Regex designing for special string

Question

Unable to implement regex for string in line starts with ? and ends with ; appended with single quotes or double quotes.

for example:

?abcdef;
'?abcdef;'
"?abcdef;"

I tried a lot, like this "^\\?([^;]+)\\;$" but it did not work.

Test code snippet:

import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class RegexTest {

    public static void main(final String[] args) {

        final String text = "This is param-start ?abcdef; param-end";

        final String patternString = "(['\"]?)\\?.*;\\1";

        final Pattern pattern = Pattern.compile(patternString, Pattern.CASE_INSENSITIVE);
        final Matcher matcher = pattern.matcher(text);

        System.out.println("lookingAt = " + matcher.lookingAt());
        System.out.println("matches   = " + matcher.matches());
    }
}

Maybe `^(['"]?)\?.*;\1$` will do? See [this regex demo](https://regex101.com/r/psUs84/2). Java: `s.matches("(['\"]?)\\?.*;\\1")`. — Wiktor Stribiżew, Apr 18 '19 at 20:14
Please add a code snippet showing how you test the regex to the question. — Wiktor Stribiżew, Apr 18 '19 at 20:20
You seem to need a partial match, so use it with `.find()`, `while (matcher.find()) { System.out.println("matches = " + matcher.group()); }`, and the regex should be `String patternString = "(['\"]?)\\?.*?;\\1"`. See [this Java demo](https://ideone.com/OKe9kW). — Wiktor Stribiżew, Apr 18 '19 at 20:34
What do you expect to get? If your string has line breaks, use the `"(?s)(['\"]?)\\?.*?;\\1"` regex, but it is not clear what you call "not working" and how to make it "working" for you. — Wiktor Stribiżew, Apr 18 '19 at 20:45
it is working Wiktor Thanks a lot, I am trying one more regex will take a help from u if not able to design thanks — Girish, Apr 18 '19 at 20:54

anubhava · Answer 1 · 2019-04-18T21:00:28.950

4

You may use this regex:

(["']?)\?.*;\1

For Java use:

final String pattern = "(['\"]?)\\?.*;\\1";

RegEx Demo

(["']?) matches an optional ' or " and captures in group #1
\1 is back-reference for the same value as in group #1

edited Apr 18 '19 at 21:00

answered Apr 18 '19 at 20:14

anubhava

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You need to call `matcher.find()` – anubhava Apr 18 '19 at 20:32
1

This works great +1. @Girish [Check this demo](https://rextester.com/MWO62634) Prints `true` – Pushpesh Kumar Rajwanshi Apr 18 '19 at 20:57

score 0 · Answer 2 · answered Apr 18 '19 at 20:18

Your description stated "starts with ? and ends with ;", but your sample string does not end with a semicolon - it's ending with wtih a double quote.

If you want to allow the string to start or end with either a single quote or double quote, your match should be: ^['"]?\?[^;]+;\['"]?$

Wiktor Stribiżew · Accepted Answer · 2019-04-18T20:59:33.543

You may use

final String text = "This is param-start ?abcdef; param-end";
final String patternString = "(['\"]?)\\?.*?;\\1";
final Pattern pattern = Pattern.compile(patternString, Pattern.DOTALL);
final Matcher matcher = pattern.matcher(text);
while (matcher.find()) {
    System.out.println("Match found: " + matcher.group());
}
// => Match found: = ?abcdef;

See the Java demo and the regex demo. Regulex graph:

Notes:

while (matcher.find()) - iterates through all potential matches in the string
matcher.group() - accesses the match value.

Regex

(['"]?) - Capturing group 1: either ' or "
\? - a ? char
.*? - any 0+ chars, as few as possible (the pattern is compiled with Pattern.DOTALL, so it will match across line breaks, too)
; - a semi-colon
\1 - the same value as captured in Group 1.

`"matches = "` is a bit confusing – Andreas Apr 18 '19 at 20:58 — Andreas, Apr 18 '19 at 20:58

Regex designing for special string

3 Answers3