Doc #64516 [Ver->Csd]: preg_quote() doesn't work to escape the replacement
Edit report at https://bugs.php.net/bug.php?id=64516&edit=1
ID: 64516
Updated by: cmb@php.net
Reported by: php at richardneill dot org
Summary: preg_quote() doesn't work to escape the replacement
-Status: Verified
+Status: Closed
Type: Documentation Problem
Package: PCRE related
Operating System: all
PHP Version: 5.4.13
Assigned To: cmb
Block user comment: N
Private report: N
Previous Comments:
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[2016-08-20 13:02:11] cmb@php.net
Automatic comment from SVN on behalf of cmb
Revision: http://svn.php.net/viewvc/?view=revision&revision=339916
Log: Fix #64516: preg_quote() doesn't work to escape the replacement
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[2016-08-20 12:41:14] cmb@php.net
In my opinion, there is no need for preg_quote_replacement() to be
in the core, because it appears to be rarely needed, and can
otherwise easily implemented in userland:
function preg_quote_replacement($str) {
return addcslashes($str, '\\$');
}
See also <https://3v4l.org/ZYue6>.
I agree, that the documentation should be improved to point out
that preg_quote() shouldn't be used on replacement strings.
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[2013-03-25 22:30:06] php at richardneill dot org
Description:
------------
preg_quote() is great for escaping a user-supplied search string.
There is no matching function to escape a user-supplied replacement string.
My wish is for preg_quote() to have an extra flag "IS_REPLACEMENT" or to have a
function preg_quote_replacement(), which would escape only backslash and dollar
signs.
Example below.
Might I also suggest this is a documentation bug, in that there is no explanation
of the correct way to work around this?
Test script:
---------------
$text = 'Invoice: You owe me *#5* !';
$user_search = "*#5*";
$user_replace = "*$5*";
$search = preg_quote($user_search, "/");
$replace_bad1 = $user_replace;
$replace_bad2 = preg_quote($user_replace);
$new_text1 = preg_replace("/$search/", "$replace_bad1", $text);
$new_text2 = preg_replace("/$search/", "$replace_bad2", $text);
echo "Input: $text\nNew1: $new_text1\nNew2: $new_text2\n";
Expected result:
----------------
I want to return the string:
Invoice: You owe me *$5* !
Actual result:
--------------
Input: Invoice: You owe me *#5* !
New1: Invoice: You owe me ** !
New2: Invoice: You owe me \*$5\* !
1. the '*' in the search string is not magic, thanks to the normal use of
preg_quote(). This is what I expect.
2. If I replace with a literal, then '$5' becomes the 5th backreference, which is
empty.
3. If I preg_quote the replacement, then we get spurious backslashes before the
'*'s.
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