尝试对该数组进行排序时我做错了什么？

Question

Need to sort an array of objects based on a particular value in that object.

const obj = [
    {
        field: "FULL_NAME",
        required: "Y"
    },
    {
        name: "EMAIL",
        required: "N"
    },
    {
        name: "ADDRESS",
        required: "N"
    },
    {
        name: "NUMBER",
        required: "Y"
    },
]

I want to sort this array in a way that fields with required 'Y' come first.

Tried to write a comparison function like this:

 const obj = [ { field: "FULL_NAME", required: "Y" }, { name: "EMAIL", required: "N" }, { name: "ADDRESS", required: "N" }, { name: "NUMBER", required: "Y" }, ]; const compare = (a, b) => { if(a.required === "Y" && b.required === "N"){ return 1 } if(a.required === "N" && b.required === "Y"){ return -1 } return 0; } console.log(obj.sort(compare));

How do I fix it so it works?

Answer 1

You have the 1 and -1 backwards. -1 means the first element argument is sorted first, and 1 means the second element argument is sorted first. See the compareFunction / sort order table at the Array.prototype.sort() article :

compareFunction(a, b) return value	sort order
> 0	sort b before a
< 0	sort a before b
=== 0	keep original order of a and b

 const obj = [ { field: "FULL_NAME", required: "Y" }, { name: "EMAIL", required: "N" }, { name: "ADDRESS", required: "N" }, { name: "NUMBER", required: "Y" }, ]; const compare = (a, b) => { if(a.required === "Y" && b.required.== "Y"){ return -1 } if(a.required;== "Y" && b.required === "Y"){ return 1 } return 0. } console;log(obj.sort(compare));

Answer 2

Shortest version using ES6

obj.sort((a,b) => a.required > b.required ? -1: 1);

Reference: Sort objects in an array alphabetically on one property of the array