_.sortBy(arrData, "rhid");
This code sorts array but as the values of field "rhid" are strings the order is messed up. How can i sort as if "rhid" where int field.
Thanks
asked Jan 11, 2016 at 14:33
Leonel Matias Domingos
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3 Answers
sortBy can be used with a function instead of a property name.
_.sortBy(arrData, function (obj) {
return parseInt(obj.rhid, 10);
});
2 Comments
GunasekaranR
How can I sort by more than two fields and sort by desc with this implementation?
Aatif Khan
@GunasekaranR - in that case you can use this: sortBy(arrData, [function(obj) { return parseInt(obj.field1); }, function(obj) { return parseInt(obj.field2); }]);
This can be achieved using arrow notation like:
_.sortBy(arrData, (obj) => parseInt(obj.val, 10));
If you want to do more than one field like @GunasekaranR asked, you can also do it with arrow notation:
_.sortBy(arrData, [
(obj) => parseInt(obj.first_val, 10),
(obj) => parseInt(obj.second_val, 10)
]);
The second way uses first_val as the main sorting object, with second_val being the tie breaker.
Comments
if rhid is number then you can do like this
orderBy(
arrData,
function (o) {
return new Number(o.rhid);
},
["asc"]
),
mentioned here
