ID: 41739 Updated by: [EMAIL PROTECTED] Reported By: trex0003 at umn dot edu -Status: Open +Status: Bogus Bug Type: Documentation problem Operating System: CentOS PHP Version: Irrelevant New Comment:
"if you try to escape any other character, the backslash will be printed too" in strings documentation. Previous Comments: ------------------------------------------------------------------------ [2007-08-17 17:16:37] trex0003 at umn dot edu Please explain the comments added to this report and how it was determined to be "bogus". The comments added by [EMAIL PROTECTED] to the bug do nothing more than reiterate the use of a single, double and quadruple backslash. My question is about why a triple backslash is necessary in this instance. The case of four backslashes in PHP makes sense -- the first and third backslashes escape the second and fourth. But when you have three backslashes in a row, doesn't the third backslash act as an escape character to whatever follows? In my example of: $needle = "/(?<!\\\)'/"; I would expect that the first backslash would escape the second, and the third would escape the right parenthesis, but that's apparently not what happens, because this particular code WORKS. ------------------------------------------------------------------------ [2007-08-16 14:21:11] [EMAIL PROTECTED] \ character \\ PCRE representation of this character '\\\\' PHP string to write this representation. ------------------------------------------------------------------------ [2007-06-19 17:34:33] trex0003 at umn dot edu Description: ------------ Documentation of how PCRE regex behaves with escaped backslashes is confusing as written in current documentation: http://php.planetmirror.com/manual/en/reference.pcre.pattern.syntax.php Perhaps this just means an expansion of the note: "Note: Single and double quoted PHP strings have special meaning of backslash. Thus if \ has to be matched with a regular expression \\, then "\\\\" or '\\\\' must be used in PHP code." While trying to use preg_split to break a string at non-escaped string quotes, I tried using a negative lookbehind assertion (see code below). With all other escaped characters in the lookbehind assertion, you only need one backwards slash before the escaped character (e.g., "(?<!\s)" works); but if you want to escape a backslash, you need to escape it twice (e.g., "(?<!\\\)" works but "(?<!\\)" does not). Based on the documentation's statement 'In particular, if you want to match a backslash, you write "\\".' it seems that (?<!\\) should work, and the extra information in the Note is too terse to explain why. Perhaps this is actually a bug, but I can't tell if it's normal behavior that could be clarified more in documentation. The closest I could find for bug reports to this issue is here: http://bugs.php.net/bug.php?id=22315 ...but this only implies the difference in behavior without explaining why. Reproduce code: --------------- $string = "This sentence is not quoted. 'But this one is, and it contains \'escaped quotes\' within!'"; $needle = "/(?<!\\)'/"; $results = preg_split($needle, $string); print_r($results); Expected result: ---------------- Array ( [0] => This sentence is not quoted. [1] => But this one is, and it contains \'escaped quotes\' within! [2] => ) // Adding a third backslash to the $needle gives this result. Actual result: -------------- Warning: preg_split() [function.preg-split]: Compilation failed: missing ) at offset 7 in /data/domains/lawweb3.law.umn.edu/public/php/regex.php on line 16 ------------------------------------------------------------------------ -- Edit this bug report at http://bugs.php.net/?id=41739&edit=1
