Code:
let names= ["Style","List","Raw"];
let results= names.find(x=> x.includes("s"));
console.log(results);
How to get the names which contain "s" from the array names, currently, I am getting only one element as a result but i need all occurrences.
You have to use filter at this context,
let names= ["Style","List","Raw"];
let results= names.filter(x => x.includes("s"));
console.log(results); //["List"]
If you want it to be case insensitive then use the below code,
let names= ["Style","List","Raw"];
let results= names.filter(x => x.toLowerCase().includes("s"));
console.log(results); //["Style", "List"]
To make it case in sensitive, we have to make the string's character all to lower case.
Use filter instead of find.
let names= ["Style","List","Raw"];
let results= names.filter(x => x.includes("s"));
console.log(results);
But you can also use forEach() method :
var names = ["Style","List","Raw"];
var results = [];
names.forEach(x => {if (x.includes("s") || x.includes("S")) results.push(x)});
console.log(results); // [ 'Style', 'List' ]
Or if you prefere :
names.forEach(x => {if (x.toLowerCase().includes("s")) results.push(x)});
