java regex char sequence

Question

I have a string with multiple "message" inside it. "message" starts with certain char sequence. I've tried:

String str = 'ab message1ab message2ab message3'
Pattern pattern = Pattern.compile('(?<record>ab\\p{ASCII}+(?!ab))');
Matcher matcher = pattern.matcher(str);
while (matcher.find()) {
    handleMessage(matcher.group('record'))
}

but \p{ASCII}+ greedy eat everything. Symbols a, b can be inside message only their sequence mean start of next message

You may try splitting string on each "ab " or whatever char sequence you have in front of it. — hrust
– hrust, Commented Feb 19, 2018 at 12:25
Try String[] res =str.split("(?!^)\\s*(?=ab)");. If the ab is always at the end of the word, add \\b after ab in the pattern (=> "(?!^)\\s*(?=ab\\b)"). — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Feb 19, 2018 at 12:26
How do you know if that charsequence is not part of the message ? Like "I am absent" ? — AxelH
– AxelH, Commented Feb 19, 2018 at 12:29
@TaherKhorshidi bad idea: this would forbid any a or b in the message. — Maurice Perry
– Maurice Perry, Commented Feb 19, 2018 at 12:33

phatfingers · Accepted Answer · 2018-02-19 13:51:44Z

2

p{ASCII}+ is the greedy regex for one or more ASCII characters, meaning that it will use the longest possible match. But you can use the reluctant quantifier if you want the shortest possible match: p{ASCII}+?. In that case, you should use a positive lookahead assertion.

The regex could become:

Pattern pattern = Pattern.compile("(?<record>ab\\p{ASCII}+?)(?=(ab)|\\z)");

Please note the (ab)|\z to match the last message...

edited Feb 19, 2018 at 13:51

phatfingers

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answered Feb 19, 2018 at 13:33

Serge Ballesta

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1 Comment

17dufa Over a year ago

Thank you. Works perfect

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