PHP refusing this regular expression

Question

So I'm trying to check for match and if match, extract a variable name out of a string. The variable name should be preceded by "$" and cannot be escaped with "\", so for example "$name" should extract "name" and "\$name" or "name" shouldn't match. Heres the command:

$match = preg_match("/^(?<!\\)(\$.*)$/", $potential, $name);

I constructed and tested it using regex101.com and it works there, however, I'm getting an error from PHP saying

"preg_match(): Compilation failed: missing ) at offset 13 in ..."

and I have no clue what its referring to.

Scuzzy · Accepted Answer · 2016-05-16 03:59:11Z

1

My thought is that you will need to escape certain characters to consume the regular expression in PHP

$match = preg_match('/^(?<!\\\\)(\$.*)$/', $potential, $name);

Edit: the backslash is the escape character in both Regex and PHP, you will need to doubly escape the slashes.

edited May 16, 2016 at 3:59

answered May 16, 2016 at 3:40

Scuzzy

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2 Comments

user81993 Over a year ago

right so escape for both regex and for the string format in php?

Scuzzy Over a year ago

Yes precisely, backslash is the escape character for both PHP and Regex

theruss · Accepted Answer · 2016-05-16 03:46:39Z

0

You've escaped a bracket:

preg_match('/^(?<!\\) <----HERE

FYI you can use several other delimiters to make your regex's more readable. Because so often we have slashes and escaped chars, then using '/' makes it hard to read. Consider using '#' or '~' or even '@' to increase readability.

Also reL your online regex tool of choice, it depends on which regular expression implementation (and version) the service uses, as to how accurate your results. I always use rubular.com (Uses PCRE) but for PHP you can use phpliveregex.com

answered May 16, 2016 at 3:46

theruss

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