sql subqueries to mongodb - mysql

Question

I am new to MongoDB and I am trying to turn SQL queries into MongoDB queries. But can't seem to find any way to turn a SQL query with a subquery to mongoDB.
for example:
SELECT article, dealer, price
FROM shop
WHERE price=(SELECT MAX(price) FROM shop);
I tried the following, but it doesn't seem to work.
db.shop.group({
"initial": {},
"reduce": function(obj, prev) {
prev.maximumvalueprice = isNaN(prev.maximumvalueprice) ? obj.price :
Math.max(prev.maximumvalueprice, obj.price);
}}).forEach(
function(data){
db.shop.find({
"price": data
},
{
"article": 1,
"dealer": 1,
"price": 1
})
})
How do I convert this SQL query into a MongoDB query?

Answer 1

If you are using MongoDB v. 3.2 or newer you can try to use $lookup.
Try to use aggregation:
$sort your collection by price by DESC;
set $limit to 1 (it will take a first document, which will be with biggest price);
then use $lookup to select the documents from the same collection by max price and set it to tmpCollection element;
$unwind tmpCollection;
$replaceRoot - change document root to $tmpCollection
Example:
db.getCollection("shop").aggregate([
{$sort: {"price":-1}},
{$limit: 1},
{$lookup: {
from: "shop",
localField: "price",
foreignField: "price",
as: "tmpCollection"
}},
{$unwind: "$tmpCollection"},
{$replaceRoot: {newRoot:"$tmpCollection"}}
]);

Answer 2

Looks like you need the aggregation framework for this task using $first within a $group pipeline stage on ordered documents. The initial pipeline step for ordering the documents in the collection is $sort:
db.shop.aggregate([
{ "$sort": { "price": -1 } }, // <-- sort the documents first in descending order
{
"$group": {
"_id": null,
"article": { "$first": "$article" },
"dealer": { "$first": "$dealer" },
"price": { "$first": "$price" }
}
}
])
or using $last
db.shop.aggregate([
{ "$sort": { "price": 1 } }, // <-- note the sort direction
{
"$group": {
"_id": null,
"article": { "$last": "$article" },
"dealer": { "$last": "$dealer" },
"price": { "$last": "$price" }
}
}
])