CODEWITHSUNDEEP
javascript
inquisitive
inquisitive

Reputation: 3974

Ends with regular expression - Javascript

"inquisitive".endsWith('ive') returns true.

I want to be able to search for something like "inquisitive".endsWith(i[a-z]e), which is like ends with 'i' and some character after that and then an 'e'. I don't want to do a substring match, since this method is going to be generic, and for most of the cases will not have regex but simple endsWith logic.

Is there a way, this can be achieved, using vanilla JS?

Upvotes: 2

Views: 1439

Answers (1)

CertainPerformance
CertainPerformance

Reputation: 370609

If you have to use only endsWith and cannot use a regular expression for whatever odd reason, you can come up with a list of all strings between 'iae' and 'ize':

const allowedStrs = Array.from(
  { length: 26 },
  (_, i) => 'i' + String.fromCharCode(i + 97) + 'e'
);
const test = str => allowedStrs.some(end => str.endsWith(end));
console.log(test('iae'));
console.log(test('ize'));
console.log(test('ife'));
console.log(test('ihe'));
console.log(test('iaf'));
console.log(test('aae'));

But that's way more complicated than it should be. If at all possible, change your code to accept the use of a regular expression instead, it's so much simpler, and should be much preferable:

const test = str => /i[a-z]e$/.test(str);
console.log(test('iae'));
console.log(test('ize'));
console.log(test('ife'));
console.log(test('ihe'));
console.log(test('iaf'));
console.log(test('aae'));

Upvotes: 4

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