Issue
I am trying to replace 2.0 to stack,
but the following code replace 2008 to 2.08
Following is my code:
$string = 'The story is inspired by the Operation Batla House that took place in 2008 ';
$tag = '2.0';
$pattern = '/(\s|^)'.($tag).'(?=[^a-z^A-Z])/i';
echo preg_replace($pattern, '2.0', $string);
Solution
Use preg_quote
and make sure you pass the regex delimiter as the second argument:
$string = 'The story is inspired by the Operation Batla House that took place in 2008 ';
$tag = '2.0';
$pattern = '/(\s|^)' . preg_quote($tag, '/') . '(?=[^a-zA-Z])/i';
// ^^^^^^^^^^^^^^^^^^^^^
echo preg_replace($pattern, '2.0', $string);
The string is not modified. See the PHP demo. The regex delimiter here is /
, thus it is passed as the 2nd parameter to preg_quote
.
Note that [^a-z^A-Z]
matches any chars but ASCII letters and ^
since you added the second ^
in the character class. I changed [^a-z^A-Z]
to [^a-zA-Z]
.
Also, the capturing group at the start may be replaced with a single lookbehind, (?<!\S)
, it will make sure your match occurs only at the string start or after a whitespace.
If you expect to also match at the end of the string, replace (?=[^a-zA-Z])
(that requires a char other than a letter immediately to the right of the current location) with (?![a-zA-Z])
(that requires a char other than a letter or end of string immediately to the right of the current location).
So, use
$pattern = '/(?<!\S)' . preg_quote($tag, '/') . '(?![a-zA-Z])/i';
Also, consider using unambiguous word boundaries
$pattern = '/(?<!\w)' . preg_quote($tag, '/') . '(?!\w)/i';
Answered By - Wiktor Stribiżew Answer Checked By - Katrina (PHPFixing Volunteer)
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