Regular Expression Repeating Pattern separated by comma

Question

I have a regular expression of the following:

.regex(/^(?!^[0-9]*$)(?!^[a-zA-Z]*$)^([a-zA-Z0-9]{4})$/)

Must be exactly 4 characters Must contain at least 1 numeric and 1 alpha

Although I rarely do regular expression, this was relatively easy. I now have a new requirement that I have tried to implement, but cannot get right.

New requirement: Be able to have a comma separated list of the same type of input as before. Cannot end with a comma. Each item must be valid per the rules above (4 characters, at least on numeric, at least one alpha)

Valid:  123F,U6Y7,OOO8
Invalid:  Q2R4,
Invalid:  Q2R4,1234
Invalid:  Q2R4,ABCD
Invalid:  Q2R4,N6

I very much appreciate your help! Thanks!

You can try this site - http://txt2re.com/ It will help you build the expression from scratch. — Vishnu Prasad Kallummel, Jan 07 '14 at 04:38

score 10 · Answer 1 · answered Jan 07 '14 at 07:08

Some of the other answers are repeating the lookahead assertions. That's not necessary.

Here's a regular expression that matches a comma-separated sequence of atoms, where each atom is four alphanumeric characters:

^[A-Z0-9]{4}(?:,[A-Z0-9]{4})*$

Of course, that's not quite what you want. You don't want atoms that are all alphabetic. Here's a negative lookahead assertion that prevents matching such an atom anywhere in the text:

(?!.*[A-Z]{4})

And you don't want atoms that are all numeric either:

(?!.*[0-9]{4})

Putting it all together:

^(?!.*[A-Z]{4})(?!.*[0-9]{4})[A-Z0-9]{4}(?:,[A-Z0-9]{4})*$

score 3 · Answer 2 · answered Jan 07 '14 at 05:04

This regex should work:

^(?:(?![0-9]*,)(?![a-zA-Z]*,)[a-zA-Z0-9]{4},)*(?![0-9]*$)(?![a-zA-Z]*$)[a-zA-Z0-9]{4}$

Or you can shorten [0-9] to \d. It doesn't really matter, though, since the expanded code can be more readable.

^(?:(?!\d*,)(?![a-zA-Z]*,)[a-zA-Z\d]{4},)*(?!\d*$)(?![a-zA-Z]*$)[a-zA-Z\d]{4}$

You can see how your regex is transformed to make this regex, when I line it up like below:

^
(?:
  (?!\d*,)(?![a-zA-Z]*,)[a-zA-Z\d]{4},
)*
  (?!\d*$)(?![a-zA-Z]*$)[a-zA-Z\d]{4}
$

Caio Oliveira · Answer 3 · 2014-01-07T05:25:53.020

2

Try using this regex (regex test):

.regex(/^(?:(?![0-9]{4}|[a-zA-Z]{4})[a-zA-Z0-9]{4})(?:(?:\b\,)(?![0-9]{4}|[a-zA-Z]{4})[a-zA-Z0-9]{4})*$/)

Explanation:

(?:\b\,) -> Match with a , at the beginning of the string only if its preceded by a word boundary
(?:(?![0-9]{4}|[a-zA-Z]{4})[a-zA-Z0-9]{4}) -> Match a string with letter and digits only if dont have 4 digits ow 4 letters

edited Jan 07 '14 at 05:25

answered Jan 07 '14 at 04:52

Caio Oliveira

1,243
13
22

you might be able to clean it up with some \w and \S and \D tags – Zach Leighton Jan 07 '14 at 05:02
@ZachLeighton: \w contains `_`, `\S` contains a bunch of space characters that may differ between engines. And it is not nice to introduce imprecision into the regex. – nhahtdh Jan 07 '14 at 05:05
Actually I think \w and \d are less effective because it match with some characters that isn't in our alphabet. See this [question](http://stackoverflow.com/questions/16621738/d-is-less-efficient-than-0-9) to understand what I'm talking about – Caio Oliveira Jan 07 '14 at 05:06
\S is any nonwhitespace and you could remove _ from \w – Zach Leighton Jan 07 '14 at 05:06
2

@ZachLeighton: There is no need to shorten the regex if it doesn't make the regex easier to read. – nhahtdh Jan 07 '14 at 05:08
If it is a comma separated list, you have to have comma separating `AAA2` from `BBB3` '-' – Caio Oliveira Jan 07 '14 at 05:16
I mean it is an invalid case, accepted by your regex. It fails the requirement of 4 characters per item. – nhahtdh Jan 07 '14 at 05:18
forget, was looking in the old regex – Caio Oliveira Jan 07 '14 at 05:21

justhalf · Answer 4 · 2014-01-07T07:15:59.550

I want to highlight the advantage of my answer as comparison to other answers as follows:

My answer intuitively encodes the requirement "at least one alpha" and "at least one numeric".
Other answers make use of the fact that considering the universal set of [a-zA-Z0-9], which after some thoughts, those two requirements translate into "not all alpha and not all digits".
In my answer, you can change the universal set to anything you want (even with a comma!), and don't need to change the regex for the other requirements.

My answer:

^(?=.{0,3}\d)(?=.{0,3}[a-zA-Z])[a-zA-Z0-9]{4}(,(?=.{0,3}\d)(?=.{0,3}[a-zA-Z])[a-zA-Z0-9]{4})*$

Regular expression visualization

Debuggex Demo

Explained as:

^
  (?=.{0,3}\d)(?=.{0,3}[a-zA-Z])[a-zA-Z0-9]{4}   # Main part
  (,
    (?=.{0,3}\d)(?=.{0,3}[a-zA-Z])[a-zA-Z0-9]{4}
  )*
$

The "main part" matches any four characters composed by alphanumerics ([a-zA-Z0-9]{4}), containing at least one digit ((?=.{0,3}\d)) and at least one letter ((?=.{0,3}[a-zA-Z])). Then just repeat that pattern preceded with a comma.

Regular Expression Repeating Pattern separated by comma

4 Answers4