CODEWITHSUNDEEP
phpregexescaping
user3021621
user3021621

Reputation: 341

regular expressions escaping rules- Perl-compatible regular expressions

why in the following code, in order to match the string, then we have to escape the '$' with two backslashes and not one?

<?php
$text = "$3.99";
preg_match_all("/\\$\d+\.\d{2}/", $text, $matches) ;
var_dump($matches) ;
?>

output: array (size=1)
  0 => 
    array (size=1)
      0 => string '$3.99' (length=5)

what is the matching rule for the pattern: "/\$\d+.\d{2}/" (one backslash)

Upvotes: 2

Views: 83

Answers (2)

user557597
user557597

Reputation:

I think php has a few options for depicting the regex string.
At the first level, they seem to want strings of some sort because
they don't have quote like operators like Perl, at least I don't think so.

Level 2, they move on to the regex delimiter (stay away from double quotes as delimiter).

Level 3, the raw regex is left bare after 1 & 2 are done.

So usually, you have reverse the process 3 - 2 -1, and present that to the php source code.

A note - Regex delimiters are a tricky business. In level 2, it is possible that you could
choose a delimiter that is un-escapable in the regular expression. In your case '$' would
not be a viable delimiter.

Some possibilities might help you... 

\$\d+\.\d{2}                 # raw regex

/\$\d+\.\d{2}/               # no quote, using '/' for delimeter
'/\$\d+\.\d{2}/'             # single quotes        ""
"/\\$\\d+\\.\\d{2}/"         # double quotes        ""

~\$\d+\.\d{2}~                # no quote, using '~' for delimeter
'~\$\d+\.\d{2}~'              # single quotes        ""
"~\\$\\d+\\.\\d{2}~"          # double quotes        ""

Upvotes: 0

exussum
exussum

Reputation: 18550

http://php.net/manual/en/language.types.string.php

From the docs

If the string is enclosed in double-quotes ("), PHP will interpret more escape sequences for special characters:

Then

\\  backslash
\$  dollar sign

So the double backslash is for the string not the regex

A single backslash would result in the $ literal which is then passed to the regex

Upvotes: 2

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