In LaTeX, the expression \o{a}{b} means the operator 'o' takes two arguments a and b. LaTeX also accepts \o{a}, and in this case treats the second argument as the empty string.
Now I try to match the regex \\\\o\{([\s\S]*?)\}\{([\s\S]*?)\} against the string \o{a}\o{a}{b}. It mistakes the whole string to be a match when it isn't. (The correct interpretation of this string is that the substring \o{a}{b} is the only match.) The point is I need to know how to tell PHP to recognise that if there is something else than { following the first }, then it is not a match.
How should I do that?
Edit: Arguments of an operator are allowed to contain the symbols \, { and }. But in this case the reason the whole string is not a match is because the curly brackets in a}\o{a do not conform to LaTeX rules (e.g. { must come before }), so that a}\o{a cannot be an argument of an operator...
Edit2: On the other hand, \o{{a}}{b} should be a match as {a} is a valid argument.
I suggest something like this:
$s = '\\o{a}\\o{a}{b}';
echo "$s\n"; # Check string
preg_match('~\\\o(\{(?>[^{}\\\]++|(?1)|\\\.)+\}){2}~', $s, $match);
print_r($match);
The regex:
[^{}\\\] and \\\.) to avoid taking literal braces for syntactical braces.\\\o # Matches \o
( # Recursive group to be
\{ # Matches {
(?> # Begin atomic group (just a group that makes the regex faster)
[^{}\\\]++ # Any characteres except braces and backslash
|
(?1) # Or recurse the outer group
|
\\\. # Or match an escaped character
)+ # As many times as necessary
\} # Closing brace
){2} # Repeat twice
The problem with your current regex is that once this part matched \\\\o\{([\s\S]*?), it will try to look for the next \} that is coming, and there, it matters not whether you are using a lazy quantifier or a greedy one. You need to somehow prevent it to match } before the actual \} comes in the regex.
That's why you have to use [^{}] and since you actually can have nested braces inside, that's the ideal situation to use recursion.