Regex values inside curly brackets

Question

I have a random variable lets call it R inside a curly brackets like so:

{R}

I have tried to regex it with this:

{(.*?)//}

I then have this error

"Caused by: java.util.regex.PatternSyntaxException: 
Syntax error U_REGEX_RULE_SYNTAX near index 1:"

Indicator targeting {(.*?)} "("

I tried doing it without the brackets same error. This time indicator targets "."

Could someone help me find an alternate solution to regex items inside a curly brackets?

score 5 · Accepted Answer · answered May 31 '12 at 15:41

5

Try escaping curly braces:

String regex = "\\{(.*?)\\}";

answered May 31 '12 at 15:41

Mikita Belahlazau

score 3 · Answer 2 · answered May 31 '12 at 15:41

3

Curly brackets are used in regexp to define specific sequence repetition.

you have to escape them in the regexp.

\{(.*?)\}

should work better

answered May 31 '12 at 15:41

dweeves

score 1 · Answer 3 · answered May 31 '12 at 15:42

1

Escape the {}s:

String regStr = "\\{.\\}";

I've found this interactive regex testing page useful in refining Java regular expressions.

answered May 31 '12 at 15:42

theglauber

score 0 · Answer 4 · answered May 31 '12 at 15:41

0

Not entirely clear what you are trying to do by

\{.*\}

should work

answered May 31 '12 at 15:41

MK.

score 0 · Answer 5 · edited May 23 '17 at 10:34

0

You can escape special characters using a backslash \. See What special characters must be escaped in regular expressions? for more information (although there is no general rule). Try escaping the curly braces {} and slashes //.

edited May 23 '17 at 10:34

Community

answered May 31 '12 at 15:42

Andrew Cheong

5 Answers5