Warning: preg_match(): Compilation failed: unrecognized character after (? or (?-

php, preg-match, regex

Solution

For a strange reason, the programmer has escaped the `>` (that is never needed).

Only these characters need to be escaped to obtain a literal character (outside a character class):

( ) ^ $ [ \ | . * + ?

{   # only in these cases: {n} {m,n} {m,}
    # where m and n are integers

+ the pattern delimiter

Most of the time an escaped character that doesn't need to be escaped (or that does not have a special meaning like `\b` `\w` `\d` ...) is simply ignored by the regex engine. But it's not the case here, because `(?>` is a fixed sequence to open an atomic group, and the sequence `(?` is not allowed except for these cases:

- a non capturing group: `(?:...)`

- an atomic group: `(?>...)`

- an inline modifier: `(?i)` `(?-i)`

- a non capturing group with inline modifiers: `(?i:...)` `(?-i:...)`

- a lookaround: `(?=...)` `(?!...)` `(?<=...)` `(?<!...)`

- a branch reset group: `(?|...|...)`

- a conditional test: `(?(condition)...|...)`

- a reference to a subpattern: `(?1)` `(?-1)` `(?R)`

- a definition group: `(?(DEFINE)...)`

- a named subpattern: `(?<John>...)` `(?P<John>...)` `(?'John'...)`

- a reference to a named subpattern: `(?&John)` `(?P>John)`

So `(?\` is not recognized as one of these sequences, this is why you obtain an error.

Perhaps the programmer wanted to write an atomic group `(?>...)` (This is probably the case, because since the group is repeated, using an atomic group reduces the backtracking if the following subpattern fails) or has forgotten the `:` for a non capturing group `(?:\\` as Avinash suggests it, but to obtain a literal backslash you need four backslashes in a double quoted string.

Problem

Original source