Regular expression to match single bracket pairs but not double bracket pairs

Question

Is it possible to make a regular expression to match everything within single brackets but ignore double brackets, so for example in:

{foo} {bar} {{baz}}


        

Answer 1

To only match foo and bar without the surrounding braces, you can use

(?<=(?<!\{)\{)[^{}]*(?=\}(?!\}))

if your language supports lookbehind assertions.

Explanation:

(?<=      # Assert that the following can be matched before the current position
 (?<!\{)  #  (only if the preceding character isn't a {)
\{        #  a {
)         # End of lookbehind
[^{}]*    # Match any number of characters except braces
(?=       # Assert that it's possible to match...
 \}       #  a }
 (?!\})   #  (only if there is not another } that follows)
)         # End of lookahead

EDIT: In JavaScript, you don't have lookbehind. In this case you need to use something like this:

var myregexp = /(?:^|[^{])\{([^{}]*)(?=\}(?!\}))/g;
var match = myregexp.exec(subject);
while (match != null) {
    for (var i = 0; i < match.length; i++) {
        // matched text: match[1]
    }
    match = myregexp.exec(subject);
}

Answer 2

In many languages you can use lookaround assertions:

(?<!\{)\{([^}]+)\}(?!\})

Explanation:

• (?<!\{): previous character is not a {
• \{([^}]+)\}: something inside curly braces, e.g. {foo}
• (?!\}): following character is not a }