JavaScript indexOf: Find a Value Without Losing Index Zero

The task

Locate exact route labels in a short timetable, then search again after the first match. This task compares primitive string values. Searching object fields by a condition is a different task for findIndex.

Runnable Node 22 example

const assert = require('node:assert/strict');

const routes = ['harbor', 'ridge', 'harbor', 'park'];
const first = routes.indexOf('harbor');
const second = routes.indexOf('harbor', first + 1);
assert.equal(first, 0);
assert.equal(second, 2);
assert.equal(routes.indexOf('airport'), -1);
assert.equal(routes.indexOf('Harbor'), -1);
assert.equal(routes.indexOf('harbor', 4), -1);
assert.equal(Boolean(first), false);
assert.equal(first !== -1, true);
console.log(`first=${first}; second=${second}; missing=${routes.indexOf('airport')}`);

Expected output and explanation

Harbor appears twice, so the first call returns zero and the second returns two. Starting at first + 1 deliberately excludes the previous match. The original array is only searched; nothing is removed or rearranged.

The capitalized label does not match the lower-case value. Decide whether exact case is the intended contract before normalizing input. The check starting at the array length demonstrates that a start position beyond the available elements cannot find a match.

Pitfall to check

Never write if (routes.indexOf(label)) to mean “found.” Inspect index !== -1 before using the number as an array position. Otherwise a missing result can silently select an unintended value when combined with other indexing logic.