MongoDB: How to get N decimals precision in a query

Question

Given the following documents in a MongoDB collection...

{ \"_id\": 1, \"amount\": { \"value\": 1.123456789999, \"rate\": 1.2 }}
{ \"_id\": 2, \"amount\": { \"va         


        

Answer 1

Starting from the answer I got from user3100115, here below is the final soluton:

db.orders.find({ "$where": function() { return this.amount.value - (this.amount.value % 0.01) === 0.03; }})

I hope it helps.

Answer 2

you can just use the regex operator

db.test.find({ value: { $regex: 1\.12[0-9]*}})

Answer 3

You can do this with the $where operator.

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

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EDIT (after this comment)

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"
        ]
     }}
])

Answer 4

You're effectively doing a range query where 1.12 <= value < 1.13, so you could do this as:

db.test.find({'amount.value': {$gte: 1.12, $lt: 1.13}})

Which is a truncation to 2 decimal places. If you want to round it, you'd be looking for 1.115 <= value < 1.125:

db.test.find({'amount.value': {$gte: 1.115, $lt: 1.125}})