Javascript: sort an array of objects with a first attribute , and when equal sort with another [duplicate]

Question

i ve an array which looks like this :

let myarray = 
[{id:1 , name: "mark" , birthday: "1990-08-18"},
{id:2 , name: "fred" , birthday: "1990-08-17"},
{id:3 , name: "franck" , birthday: "1990-08-16"},
{id:4 , name: "mike" , birthday: "1990-08-15"},
{id:5 , name: "jean" , birthday: "1990-08-17"}]

i'm sorting thos object by "birthday"

myarray.sort((a, b) => new Date(b.birthday).getTime() - new Date(a.birthday).getTime());

but sometimes , i ve the same birthday value for two objects (birthday is equal) , in that case i want that the sort compares those two objects (not all the objects) with the higher "id" (which have the higher id is prior)

Suggestions ?

btw, with ISO 8601 dates, you could sort with String#localeCompare. — Nina Scholz
– Nina Scholz, Commented Aug 18, 2020 at 11:06

Majed Badawi · Accepted Answer · 2020-08-18 11:07:02Z

1

You need to add a separate case in the sort function as follows:

let myarray = [
{id:1 , name: "mark" , birthday: "1990-08-18"},
{id:2 , name: "fred" , birthday: "1990-08-17"},
{id:3 , name: "franck" , birthday: "1990-08-16"},
{id:4 , name: "mike" , birthday: "1990-08-15"},
{id:5 , name: "jean" , birthday: "1990-08-17"}
]

myarray.sort((a, b) => {
     let dif = new Date(b.birthday).getTime() - new Date(a.birthday).getTime();
     if(dif != 0) return dif;
     else return b.id - a.id;
});
console.log(myarray);

answered Aug 18, 2020 at 11:07

Majed Badawi

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