find all the indexes of certain character in string javascript

Question

i have certain string

const str = "・Welcome to ・StackOverFlow ・Best Regards";

i want to get all the indexes of the character "・"

unfornately the indexOf only gives the first instance

const str = "・Welcome to ・StackOverFlow ・Best Regards";
let bullet_ndx = [];
console.log(str.indexOf("・"));

is there a way i could find all the indexes of the certain character?

If you want a one-liner `bullet_ndx = str.split("").map((c,i)=> c == "・" ? i : -1 ).filter(i => i >= 0 )` — Ricky Mo, Aug 03 '21 at 04:00
`bullet_ndx = str.split('').reduce((a, c, i) => c === '・' ? [...a, i] : a, [])` — Yosvel Quintero, Aug 03 '21 at 04:14

score 0 · Answer 1 · answered Aug 03 '21 at 03:51

0

You can use a simple for loop, and an if condition for this.

const str = "・Welcome to ・StackOverFlow ・Best Regards";

let bulletIndex = []

for(var i = 0; i<str.length; i++){
  if(str[i] == "・"){
    bulletIndex.push(i)
  }
}

console.log(bulletIndex)

answered Aug 03 '21 at 03:51

TechySharnav

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i was hoping if there is one function to address this without looping but this is the best possible way. – Stacks Queue Aug 03 '21 at 03:53
@StacksQueue Unfortunately, there is not. There is `indexOf` and `lastIndexOf` that gives index of the char starting from forward and backwards respectively. You can write some user defined function though. – TechySharnav Aug 03 '21 at 03:57

score 0 · Answer 2 · answered Aug 03 '21 at 03:55

indexOf takes a second argument, as a start position to look from

const str = "・Welcome to ・StackOverFlow ・Best Regards";
let bullet_ndx = [];
let index = 0;
while((index = str.indexOf("・", index))>=0) bullet_ndx.push(index++);
console.log(bullet_ndx);

find all the indexes of certain character in string javascript

2 Answers2