Regular expression in java (java String)

Question

from this -> contractor:"Hi, this is \"Paul\", how are you?" client:"Hi ...." <-

I want to get just -> Hi, this is \"Paul\", how are you? <-

I need a regular expression in java to do that I try it but I m struggle with the inner quotation (\") is driving me mad.

Thanks for any hint.

Do you have to use regex? You could just strip out the redundant text around the keywords by using `.indexOf()` and `.substring()` — Spectric, Oct 29 '20 at 15:11
So briefly - you need to get the text inside a pair of double quotes `"the text to be matched"`, and this text may contain other double quotes preceded with a backslash `\"`? — Nowhere Man, Oct 29 '20 at 15:20
@AlexRudenko . Yes Alex , I edit the question because editor doesn't skip inner quotations — Xenione, Oct 29 '20 at 15:23

sp00m · Accepted Answer · 2020-10-29T16:11:27.817

1

Java supports lookbehinds, so vanilla regex:

"(.*?(?<!\\))"

Inside a Java string (see https://stackoverflow.com/a/37329801/1225328):

\"(.*?(?<!\\\\))\"

The actual text will be contained inside the first group of each match.

Demo: https://regex101.com/r/8OXujX/2

For example, in Java:

String regex = "\"(.*?(?<!\\\\))\"";
String input = "contractor:\"Hi, this is \\\"Paul\\\", how are you?\" client:\"Hi ....\"";
Pattern pattern = Pattern.compile(regex);
Matcher matcher = pattern.matcher(input);
if (matcher.find()) { // or while (matcher.find()) to iterate through all the matches
    System.out.println(matcher.group(1));
} else {
    System.out.println("No matches");
}

Prints:

Hi, this is \"Paul\", how are you?

edited Oct 29 '20 at 16:11

answered Oct 29 '20 at 15:26

sp00m

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That would include the final "Hi...." though. OP doesn't want that. – Spectric Oct 29 '20 at 15:27
Yeah, first match should be taken I believe, at least in OP's example. – sp00m Oct 29 '20 at 15:28
works fine with online regexp editor but no on java code – Xenione Oct 29 '20 at 15:49

score 1 · Answer 2 · answered Oct 29 '20 at 15:31

1

The regexp should be like this: "(?:\\.|[^"\\])*"

Online demo

It uses non-capturing group ?:, matching any character . or a single character NOT in the list of double quote and backslash.

answered Oct 29 '20 at 15:31

Nowhere Man

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Not the OP's requirements. – Spectric Oct 29 '20 at 15:37

score 0 · Answer 3 · answered Oct 29 '20 at 15:48

var text1 = "contractor:\"Hi, this is \\\"Paul\\\", how are you?\" client:\"Hi ....\" <-";
    var regExWithQuotation = "contractor:(.+\".+\".+) client:";
    Pattern p = Pattern.compile(regExWithQuotation);

    var m = p.matcher(text1);
    ;
    if (m.find()) {
        var res = m.group(1);
        System.out.println(res);
    }

    var regExWithoutQuotation = "contractor:\"(.+\".+\".+)?\" client:";
    p = Pattern.compile(regExWithoutQuotation);
    m = p.matcher(text1);

    if (m.find()) {
        var res = m.group(1);
        System.out.println(res);
    }

Output is:

"Hi, this is "Paul", how are you?"

Hi, this is "Paul", how are you?

score 0 · Answer 4 · answered Oct 29 '20 at 16:12

You can use the regex, (?<=contractor:\").*(?=\" client:)

Description of the regex:

(?<=contractor:\") specifies positive lookbehind for contractor:\"
.* specifies any character
(?=\" client:) specifies positive lookahead for \" client:

In short, anything preceded by contractor:\" and followed by \" client:

Demo:

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

public class Main {
    public static void main(String[] args) {
        String str = "contractor:\"Hi, this is \\\"Paul\\\", how are you?\" client:\"Hi ....\"";
        String regex = "(?<=contractor:\").*(?=\" client:)";
        Pattern pattern = Pattern.compile(regex);
        Matcher matcher = pattern.matcher(str);
        while (matcher.find()) {
            System.out.println(matcher.group());
        }
    }
}

Output:

Hi, this is \"Paul\", how are you?

Regular expression in java (java String)

4 Answers4