Javascript RegEx for Pattern Not Working

Question

This should be easy. I have the following code:

var patt = new RegExp("\d{3}[\-]\d{1}");
var res = patt.test(myelink_account_val);
if(!res){
  alert("Inputs must begin with something like XXX-X of numbers and a dash!");
  return;
}

Basically, forcing users to enter something like 101-4 . The code is borrowed from Social Security Number input validation . And I can confirm that my inputs are indeed like 101-4; only the first five characters need to fit the pattern.

But running my code always gives the alert--the condition is never matched.

Must be something simple?!

Thanks.

Use var patt = /^\d{3}-\d$/ if the whole input should match the pattern. If it should only start with it, use var patt = /^\d{3}-\d/ — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Feb 16, 2017 at 14:09
If you create a regexp using a string then you need to escape the \ so you have to write \\d — t.niese
– t.niese, Commented Feb 16, 2017 at 14:11
@t.niese solution works--by escaping with another \ . Thanks.! — IrfanClemson
– IrfanClemson, Commented Feb 16, 2017 at 14:14
Possible duplicate of Why do regex constructors need to be double escaped? — t.niese
– t.niese, Commented Feb 16, 2017 at 14:16

dquijada · Accepted Answer · 2017-02-16 14:13:42Z

1

When you use "new RegExp" you are passing it a string.

Two solutions here:

1) Don't use "new RegExp()", but a regexp pattern:

var patt = /\d{3}[\-]\d{1}/

2) If you want to use it, remember you will have to escape the escapes:

var patt = new RegExp("\\d{3}[\\-]\\d{1}");

Also, remember, if a '-' is the only symbol (or first, or last) on a [], you can skip the escape:

var patt = new RegExp("\\d{3}[-]\\d{1}");

answered Feb 16, 2017 at 14:13

dquijada

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Tamas Hegedus · Accepted Answer · 2017-02-16 14:12:54Z

1

var patt = new RegExp("^\\d{3}[\\-]\\d{1}");
console.log(patt.test("123-4"));
console.log(patt.test("123-456"));
console.log(patt.test("12-4"));
console.log(patt.test("abc-d"));

answered Feb 16, 2017 at 14:12

Tamas Hegedus

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