为什么相同的一次性正则表达式在两个平台上表现不同？

Question

I'm attempting to validate an RFC (Mexican tax id) code in a form with JS and JQuery. I only require to validate whether the following Regex match returns null :

 function validateRFC(rfc) { const rgxFisica = "^(([A-ZÑ&]{4})([0-9]{2})([0][13578]|[1][02])(([0][1-9]|[12][\\d])|[3][01])([A-Z0-9]{3}))|" + "(([A-ZÑ&]{4})([0-9]{2})([0][13456789]|[1][012])(([0][1-9]|[12][\\d])|[3][0])([A-Z0-9]{3}))|" + "(([A-ZÑ&]{4})([02468][048]|[13579][26])[0][2]([0][1-9]|[12][\\d])([A-Z0-9]{3}))|" + "(([A-ZÑ&]{4})([0-9]{2})[0][2]([0][1-9]|[1][0-9]|[2][0-8])([A-Z0-9]{3}))$"; return rfc.match(rgxFisica);= null } const foo = validateRFC("XXXX901212UDX"). console;log(foo);

The pattern is basically XXXXYYMMDDYYY where X must be a letter, YMD are date values and Y can be any alphanumeric character.

I did a small unit test on codepen with my real id and it works. However, when testing it in my test environment it always returns null. Here's a screenshot. In case it matters, I'm rendering the entire page with Flask in Python 3.7, and the validation is inside document.ready

Is it possible that my Javascript is interpreting regular expressions differently?

EDIT:

It seems the embedded snippet also returns true . I'm using VS2019 16.3.5, Python 16.3.19252.1 in a Python 3.7 64-bit environment.

Answer 1

In some cases of javascript, like ECMAScript, if the match fails, then the call rfc.match(rgxFisica) will return an array. If the match passes, it will return a boolean true .

see example ( https://jsfiddle.net/coq3Lugw/ ):

function validateRFC(rfc) {
  const rgxFisica = "^(([A-ZÑ&]{4})([0-9]{2})([0][13578]|[1][02])(([0][1-9]|[12][\\d])|[3][01])([A-Z0-9]{3}))|" +
    "(([A-ZÑ&]{4})([0-9]{2})([0][13456789]|[1][012])(([0][1-9]|[12][\\d])|[3][0])([A-Z0-9]{3}))|" +
    "(([A-ZÑ&]{4})([02468][048]|[13579][26])[0][2]([0][1-9]|[12][\\d])([A-Z0-9]{3}))|" +
    "(([A-ZÑ&]{4})([0-9]{2})[0][2]([0][1-9]|[1][0-9]|[2][0-8])([A-Z0-9]{3}))$";
    console.log(rfc.match(rgxFisica))

    console.log("hello 0670 world".match(rgxFisica))

  return rfc.match(rgxFisica) != null
}

const foo = validateRFC("XXXX901212UDX");
//console.log(foo);

what I would do is set an if-then statement and return that way:

function validateRFC(rfc) {
  const rgxFisica = "^(([A-ZÑ&]{4})([0-9]{2})([0][13578]|[1][02])(([0][1-9]|[12][\\d])|[3][01])([A-Z0-9]{3}))|" +
    "(([A-ZÑ&]{4})([0-9]{2})([0][13456789]|[1][012])(([0][1-9]|[12][\\d])|[3][0])([A-Z0-9]{3}))|" +
    "(([A-ZÑ&]{4})([02468][048]|[13579][26])[0][2]([0][1-9]|[12][\\d])([A-Z0-9]{3}))|" +
    "(([A-ZÑ&]{4})([0-9]{2})[0][2]([0][1-9]|[1][0-9]|[2][0-8])([A-Z0-9]{3}))$";

  if (rfc.match(rgxFisica)) == null{
      return True
  } else {
      return False
  }
}

const foo = validateRFC("XXXX901212UDX");
//console.log(foo);

An example of a TRUE regex match:

function validateRFC(rfc) {
  const rgxFisica = "^(([A-ZÑ&]{4})([0-9]{2})([0][13578]|[1][02])(([0][1-9]|[12][\\d])|[3][01])([A-Z0-9]{3}))|" +
    "(([A-ZÑ&]{4})([0-9]{2})([0][13456789]|[1][012])(([0][1-9]|[12][\\d])|[3][0])([A-Z0-9]{3}))|" +
    "(([A-ZÑ&]{4})([02468][048]|[13579][26])[0][2]([0][1-9]|[12][\\d])([A-Z0-9]{3}))|" +
    "(([A-ZÑ&]{4})([0-9]{2})[0][2]([0][1-9]|[1][0-9]|[2][0-8])([A-Z0-9]{3}))$";

 console.log(rfc.match(rgxFisica));

  if (rfc.match(rgxFisica) == null){
      return true
  } else{
      return false
  }
}

const foo = validateRFC("hello 0670 world");
console.log(foo);