mongodb foreach doc find min of subset and update it - node.js

Question

Product and variants price
Find Minimum of all products.Variants.Price where size is small and update it by 15%
{
"_id" : 23,
"name" : "Polo Shirt",
"Variants" : [
{
"size" : "Large",
"Price" : 82.42
},
{
"size" : "Medium",
"Price" : 20.82 // this should get increased by 15%
},
{
"size" : "Small",
"Price" : 42.29
}
]
},
{
"_id" : 24,
"name" : "Polo Shirt 2",
"Variants" : [
{
"size" : "Large",
"Price" : 182.42
},
{
"size" : "Medium",
"Price" : 120.82 // this should get increased by 15%
},
{
"size" : "Small",
"Price" : 142.29
}
]
}
I started something like this. Not sure if this is the right start
db.products.find().forEach(function(product){
var myArr = product.Variants;
print(myArr.min());
});

Answer 1

There is a problem here in that you cannot in a single update statement identify the "minimum" value in an array to use with a positional update, so you are right in a way with your current approach.
It is arguable that a better approach would be to pre-determine which element is the minimal element and this pass that to the update. You can do this using .aggregate():
var result = db.products.aggregate([
{ "$unwind": "$Variants" },
{ "$sort": { "_id": 1, "Variants.price" } }
{ "$group": {
"_id": "$_id",
"size": { "$first": "$Variants.size" },
"price": { "$first": "$Variants.price" }
}},
{ "$project": {
"size": 1,
"price": 1,
"adjusted": { "$multiply": [ "$price", 1.15 ] }
}}
])
So of course that is yet only a result with simply the lowest Variant item details for each product but then you could use the results like this:
result.result.forEach(function(doc) {
db.products.update(
{
"_id": doc._id,
"Variants": { "$elemMatch": {
"size": doc.size,
"price": doc.price
}}
},
{
"$set": {
"Variants.$.price": doc.adjusted
}
}
}
})
That is not the best form but it does at least remove some of the overhead with iterating an array and allows a way to do the calculations on the server hardware, which is possibly of a higher spec from the client.
It still doesn't really look like too much though until you take in some features available for MongoDB 2.6 and upwards. Notably that aggregate gets a cursor for a response and that you can now also do "bulk updates". So the form can be changed like so:
var cursor = db.products.aggregate([
{ "$unwind": "$Variants" },
{ "$sort": { "_id": 1, "Variants.price" } }
{ "$group": {
"_id": "$_id",
"size": { "$first": "$Variants.size" },
"price": { "$first": "$Variants.price" }
}},
{ "$project": {
"size": 1,
"price": 1,
"adjusted": { "$multiply": [ "$price", 1.15 ] }
}}
]);
var batch = [];
while ( var doc = cursor.next() ) {
batch.push({
"q": {
"_id": doc._id,
"Variants": { "$elemMatch": {
"size": doc.size,
"price": doc.price
}}
},
"u": {
"$set": {
"Variants.$.price": doc.adjusted
}
}
});
if ( batch.length % 500 == 0 ) {
db.command({ "update": "products", "updates": batch });
}
}
db.command({ "update": "products", "updates": batch });
So that is really nice in that while you are still iterating over a list the traffic and waiting for responses over the wire has really been minimized. The best part is the batch updates which are occurring ( by the math usage ) only once per 500 items. The maximum size of the batch items is actually the BSON limit of 16MB so you can tune that as appropriate.
That gives a few good reasons if you are currently developing a product to move to the 2.6 version.
The only final footnote I would add considering you are dealing with "prices" is try not to use floating point math for this and look for a form using whole integers instead as it avoids a lot of problems.

Answer 2

This is how I pulled it off.
var result = db.Products.aggregate(
[ { "$unwind":"$Variants" },{"$match":{"Variants.size":"small"}},
{ "$group":
{"_id":"$_id","minprice":
{"$min":"$Variants.price" }}},
{$sort:{ _id: 1}} ] )
result.result.forEach(function(doc) {
db.Products.update( { "_id": doc._id },
{ "$pull": { "Variants" : {
"price":doc.minprice,
"size":"small"
} } } ,
{ $addToSet: { "Variants":{
"price":doc.minprice*1.15,
"size":"small"
} }
);
});