Validate string in regular expression

Just use

/^[a-z-]+$/

Explanation

• ^ : Match from beginning string.
• [a-z-] : Match all character between a-z and -.
  • [] : Only characters within brackets are allowed.
  • a-z : Match all character between a-z. Eg: p,s,t.
  • - : Match only strip (-) character.
• + : The shorthand of {1,}. It's means match 1 or more.
• $: Match until the end of the string.

Example

const regex= /^[a-z-]+$/

console.log(regex.test("abc")) // true
console.log(regex.test("aBcD")) // false
console.log(regex.test("a-c")) // true

Try this:

var regex = /^[-a-z]+$/;

var regex = /^[-a-z]+$/;
var strs = [
  "a",
  "aB",
  "abcd",
  "abcde-",
  "-",
  "-----",
  "a-b-c",
  "a-D-c",
  " "
];
strs.forEach(str=>console.log(str, regex.test(str)));

Try this

/^[a-z-]*$/

it should match the letters a-z or - as many times as possible.

What you regex does is trying to match a-z one single time, followed by any of -, whitespace or dot one single time. Then expect the string to end.

Use this regular expression:

let regex = /^[a-z\-]+$/;

Then:

regex.test("abcd") // true
regex.test("ab-d") // true
regex.test("ab3d") // false
regex.test("") // false

PS: If you want to allow empty string "" to pass, use /^[a-z\-]*$/. Theres an * instead of + at the end. See Regex Cheat Sheet: https://www.rexegg.com/regex-quickstart.html

I hope this helps

var str = 'asdadWW--asd';

console.log(str.match(/[a-z]|\-/g));

This will work:

var regex = /^[a-z|\-|\s]+$/ //For this regex make use of the 'or' | operator

str = 'test- ';

str.match(regex); //["test- ", index: 0, input: "test- ", groups: undefined]

str = 'testT- ' // string now contains an Uppercase Letter so it shouldn't match anymore

str.match(regex) //null