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?
>Solution :
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));