1.2 }}
{ "_id": 3, "amount": { "value": 1.123876, "rate": 1.2 }}
{ "_id": 4, "amount": { "value": 3.3557886, "rate": 1.2 }}
{ "_id": 5, "amount": { "value": 1.12599976, "rate": 1.2 }}

...是否可以选择 amount 为 1.12 的所有文档(即在查询中获得 2 位小数精度)?

... is it possible to select all those documents where amount is 1.12 (i.e. get 2 decimals precision in the query)?

推荐答案

您可以使用 $where 运算符.

You can do this with the $where operator.

db.collection.find({ "$where": function() { 
    return Math.round(this.amount.value * 100)/ 100 === 1.12; 
    }
})

<小时>

编辑(在此评论之后)

在这种情况下，您应该使用聚合框架，尤其是 $redact 运算符，这比使用 $where

In this case you should use the aggregation framework especially the $redact operator and this is much faster than the solution using $where

db.collection.aggregate([
    { "$redact": { 
        "$cond": [
            { "$eq": [
                { "$subtract": [ 
                    "$amount.value", 
                    { "$mod": [ "$amount.value",  0.01 ] }
                ]}, 
                0.03
            ]}, 
            "$$KEEP", 
            "$$PRUNE"
        ]
     }}
])

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