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JavaScript compare strings alphabetically | Example code

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Use localeCompare method compare strings alphabetically, It returns -1 since "a" < "b", 1 or 0 otherwise.


Also, if what you are sorting contains numbers, you may want:

"a5b".localeCompare("a21b", undefined, { numeric: true })

This returns -1, recognizing that 5 as a number is less than 21. Without { numeric: true } it returns 1, since “2” sorts before “5”. In many real-world applications, users expect “a5b” to come before “a21b”.


JavaScript compares strings alphabetically

Simple example code.

<!DOCTYPE html>

   var res = "a".localeCompare("b");



JavaScript compare strings alphabetically

Compare 2 Strings Alphabetically for Sorting Purposes with JavaScript with the localeCompare Method. It’ll return -1 is a before b alphabetically, 0 if they’re the same, and 1 otherwise.

const arr = ['foo', 'bar', 'baz']
const sorted = arr.sort((a, b) => a.localeCompare(b))

Output: [“bar”, “baz”, “foo”]

Do comment if you have any doubts or suggestions on this JS string topic.

Note: The All JS Examples codes are tested on the Firefox browser and the Chrome browser.

OS: Windows 10

Code: HTML 5 Version

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