Restrict part of pattern based on first character or previous occurance of character

Question

I have the following regexp:

/^(?:(?:>|<)?[a-zA-Z]+(?:(?:\+|-)\d*\.?\d*(?:em)?)?)(?:<[a-zA-Z]+(?:(?:\+|-)\d*\.?\d*(?:em)?)?)?$/

Which you can think about like this:

^
  (?:
    (?:>|<)?
    [a-zA-Z]+
    (?:(?:\+|-)\d*\.?\d*(?:em)?)?
  )
  (?:
    <
    [a-zA-Z]+
    (?:(?:\+|-)\d*\.?\d*(?:em)?)?
  )?
$

It is effectively the same pattern repeated once or twice with a small difference. The core of each pattern is one or more letter [a-zA-Z] followed by an optional minus or plus and a numeric value possibly followed by em. The first instance can start with either < or > and the second instance can only start with <.

So the following are all valid:

  `alpha`,
  `alphaBravo`,
  `alphaBravoCharlie`,
  `>alpha`,
  `<alpha`,
  `>alpha+10`,
  `<alpha+10`,
  `>alpha+1.5`,
  `<alpha+1.5`,
  `>alpha-10`,
  `>alpha-10`,
  `>alpha-1.5`,
  `>alpha-1.5`,
  `>alpha+10em`,
  `<alpha+10em`,
  `>alpha+1.5em`,
  `<alpha+1.5em`,
  `>alpha-1.5em`,
  `>alpha-1.5em`,
  `alpha-50em<delta-100em`,
  `alpha-50em<delta+100em`,
  `>alpha-50em<delta+100em`,

My problem is that if the first instance starts with a < then the second instance shouldn't be allowed, so the following should be invalid:

<alpha<bravo

Is it possible to add this restriction to the regexp?

The two approaches I can think of are:

Check the first character and make the second instance invalid if it is <
Check if < has already ocurred in the string (or if < occurs again in the string) and if so, make the second instance invalid.

However I'm not sure how to implement either of these approaches here.

I removed extra `?` in extended version which made `<` optional and my previous solution was based on it. It seems your first regex is the correct one. — revo, Jun 03 '18 at 18:32

revo · Accepted Answer · 2018-06-03T18:09:30.993

3

You could use a very early negative lookahead right after caret ^:

(?!<[^<\s]*<)

Live demo

You also don't need to use alternations to match a single character at a time i.e. (?:>|<) should be [<>] or (?:\+|-) should be [+-].

Extended mode:

^
  (?!<[^<\s]*<) # We have this extra one
  (?:
    [<>]?
    [a-zA-Z]+
    (?:[-+]\d+(?:\.\d+)?(?:em)?)?
  )
  (?:
    <
    [a-zA-Z]+
    (?:[-+]\d+(?:\.\d+)?(?:em)?)?
  )?
$

In a line:

^(?!<[^<\s]*<)(?:[<>]?[a-zA-Z]+(?:[-+]\d+(?:\.\d+)?(?:em)?)?)(?:<[a-zA-Z]+(?:[-+]\d+(?:\.\d+)?(?:em)?)?)?$

edited Jun 03 '18 at 18:09

answered Jun 03 '18 at 16:35

revo

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2

The second approach does not seem to work in JS engines due to the back-reference: https://regex101.com/r/NkR12Z/3. – JJWesterkamp Jun 03 '18 at 16:42
Good point. I removed the second approach. It seems a non-captured capturing group in JS is initialized as a zero-length string in its back-reference. – revo Jun 03 '18 at 16:58
This definitely solves the original problem but now allows: `alpha+100px` which the original doesn't. – Undistraction Jun 03 '18 at 18:00
3

You can use the second approach, if you capture the first character (whatever it is) in a lookahead. https://regex101.com/r/NkR12Z/4 . This way the capture is never empty. – Casimir et Hippolyte Jun 03 '18 at 18:02
@Undistraction In your own regex second `<` is not optional but in your extended version it is and I worked with extended version. So it produces the same result I think. I updated now and made it mandatory. – revo Jun 03 '18 at 18:05
@CasimiretHippolyte Interesting. Is this behavior documented somewhere? I don't remember if I had faced with it. – revo Jun 03 '18 at 18:11
@revo: not really, `(?!\1)` when the first capture group is empty (or in an unused branch) gives `(?!)` that is an always failing subpattern. – Casimir et Hippolyte Jun 03 '18 at 18:16
@CasimiretHippolyte Right. A mistake I'm still in is thinking about an optional `<`. OP's both Regular Expressions define this character differently. – revo Jun 03 '18 at 18:30
@revo For a bonus round: The above works perfectly, but if I need to add a second character that should disallow the second instance (so if either `<` or `@` start the string the second instance should be disallowed), can I simplify this: `^(?!<[^<\s]*<)(?!@[^@\s]*<)(?:[@<>]?[a-zA-Z]+(?:(?:[+-])\d*\.?\d*(?:em)?)?)(?:<[a-zA-Z]+(?:(?:\+|-)\d*\.?\d*(?:em)?)?)?$` (I'm repeating your initial negative lookahead but I'm wondering if the two negative lookaheads can be combined)? – Undistraction Jun 04 '18 at 09:29
1

@Undistraction Yes, you only need one negative lookahead `(?![<@][^<\s]*<)` – revo Jun 04 '18 at 09:32
@revo Thank you. I think I finally understand negative lookaheads. – Undistraction Jun 04 '18 at 09:35
1

@Undistraction You're welcome. You may also want to have a look at [Reference - What does this regex mean?](https://stackoverflow.com/questions/22937618/reference-what-does-this-regex-mean/22944075) for a better understanding of regular expressions. – revo Jun 04 '18 at 09:37

Ωmega · Answer 2 · 2018-06-03T17:15:46.240

2

Just replace (?:(?:>|<)? with (?:(?:>|<(?!.*<))? to get desired results.

Test it here.

If you want to extend this feature from < character to > character as well, you can replace same part of the pattern (?:(?:>|<)? with (?:([<>])(?!.*\1))? and replace <? with [<>]? in the second part of your pattern.

Test it here.

edited Jun 03 '18 at 17:15

answered Jun 03 '18 at 16:54

Ωmega

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I know the answer now, but to make the answer more useful to others, please add an explaination of what `(?!.*<)` brings to the table? – Undistraction Jun 03 '18 at 18:05
@Undistraction - `(?!.*<)` is a negative lookahead (see https://www.regular-expressions.info/lookaround.html), saying that there is no `<` anywhere ahead, regardless of its position in the remaining string. – Ωmega Jun 04 '18 at 19:10

Restrict part of pattern based on first character or previous occurance of character

2 Answers2