and so this must pass:

454555, 939999 , 019999    ,727663

its for a user entering 6 digit invoice numbers. it should fail if a number is 5 or 7 digit and not 6. so 1234567, 123456 should fail, as one set is more than 6 numbers.

So far I have :

[0-9]{6}(\s*,*,\s*[0-9]{6})*

which only draw back is that it accepts 7 or more digit numbers. cant figure out if its even possible at this point to do both, test for 6 digits separated by a ma and one or more space, and all the digits have to be only 6 digits and fail if one is not.

any help appreciated. regular expressions are not my forte.

thanks

Norman

and so this must pass:

454555, 939999 , 019999    ,727663

its for a user entering 6 digit invoice numbers. it should fail if a number is 5 or 7 digit and not 6. so 1234567, 123456 should fail, as one set is more than 6 numbers.

So far I have :

[0-9]{6}(\s*,*,\s*[0-9]{6})*

which only draw back is that it accepts 7 or more digit numbers. cant figure out if its even possible at this point to do both, test for 6 digits separated by a ma and one or more space, and all the digits have to be only 6 digits and fail if one is not.

any help appreciated. regular expressions are not my forte.

thanks

Norman

• javascript
• regex

• it should also pass if just one number is entered: 333222 – Norman Bird Commented Aug 30, 2013 at 15:17
• Did you tried adding anchors - ^ and $ at the ends? – Rohit Jain Commented Aug 30, 2013 at 15:17
• @mvw That works just fine; I remend that you delete your misleading ment. – Phrogz Commented Aug 30, 2013 at 15:32
• You have ,* in your regex; do you really mean to allow 123456,,,,,123456 as valid? How about input with only whitespace, e.g. 123456 123456; is that valid? – Phrogz Commented Aug 30, 2013 at 15:33

7 Answers 7

You can write it using regex like the function below.

const isPassword = (password: string) => /^\d{6}$/gm.test(password);

And here is an example test file below.

  test('should recognize a valid password', () => {
    expect(isPassword('123456')).toBe(true);
    expect(isPassword('000000')).toBe(true);
  });

  test('should recognize an invalid password', () => {
    expect(isPassword('asdasda1234')).toBe(false);
    expect(isPassword('1234567')).toBe(false);
    expect(isPassword('a123456a')).toBe(false);
    expect(isPassword('11.11.11')).toBe(false);
    expect(isPassword('aaaaaa')).toBe(false);
    expect(isPassword('eeeeee')).toBe(false);
    expect(isPassword('......')).toBe(false);
    expect(isPassword('werwerwerwr')).toBe(false);
  });

In order to validate the full string you can use this regex.

^(\s*\d{6}\s*)(,\s*\d{6}\s*)*,?\s*$

It works with six digits only, and you have to enter at least one 6 digit number. It also works if you have a trailing ma with whitespaces.

It's accepting more than six digit numbers because you're not anchoring the text, and for some odd reason you're optionally repeating the ma. Try something like this:

^[0-9]{6}(?:\s*,\s*[0-9]{6})*$

Also note that [0-9] is equivalent to \d, so this can be rewritten more concisely as:

^\d{6}(?:\s*,\s*\d{6})*$

Your regex does not match 7 digits in a row, but it also doesn't enforce that it matches the whole string. It just has to match some substring in the string, so it would also match each of these:

"1234512345612345612345"
"NaNaNaN  123456,    123456 BOOO!"
"!@#$%^&*({123456})*&^%$#@!"

Just add the start of string (^) and end of string ($) anchors to enforce that the whole string matches and it will work correctly:

^[0-9]{6}(\s*,*,\s*[0-9]{6})*$

Also note that ,*, could be shortened to ,+, and if you only want one ma in a row, just use ,, not ,* or ,+.

You can also replace [0-9] with \d:

^\d{6}(\s*,\s*\d{6})*$

Using only regex:

var maSeparatedSixDigits = /^(?:\d{6}\s*,\s*)*\d{6}$/;
if (myInput.test(maSeparatedSixDigits)) console.log( "Is good!" );

This says:

• ^ - Starting at the beginning of the string
• (?:…)* - Find zero or more of the following:
  • \d{6} - six digits
  • \s* - maybe some whitespace
  • , - a literal ma
  • \s* - maybe some whitespace
• \d{6} - Followed by six digits
• $ - Followed by the end of the string

Alternatively:

var maSeparatedSixDigits = /^\s*\d{6}(?:\s*,\s*\d{6})*\s*$/;

I leave it as an exercise to you to decipher what's different about this.

Using JavaScript + regex:

function isOnlyCommaSeparatedSixDigitNumbers( str ){
  var parts = srt.split(/\s*,\s*/);
  for (var i=parts.length;i--;){
    // Ensure that each part is exactly six digit characters
    if (! /^\d{6}$/.test(parts[i])) return false;
  }
  return true;
}

I see a lot of plication here. Sounds to me like what you want is pretty simple:

/^(\d{6},)*\d{6}$/

Then we account for whitespace:

/^\s*(\d{6}\s*,\s*)*\d{6}\s*$/

But as others have noted, this is actually quite simple in JavaScript without using regex:

function check(input) {
    var parts = input.split(',');
    for (var i = 0, n = parts.length; i < n; i++) {
        if (isNaN(+parts[i].trim())) {
            return false;
        }
    }
    return true;
}

Tested in the Chrome JavaScript console.

There isn;t any real need for a regexp. Limit the input to only 6 characters, only accept numbers and ensure that the input has 6 digits (not show here). So you would need:

HTML

<input type='text' name='invoice' size='10' maxlength='6' value='' onkeypress='evNumersOnly(event);'>

JavaScript

<script>
function evNumbersOnly( evt ) {
  //---  only accepts numbers
  //--- this handles inpatabilities between browsers
  var theEvent = evt || window.event;
  //--- this handles inpatabilities between browsers
  var key = theEvent.keyCode || theEvent.which;
  //--- convert key number to a letter
  key = String.fromCharCode( key );
  var regex = /[0-9]/;      // Allowable characters 0-9.+-,
  if( !regex.test(key) ) {
    theEvent.returnValue = false;
    //--- this prevents the character from being displayed
    if(theEvent.preventDefault) theEvent.preventDefault();
  }
}
</script>