javascript - 如果使用过滤器找不到牙医，则返回null(Return null if no dentist found using filter)

Question

I have the following code:(我有以下代码：)

function findFirstDentist(people) { let firstDentistFound = people.filter(dentist => person.isDentist) } This is been run against:(这是针对：) const dentists = [ { name: "Johnny Karate", isDentist: false }, { name: "Lucy Hobbs Taylor", isDentist: true }, { name: "Pierre Fauchard", isDentist: true }, { name: "Bert Macklin", isDentist: false }, { name: "Orin Scrivello", isDentist: true }, { name: "Kip Hackman", isDentist: false } ]; I am struggling to make this return null if there are no dentists showing true.(如果没有牙医证明我是真的，我正在努力使此返回为空。)   ask by JJSR23 translate from so

Answer 1

You could check the length of the new array and return either the array or null .(您可以检查新数组的length ，然后返回array或null 。)

function findFirstDentist(people) { let firstDentistFound = people.filter(dentist => dentist.isDentist); return firstDentistFound.length ? firstDentistFound : null; }