Why special characters like = or " break PHP regexp when using \b word boundary?

Question

this is a follow up after reading How to specify "Space or end of string" and "space or start of string"?

From there, it states means to match a word in a phrase. I can even add a few other solutions. But as soon as a = or " is added, it quit working. Why?

i am going to search for stackoverflow and replace it with OK using preg_replace()

preg_replace('/\bstackoverflow\b/', 'OK', $input_line)

input:
1: stackoverflow xxx
2: xxx stackoverflow xxx
3: xxx stackoverflow
result:
1: OK xxx
2: xxx OK xxx
3: xxx OK

now, if i change it to match stackoverflow="", it stops working.

preg_replace('/\bstackoverflow=""\b/', 'OK', $input_line)

input:
1: stackoverflow="" xxx
2: xxx stackoverflow="" xxx
3: xxx stackoverflow=""
result:
1: stackoverflow="" xxx
2: xxx stackoverflow="" xxx
3: xxx stackoverflow=""

the same will happen if i use on my regex: /\bstackoverflow=\b/ or /\bstackoverflow"\b/. I already checked the manual if = or " are special chars, they are not. but i even tried /\bstackoverflow\=\"\"\b/

Why is that?

in that example removing \b will also solve it, but it will also match nostackoverflow=""not which i do not want.

i also tried alternatives to \b such as [ ^] and ( |^). Interestingly [ ^] (space or beginning of line) will not work for beginning of line, only space. But ( |^) will work fine for both.

Answer 1

The problem is your use of \b which is a "word boundary." It's a placeholder for (^\w|\w$|\W\w|\w\W), where \w is a "word" character [A-Za-z0-9_] and \W is the opposite. The problem is that a " doesn't match the "word" characters, so the boundary condition is not met.

Try using a \s instead, which will match any whitespace character.

(?:^|\s)stackoverflow=""(?:\s|$)

Characters inside a class are not interpreted, except for ^ used as a negation operator at the beginning of a class, and - as a range operator. This is why [ ^] wouldn't work for you. It was searching for a literal ^.

$ php -a
Interactive shell

php > $input_line='
php ' stackoverflow="" xxx
php ' xxx stackoverflow="" xxx
php ' xxx stackoverflow=""
php ' ';
php > echo preg_replace('/(?:^|\s)stackoverflow=""(?:\s|$)/', 'OK', $input_line);
OKxxx
xxxOKxxx
xxxOK

https://regex101.com/r/nP2aB8/1

Answer 2

Background

From the regular-expressions.info Word boundaries page:

The metacharacter \b is an anchor like the caret and the dollar sign. It matches at a position that is called a "word boundary". This match is zero-length.

There are three different positions that qualify as word boundaries:
- Before the first character in the string, if the first character is a word character.
- After the last character in the string, if the last character is a word character.
- Between two characters in the string, where one is a word character and the other is not a word character.

A very good explanation from nhahtdh post:

A word boundary \b is equivalent to:

(?:(?<!\w)(?=\w)|(?<=\w)(?!\w))

Which means:

• Right ahead, there is (at least) a character that is a word character, and right behind, we cannot find a word character (either the character is not a word character, or it is the start of the string).

  OR

• Right behind, there is (at least) a character that is a word character, and right ahead, we cannot find a word character (either the character is not a word character, or it is the end of the string).

What's wrong with your regex

The reason why \b is not suitable is because it requires a word/non-word character to appear after/before it which depends on the immediate context on both sides of \b. When you build a regex dynamically, you do not know which one to use, \B or \b. For your case, you could use '/\bstackoverflow=""\B/', but it would require a smart word/non-word boundary appending. However, there is an easier way: use negative lookarounds.

Solution

(?<!\w)stackoverflow=""(?!\w)

See regex demo

The regex contains negative lookarounds instead of word boundaries. The (?<!\w) lookbehind fails the match if there is a word character before stackoverflow="", and (?!\w) lookahead fails the match if stackoverflow="" is followed by a word character.

What a word shorthand character class \w matches depends if you enable the Unicode modifier /u. Without it, a \w matches just [a-zA-Z0-9_]. You can lay further restrictions using the lookarounds.

Demo

PHP demo:

$re = '/(?<!\w)stackoverflow=""(?!\w)/'; 
$str = ",stackoverflow=\"\" xxx\nxxx stackoverflow=\"\" xxx\nxxx stackoverflow=\"\"\nstackoverflow=\"\" xxx"; 
echo preg_replace($re, "NEW=\"\"", $str);

NOTE: If you pass your string as a variable, remember to escape all special characters in it with preg_quote:

$re = '/(?<!\w)' . preg_quote($keyword, '/') . '(?!\w)/'; 

Here, notice the second argument to preg_quote, which is /, the regex delimiter char.