Reputation: 899
I was looking for a way to sort array of object by an arbitrary list. Lets assume I have this array of objects.
[
{
"_id": "4JEEuhNIae",
"category": "grocery"
},
{
"_id": "4JW7miNITl",
"category": "food"
},
{
"_id": "4Je4kmrrbZ",
"category": "coffee"
},
{
"_id": "4JgAh3N86x",
"category": "coffee"
}
]
This is the array that I would like to use as sorting criteria. Records with food
should come first, then coffee
and grocery
.
['food','coffee','grocery']
Result should be:
[
{
"_id": "4JW7miNITl",
"category": "food"
},
{
"_id": "4Je4kmrrbZ",
"category": "coffee"
},
{
"_id": "4JgAh3N86x",
"category": "coffee"
},
{
"_id": "4JEEuhNIae",
"category": "grocery"
},
]
How can I do this type of sorting on mongodb by using mongoose? I really don't want to make any operations on the code after fetching data.
Upvotes: 3
Views: 1175
Reputation: 103335
You could try running a custom comparator function with the native JavaScript sort()
method on the array returned from the cursor.toArray()
method:
var order = ["food", "coffee", "grocery"];
var docs = db.collection.find().toArray().sort(function(a, b) {
return order.indexOf(a.category) - order.indexOf(b.category);
});
printjson(docs);
Sample Output
[
{
"_id" : "4JW7miNITl",
"category" : "food"
},
{
"_id" : "4Je4kmrrbZ",
"category" : "coffee"
},
{
"_id" : "4JgAh3N86x",
"category" : "coffee"
},
{
"_id" : "4JEEuhNIae",
"category" : "grocery"
}
]
With the new MongoDB 3.4 version, you should be able to leverage the use of the native MongoDB operators $addFields
and $indexOfArray
in the aggregation framework.
$addFields
pipeline step allows you to $project
new fields to existing documents without knowing all the other existing fields. $indexOfArray
expression returns position of particular element in a given array.So putting that altogether you could try the following aggregate operation (with MongoDB 3.4):
var order = ["food", "coffee", "grocery"],
projection = {
"$addFields" : {
"__order" : { "$indexOfArray" : [ order, "$category" ] }
}
},
sort = { "$sort" : { "__order" : 1 } };
db.collection.aggregate([ projection, sort]);
Upvotes: 2