I would like to trim all the punctuation and leave only letters or numbers at the beginning and at the end of the string. Any punctuation between letters and numbers should be retained.
This is what I tried from here PHP preg_replace: remove punctuation from beginning and end of string:
$str = '£££2343423 34234238& ';
$new = preg_replace('/^\PL+|\PL\z/', '', $str);
echo $new;
Kindly any recommendations, please?
You can use
$new = preg_replace('/^[^\p{L}0-9]+|[^\p{L}0-9]+\z/u', '', $str);
The regex matches
^[^\p{L}0-9]+ - any one or more chars other than Unicode letters and ASCII digits at the start of string
| - or
[^\p{L}0-9]+\z - any one or more chars other than Unicode letters and ASCII digits at the end of string.
See the PHP demo online and a regex demo.
Related
I am allowing only alpha-numeric, _ & - values in string and removing all other characters. Its working fine but when string size 1 character (does not matter its alphabet or numeric or _ or -), I got empty value instead of single charter.
Here is sample code
$str = 1;
$str = preg_replace('/^[a-zA-Z0-9_-]$/', '', $str);
var_dump($str);
or
$str = 'a';
$str = preg_replace('/^[a-zA-Z0-9_-]$/', '', $str);
var_dump($str);
I have tested this multiple versions of PHP as well
You are removing any chars other than ASCII letters, digits, _ and - anywhere inside the string. You need to remove anchors and convert the positive character class into a negated one:
$str = preg_replace('/[^\w-]+/', '', $str);
See the PHP demo online and a regex demo.
Details
[^ - start of a negated character class
\w - a word char: letter, digit or _
- - a hyphen
] - end of the character class
+ - a quantifier: 1 or more repetitions.
I'm trying to add a special string '|||' after newlines, blankspaces and other characters. I'm doing this because I want to split my text into an array. So I was thinking to do it like this:
$result = preg_replace("/<br>/", "<br>|||", preg_replace("/\s/", " |||", preg_replace("/\r/", "\r|||", preg_replace("/\n/", "\n|||", preg_replace("/’/", "’|||", preg_replace("/'/", "'|||", $text))))));
$result = preg_split("/[|||]+/", $result);
It works with every word but words which contain à char. It is replaced by �.
I'm sure the problem is here because my string $text shows the char à.
Since your pattern deals with a Unicode string, pass the /u modifier.
Also, you do not need so many chained regex replacements, group the first patterns and use a backreference in the replacement.
Use
preg_replace("/(<br>|[\s’'])/u", "$1|||", $text)
Note that \s matches spaces, carriage returns and newlines.
Details:
(<br>|[\s’']) - Group 1 capturing either a
<br> - character sequence
| - or
[\s’'] - a whitespace, ’ or '.
See the PHP demo:
$text = "Voilà. C'est vrai.";
echo preg_replace("/(<br>|[\s’'])/u", "$1|||", $text);
I need a regex to remove all non-alphanumeric and space characters, I have this
$page_title = preg_replace("/[^A-Za-z0-9 ]/", "", $page_title);
but it doesn't remove space characters and replaces some non-alphanumeric characters with numbers.
I need the special characters like puntuation and spaces removed.
If all you want to leave all of the alphanumeric bits you would use this:
(\W)+
Here is some test code:
$original = "Match spaces and {!}#";
echo $original ."<br>";
$altered = preg_replace("/(\W)+/", "", $original);
echo $altered;
Here is the output:
Match spaces and {!}#
Matchspacesand
Here is the explanation:
1st Capturing group: (\W) matches any non-word character [^a-zA-Z0-9_]
Quantifier: + Between one and unlimited times, as many times as possible, giving back as needed [greedy]
I need the special characters like puntuation and spaces removed.
