RegExp (Options)

JavaScript
1

I have created a RegExp but need to add some other variables

Valid exp should allow an empty string - if not an empty string must be a number only or the following 000000-A-000000-000 as below.
I’m not sure how to add ‘conditionals’ in exp’s

/\d{6}-[A-Za-z]{1}-\d{6}-\d{3}/gm
2

Hi,

the | operator allows the regular expression to match if any one of a number of conditions is satisfied. E.g. /foo|bar/ will match “foo” or “bar”, but nothing else.

In your case you need to check for:

  • An empty string.
  • A string that is only numbers.
  • The specific format 000000-A-000000-000.

So you can do this:

/^$|^\d+$|^\d{6}-[A-Za-z]{1}-\d{6}-\d{3}$/gm

To test:

const regex = /^$|^\d+$|^\d{6}-[A-Za-z]{1}-\d{6}-\d{3}$/;

console.log(regex.test("")); // true
console.log(regex.test("1234567890")); // true
console.log(regex.test("123456-A-123456-789")); // true
console.log(regex.test("789-A-123456-123456")); // false
console.log(regex.test("JavaScript")); // false

Alternatively, test here:

1 Like
3

Throw your stones, but I would never use regex for such a scenario. If I have exact three states to check I would always use an if with three conditions.

Regex is great if you need to match patterns but not for fix values

4

How do you want to do either of those without a regex?

I understood the second requirement to be a string in this format:

six digits
minus
uppercase A-Z
six digits
minus
three digits

Not just the string “000000-A-000000-000”

–

Edit:

Or do you mean:

const val = ...

if (val === '') { 
  ...
} elseif (/^\d+$/.test(val)) {
  ...
} elseif (/^\d{6}-[A-Za-z]{1}-\d{6}-\d{3}$/.test(val)) {
  ...
}
5

Sorry you are right. I thought the value is a static one but his regex showed that it looks like a pattern

2 Likes