CODEWITHSUNDEEP
regexregular-language
Carol.Kar
Carol.Kar

Reputation: 5355

Match middle and end of length x width x height

I am trying to match, the following cases:

1. Get either between or if only one x exists the end

Example:

| Matches/Cases     | Result |
|-------------------|--------|
| 200 x 90 x 14     | 90     |
| 90x200            | 200    |
| 200 x 90x20       | 90     |
| 60,4 x46,5 x 42,6 | 46,5   |
| 90x190,9          | 190,9  |

2. Get if two x exist the final one, and if only one exist no result

Examples:

| Matches/Cases     | Result |
|-------------------|--------|
| 200 x 90 x 14     | 14     |
| 90x200            | -      |
| 200 x 90x20       | 20     |
| 60,4 x46,5 x 42,6 | 42,6   |
| 90x190,9          | -      |

I stuck at getting one specific case! I tried to match with the following regex x\s?((\d+(?:,\d+)?))\s?, but I still get only the last part of the cases like for 90x200 I get 200, but for 200 x 90 x 14 I get 90 x 14.

Any suggestions of two regex that works for case 1 or case 2?

I appreciate your replies!

Upvotes: 1

Views: 691

Answers (2)

revo
revo

Reputation: 48751

I tried to match with the following regex x\s?((\d+(?:,\d+)?))\s?, but I still get only the last part of the cases.

Actually by your own RegEx you are going to capture all digits or floats followed by a x. So it's not only last part but all similar occurrences.

Solution (main regex):

(?: *(\d+(?:,\d+)?) *(?:x|$))

If you want it for case #1 then append quantifier {2}

(?: *(\d+(?:,\d+)?) *(?:x|$)){2}

Live demo

If you want it for case #2 append quantifier {3}

(?: *(\d+(?:,\d+)?) *(?:x|$)){3}

Live demo

m modifier should be set in both cases

Upvotes: 1

Jan
Jan

Reputation: 43169

Just turning the comments into an answer.
For the first case, you could use:

[\d,]+\h*x\h*([\d,]+)(?:\h*x*[\d,]+)?

See a demo on regex101.com.

And the second:

[\d,]+\h*x\h*[\d,]+\h*x\h*([\d,]+)

Another demo on regex101.com.

Hint: Replace \h with either [ ]* or \s* if it is not supported.

Upvotes: 1

Related Questions