How to get multiple counts in one query for one field? - mongodb

Question

So what i'm trying to do is to get multiple counts of one field depending on a min max value.
Collection holds something like
{name:'hi',price:100},
{name:'hi',price:134},
{name:'hi',price:500}
What i want to get is for example the count of items that are between price 100-200, 200-300, 300-400, 400-500.
Is there a way to do this in mongoDB with one query? Is there a way to get the query without knowing min max?

Answer 1

You want .aggregate() here with the $cond ternary operator to determine the grouping id withing $group:
db.collection.aggregate([
{ "$match": {
"price": { "$gte": 100, "$lte" 500 }
}},
{ "$group": {
"_id": {
"$cond": [
{ "$lte": [ "$price", 200 ] },
"100-200",
{ "$cond": [
{ "$lte": [ "$price", 300 ] },
"200-300",
{ "$cond": [
{ "$lte": [ "$price", 400 ] },
"300-400",
"400-500"
]}
]}
]
},
"count": { "$sum": 1 }
}}
])
As a "ternary" if/then/else the $cond will evaluate the expression in the first argument "if" and then either return the second argument "then" where true or the third "else" where false.
The cascading logic means that you "nest" each ternary operation inside the false assertion till you reach an eventual result.
With the grouping _id value provided by conditions, you then just use $sum with an argument of 1 to "count" the matches in the group.
This gives you a response on the sample as:
{ "_id": "100-200", "count": 2 }
{ "_id": "400-500", "count": 1 }
The $match is making sure that all results are in the "ranges" that wil be tested. If you exclude that then you likely want a last $cond "else" condition to return another value if the "price" was outside of an expected "range".
If you are looking to return "each" range, then you are better off inspecting the result and inserting a 0 count entry for every range that is not returned.