In the following JavaScript code, I am trying to get names that have the letter "Y" or "y". When I looked up how to do that, I saw .filter and .find methods. But this code only returns containsY = ['Tommy']. I tried for loop, but .find only returns the value of the first element in the provided condition, and it doesn't return all names with "y"s. Is there any better way to get all of "Y" and "y" names?
const students = ["Tommy","Noah","Xander","Adil","Bradley","Nicholas","Damien"]
let containsY = students.filter((student) => student.includes("Y"))
//Look for name with "y", and push it to containsY
containsY.push(students.find((student) => student.includes("y")))
console.log(containsY)
Your filter approach is fine. It's giving incorrect results because
includes is case sensitive.
Instead, inside the filter callback, convert student to lowercase, then check whether it includes "y":
const students = ["Tommy","Noah","Xander","Adil","Bradley","Nicholas","Damien"]
const res = students.filter((student) => student.toLowerCase().includes("y"))
console.log(res)Either use a regular expression (which has the benefit of only requiring a single test)
const students = ["Tommy", "Noah", "Xander", "Adil", "Bradley", "Nicholas", "Damien"]
const result = students.filter(student => /y/i.test(student))
console.log(result)Or test for both .includes('y') and .includes('Y') in (a single) callback.
const students = ["Tommy", "Noah", "Xander", "Adil", "Bradley", "Nicholas", "Damien"]
const result = students.filter(student => student.includes('y') || student.includes('Y'));
console.log(result)