Then use:
$page_title = preg_replace('/[\p{P}\p{Zs}]+/u', "", $page_title);
\p{P} matches any punctuation character
\p{Zs} matches any space character
/u - To support unicode
Try this
preg_replace('/[^[:alnum:]]/', '', $page_title);
[:alnum:] matches alphanumeric characters
Works good for me on Sublime and PHP Regex Tester
$page_title = preg_replace("/[^A-Za-z0-9]/", "", $page_title);
I'm trying to remove all words of less than 3 characters from a string, specifically with RegEx.
The following doesn't work because it is looking for double spaces. I suppose I could convert all spaces to double spaces beforehand and then convert them back after, but that doesn't seem very efficient. Any ideas?
$text='an of and then some an ee halved or or whenever';
$text=preg_replace('# [a-z]{1,2} #',' ',' '.$text.' ');
echo trim($text);
Removing the Short Words
You can use this:
$replaced = preg_replace('~\b[a-z]{1,2}\b\~', '', $yourstring);
In the demo, see the substitutions at the bottom.
Explanation
\b is a word boundary that matches a position where one side is a letter, and the other side is not a letter (for instance a space character, or the beginning of the string)
[a-z]{1,2} matches one or two letters
\b another word boundary
Replace with the empty string.
Option 2: Also Remove Trailing Spaces
If you also want to remove the spaces after the words, we can add \s* at the end of the regex:
$replaced = preg_replace('~\b[a-z]{1,2}\b\s*~', '', $yourstring);
Reference
Word Boundaries
You can use the word boundary tag: \b:
Replace: \b[a-z]{1,2}\b with ''
Use this
preg_replace('/(\b.{1,2}\s)/','',$your_string);
As some solutions worked here, they had a problem with my language's "multichar characters", such as "ch". A simple explode and implode worked for me.
$maxWordLength = 3;
$string = "my super string";
$exploded = explode(" ", $string);
foreach($exploded as $key => $word) {
if(mb_strlen($word) < $maxWordLength) unset($exploded[$key]);
}
$string = implode(" ", $exploded);
echo $string;
// outputs "super string"
To me, it seems that this hack works fine with most PHP versions:
$string2 = preg_replace("/~\b[a-zA-Z0-9]{1,2}\b\~/i", "", trim($string1));
Where [a-zA-Z0-9] are the accepted Char/Number range.
I want to change a specific character, only if it's previous and following character is of English characters. In other words, the target character is part of the word and not a start or end character.
For Example...
$string = "I am learn*ing *PHP today*";
I want this string to be converted as following.
$newString = "I am learn'ing *PHP today*";
$string = "I am learn*ing *PHP today*";
$newString = preg_replace('/(\w)\*(\w)/', '$1\'$2', $string);
// $newString = "I am learn'ing *PHP today* "
This will match an asterisk surrounded by word characters (letters, digits, underscores). If you only want to do alphabet characters you can do:
preg_replace('/([a-zA-Z])\*([a-zA-Z])/', '$1\'$2', 'I am learn*ing *PHP today*');
The most concise way would be to use "word boundary" characters in your pattern -- they represent a zero-width position between a "word" character and a "non-word" characters. Since * is a non-word character, the word boundaries require the both neighboring characters to be word characters.
No capture groups, no references.
Code: (Demo)
$string = "I am learn*ing *PHP today*";
echo preg_replace('~\b\*\b~', "'", $string);
Output:
I am learn'ing *PHP today*
To replace only alphabetical characters, you need to use a [a-z] as a character range, and use the i flag to make the regex case-insensitive. Since the character you want to replace is an asterisk, you also need to escape it with a backslash, because an asterisk means "match zero or more times" in a regular expression.
$newstring = preg_replace('/([a-z])\*([a-z])/i', "$1'$2", $string);
To replace all occurances of asteric surrounded by letter....
$string = preg_replace('/(\w)*(\w)/', '$1\'$2', $string);
AND
To replace all occurances of asteric where asteric is start and end character of the word....
$string = preg_replace('/*(\w+)*/','\'$1\'', $string);