I would like to know how to strip all non-alphanumeric characters from a string except for underscores and dashes in PHP.
Use preg_replace with /[^a-zA-Z0-9_\-]/ as the pattern and '' as the replacement.
$string = preg_replace('/[^a-zA-Z0-9_\-]/', '', $string);
EDIT
As skippy said, you can use the i modifier for case insensitivity:
$string = preg_replace('/[^a-z0-9_\-]/i', '', $string);
Use preg_replace:
$str = preg_replace('/[^\w-]/', '', $str);
The first argument to preg_replace is a regular expression. This one contains:
/ - starting delimiter -- start the regex
[ - start character class -- define characters that can be matched
^ - negative -- make the character class match only characters that don't match the selection that follows
\w - word character -- so don't match word characters. These are A-Za-z0-9 and _ (underscore)
- - hyphen -- don't match hypens either
] - close the character class
/ - ending delimiter -- close the regex
Note that this only matches hyphens (i.e. -). It does not match genuine dash characters (– or —).
Accepts a-z, A-Z, 0-9, '-', '_' and spaces:
$str = preg_replace("/[^a-z0-9\s_-]+/i", '', $tr);
No spaces:
$str = preg_replace("/[^a-z0-9_-]+/i", '', $tr);
Related
I have string which contain space in its html tags
$mystr = "＆lt; h3＆gt; hello mom ？＆lt; / h3＆gt;"
so i wrote regex expression for it to detect the spaces in it
$pattern = '/(?<=<)\s\w+|\s\/\s\w+|\s\/(?=>)/mi';
so next i want to modify the matches by removing space from it and replace it, so any idea how it can be done? so that i can fix my string like
"＆lt;h3＆gt; hello mom ？＆lt;/h3＆gt;"
i know there is php function pre_replace but not sure how i can modify the matches
$result = preg_replace( $pattern, $replace , $mystr );
For the specific tags like you showed, you can use
preg_replace_callback('/＆lt;(?:\s*\/)?\s*\w+\s*＆gt;/ui', function($m) {
return preg_replace('/\s+/u', '', $m[0]);
}, $mystr)
The regex - note the u flag to deal with Unicode chars in the string - matches
＆lt; - a literal string
(?:\s*\/)? - an optional sequence of zero or more whitespaces and a / char
\s* - zero or more whitespaces
\w+ - one or more word chars
\s* - zero or more whitespaces
＆gt; - a literal string.
The preg_replace('/\s+/u', '', $m[0]) line in the anonymous callback function removes all chunks of whitespaces (even those non-breaking spaces).
You could keep it simple and do:
$output = str_replace(['＆lt; / ', '＆lt; ', '＆gt; '],
['＆lt;/', '＆lt;', '＆gt;'], $input);
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 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 need a regular expression (php) to remove the forward slash, the dot and eveything after the dot in my string so that
$str = "ab/12c.3de";
becomes
$newstr = "ab12c";
You can use alternation in regex:
$str = "ab/12c.3de";
$newstr = preg_replace('~/|\..*~', '', $str);
//=> ab12c
Regex: /|\..*
/ matches literal /
| OR (alternation)
\..* matches a dot and everything after it
Replacement is just by empty string.
I am looking to find and replace words in a long string. I want to find words that start looks like this: $test$ and replace it with nothing.
I have tried a lot of things and can't figure out the regular expression. This is the last one I tried:
preg_replace("/\b\\$(.*)\\$\b/im", '', $text);
No matter what I do, I can't get it to replace words that begin and end with a dollar sign.
Use single quotes instead of double quotes and remove the double escape.
$text = preg_replace('/\$(.*?)\$/', '', $text);
Also a word boundary \b does not consume any characters, it asserts that on one side there is a word character, and on the other side there is not. You need to remove the word boundary for this to work and you have nothing containing word characters in your regular expression, so the i modifier is useless here and you have no anchors so remove the m (multi-line) modifier as well.
As well * is a greedy operator. Therefore, .* will match as much as it can and still allow the remainder of the regular expression to match. To be clear on this, it will replace the entire string:
$text = '$fooo$ bar $baz$ quz $foobar$';
var_dump(preg_replace('/\$(.*)\$/', '', $text));
# => string(0) ""
I recommend using a non-greedy operator *? here. Once you specify the question mark, you're stating (don't be greedy.. as soon as you find a ending $... stop, you're done.)
$text = '$fooo$ bar $baz$ quz $foobar$';
var_dump(preg_replace('/\$(.*?)\$/', '', $text));
# => string(10) " bar quz "
Edit
To fix your problem, you can use \S which matches any non-white space character.
$text = '$20.00 is the $total$';
var_dump(preg_replace('/\$\S+\$/', '', $text));
# string(14) "$20.00 is the "
There are three different positions that qualify as word boundaries \b:
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.
$ is not a word character, so don't use \b or it won't work. Also, there is no need for the double escaping and no need for the im modifiers:
preg_replace('/\$(.*)\$/', '', $text);
I would use:
preg_replace('/\$[^$]+\$/', '', $text);
You can use preg_quote to help you out on 'quoting':
$t = preg_replace('/' . preg_quote('$', '/') . '.*?' . preg_quote('$', '/') . '/', '', $text);
echo $t;
From the docs:
This is useful if you have a run-time string that you need to match in some text and the string may contain special regex characters.
The special regular expression characters are: . \ + * ? [ ^ ] $ ( ) { } = ! < > | : -
Contrary to your use of word boundary markers (\b), you actually want the inverse effect (\B)-- you want to make sure that there ISN'T a word character next to the non-word character $.
You also don't need to use capturing parentheses because you are not using a backreference in your replacement string.
\S+ means one or more non-whitespace characters -- with greedy/possessive matching.
Code: (Demo)
$text = '$foo$ boo hi$$ mon$k$ey $how thi$ $baz$ bar $foobar$';
var_export(
preg_replace(
'/\B\$\S+\$\B/',
'',
$text
)
);
Output:
' boo hi$$ mon$k$ey $how thi$ bar '