Extra backslash needed in PHP regexp pattern

Question

When testing an answer for another user\'s question I found something I don\'t understand. The problem was to replace all literal \\t \\n \\r

Answer 1

Its works in perl because you pass that directly as regex pattern /(?:\\[trn])+/

but in php, you need to pass as string, so need extra escaping for backslash itself.

"/(?:\\\\[trn])+/"

The regex \ to match a single backslash would become '/\\\\/' as a PHP preg string

Answer 2

You need 4 backslashes to represent 1 in regex because:

• 2 backslashes are used for unescaping in a string ("\\\\" -> \\)
• 1 backslash is used for unescaping in the regex engine (\\ -> \)

From the PHP doc,

escaping any other character will result in the backslash being printed too¹

Hence for \\\[,

• 1 backslash is used for unescaping the \, one stay because \[ is invalid ("\\\[" -> \\[)
• 1 backslash is used for unescaping in the regex engine (\\[ -> \[)

Yes it works, but not a good practice.

Answer 3

The regular expression is just /(?:\\[trn])+/. But since you need to escape the backslashes in string declarations as well, each backslash must be expressed with \\:

"/(?:\\\\[trn])+/"
'/(?:\\\\[trn])+/'

Just three backspaces do also work because PHP doesn’t know the escape sequence \[ and ignores it. So \\ will become \ but \[ will stay \[.

Answer 4

Use str_replace!

$code = str_replace(array("\t","\n","\r"),'',$code);

Should do the trick