Java getting substring from square brackets

Question

I am trying to get a substring from square brackets in a string, I tried multiple regexes but nothing works.

I tried multiple regexes and code from multiple stackoverflow posts but nothing works.

String mydata = "some string with [the data i want] inside";
Pattern pattern = Pattern.compile("[(.*?)]");
Matcher matcher = pattern.matcher(mydata);
if (matcher.find())
{
    System.out.println(matcher.group(1));
    output = output.replace("%item%", matcher.group(1));
}

It should return "the data i want" but instead it says that no pattern was found.

`[` must be escaped to match a literal `[` char. See https://ideone.com/8FYZbu — Wiktor Stribiżew, Feb 04 '19 at 15:45
@WiktorStribiżew a closing bracket is automatically escaped if there is no opening bracket before? — Cid, Feb 04 '19 at 15:49
@Cid There is no need to escape `]` as the regex engine knows if it closes the character class or not. Inside a character class in Java regexes though, both `[` and `]` must be escaped. — Wiktor Stribiżew, Feb 04 '19 at 15:53
@WiktorStribiżew nice, something new I've learnt. RegEx engines are really smart :) — Cid, Feb 04 '19 at 15:54

score 0 · Accepted Answer · answered Feb 04 '19 at 15:48

0

You need to escape the [] characters so they are seen as literal square brackets.

Using this as your regex should work: \\[(.*?)\\]

answered Feb 04 '19 at 15:48

Bill

1,407
10
18

Thank you so much! it worked – DA - The Anonymous Feb 04 '19 at 15:57

Java getting substring from square brackets

1 Answers1