如何在不改变其他元素的情况下替换数组中的元素？

Question

If I have this array ['Jan', 'March', 'April', 'June'] . How to replace 'April' with 'newVal' and keep all other elements unchanged. I tried

const months = ['Jan', 'March', 'April', 'June'];
months.splice(3, 1, 'newVal');
console.log(months);

I want it to return ['Jan', 'March', 'newVal', 'June'] but it returns ["Jan", "March", "April", "newVal"]

Answer 1

Why not assign directly?

 const months = ['Jan', 'March', 'April', 'June']; months[2] = 'newVal'; console.log(months);

Answer 2

You can do this:

a.splice(2,1,"newVal")

It deletes 1 element at position 2 and adds the new value "newVal"

Answer 3

You could first find the Arrray.prototype.indexOf() some "string" , than you can directly replace it:

const arr = ["a", "b", "c", "d"];
const idx = arr.indexOf("c");       // 0..N or -1 if not found

// Make sure the index is found, than replace it
if (idx > -1) arr.splice(idx, 1, "newVal");

A nifty reusable function:

 const replaceInArray = (arr, oldVal, newVal) => { const idx = arr.indexOf(oldVal); if (idx > -1) arr.splice(idx, 1, newVal); }; const months = ['Jan', 'March', 'April', 'June']; replaceInArray(months, "April", "newVal"); console.log(months)

Answer 4

如果您知道要替换的值（即：您总是想替换 'April'）：

months[months.indexOf("April")] = 'newVal';

Answer 5

永远记住数组的第一个索引总是 0，所以只需这样做：

months.splice(2,1,"newVal")

Answer 6

The function which you used was correct but You mistook the index value of element april of the array months

The correct index value of april is 2

So your code will be like this

 const months = ['Jan', 'March', 'April', 'June']; months.splice(2, 1, 'newVal'); console.log(months);