Using Regex to with fixed character count and delimiters

Question

So I have a case where i want to get every 5 character code which is delimited by / Problem is the codes itself can contain the delimiter so I can't use something like ([^\/]+)

E.g.

07070 should give me 07070
07070/01010 should give me 07070 01010
0/0/0/BBBBB should give me 0/0/0 BBBBB
AAAAA/BBBBB/CCCCC should give me AAAAA BBBBB CCCCCC

Any ideas how i can use a regex to achieve this?

Your middle examples aren't "delimited" by `/` (at least not *only* `/`). — Scott Hunter, Jan 15 '20 at 15:13
Please clarify your examples with " * " ? Is this a literal * character or something else. How do we know the code is "70707" and not "707**"? — Gem Taylor, Jan 15 '20 at 15:14

Amjad Abdul-Ghani · Answer 1 · 2020-01-15T15:33:23.167

use : (?<yourToken>.{5})\/?

yourToken : is the name of match group you want. while iterating on your code get match group with name yourToken

Note that : 07/01010 will get 07/01 only

in java :

Pattern pattern = Pattern.compile("(?<yourToken>.{5})\\/?");
Matcher matcher = pattern.matcher("AAAAA/BBBBB/CCCCC");
while (matcher.find()){
  System.out.println(matcher.group("yourToken"));
}

check here: https://regex101.com/r/4vBAL9/1

score 0 · Answer 2 · answered Jan 15 '20 at 15:36

0

Since you know your codes are always 5 chars followed by a delimiting / you can just count 5 chars:

(.{5})(\/?)

https://regex101.com/r/paLxPv/1

Your codes will be in \1

answered Jan 15 '20 at 15:36

MonkeyZeus

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score 0 · Answer 3 · answered Jan 15 '20 at 15:55

0

Find: /(?=[^/]{5}(?:/| |$))
Replace: # (a space)

Demo & explanation

answered Jan 15 '20 at 15:55

Toto

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Using Regex to with fixed character count and delimiters

3 Answers3