I want to remove all non-alphanumeric signs from left and right of the string, leaving the ones in middle of string.
I've asked similar question here, and good solution is:
$str = preg_replace('/^\W*(.*\w)\W*$/', '$1', $str);
But it does remove also some signs like ąĄćĆęĘ etc and it should not as its still alphabetical sign.
Above example would do:
~~AAA~~ => AAA (OK)
~~AA*AA~~ => AA*AA (OK)
~~ŚAAÓ~~ => AA (BAD)
Make sure you use u flag for unicode while using your regex.
Following works with your input:
$str = preg_replace('/^\W*(.*\w)\W*$/u', '$1', '~~ŚAAÓ~~' );
// str = ŚAAÓ
But this won't work: (Don't Use it)
$str = preg_replace('/^\W*(.*\w)\W*$/', '$1', '~~ŚAAÓ~~' );
You can pass in a list of valid characters and tell the function to replace any character that is not in that list:
$str = preg_replace('/[^a-zA-Z0-9*]+/', '', $str);
The square brackets say select everything in this range. The carat (^) is the regex for not. We then list our valid characters (lower case a to z, uppercase a to z, numbers from 0 to 9, and an asterisks). The plus symbol on the end of the square bracket says select 0 or more characters.
Edit:
If this is the list of all characters you want to keep, then:
$str = preg_replace('/[^ĄąĆ毿ŹźŃńŁłÓó*]+/', '', $str);
Related
I Have one string like below.
$string = "2346#$ABSC$%###234567";
Now I want last character from this string that is not numeric or special character, It should be only A-a to Z-z.
Means, I need only "C" from this string.
I have try this formula:
substr($string, -1);
You should look into regular expressions using something like preg_match()
An expression like this would match:
/([a-z])[^a-z]*$/i
It means:
([a-z]) Capture an a-z character (the i at the end makes it case-insensitive)
[^a-z]*$ followed by 0 or more non a-z characters until the end of the string
See an example.
This should work for you:
(Here I just replace everything expect a-zA-Z with an empty string. After this I just access the last character)
<?php
$string = '2346#$ABSC$%###234567';
$string = preg_replace("/[^a-zA-Z]/", "", $string);
echo $string[strlen($string)-1];
?>
output:
C
The proper regex is: ([a-z])[^a-z]*$
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.
So I'm trying to do a string replace and something is happening that I wouldn't expect to happen and wanted to see if someone could shed some light on it.
I'm trying to do a regex replace where I replace '| ' if it is present. I'm using a group matching and the question mark to get it done, but for some reason it's replacing just spaces as well.
$str = 'x x';
$str = preg_replace('/(| )?/','',$str);
echo $str; // Echoes out 'xx' whereas it should return 'x x'
But when I replace a space with a carret I get:
$str = 'x^x';
$str = preg_replace('/(|^)?/','',$str);
echo $str; // Echoes out 'x^x' as expected
Is there some special thing with spaces that I'm not remembering? Or should this just work?
I tried the following:
$str = preg_replace('/(|\s)?/','',$str);
$str = preg_replace('/(|[ ])?/','',$str);
And both of them are also giving the inaccurate results. Thoughts?
Oh, didn't know you were waiting for me xD
As per comment, you should escape the pipe with a backslash: \|.
The | (pipe) is a special character in regex and means 'or', so that your regex were matching either 'nothing' or 'space' in the first one and either nothing or caret ^ in the second one.
I have a set of strings like this:
Pants [+$50]
Shirts [+$10]
Jeans [+$5]
Jackets [+$100]
How can I remove the ' [xxx]' in these lines and leaving just the item name (without the trailing space)? I was told to define a regular expression, not sure how that works...
That's actually a bit of a confusing regex, since [ and ] are special characters:
$str = 'Pants [+$50]';
$str = rtrim(preg_replace('/\[[^\]]*\]/', '', $str));
// 'Pants'
Basically the partern \[[^\]]*\] means to match a literal [ followed by 0 or more characters that are not ] followed by a ]. The second string in preg_replace is what it gets replaced with. In this case the empty string since we want to remove it. Then we use rtrim to trim any trailing whitespace.
Try this one:
The RegEx
(?im)[ \t]*\[[^\]\[]+\][ \t]*$
Code
$result = preg_replace('/^(.+?)[ \t]*\[[^\][]+\][ \t]*$/im', '$1', $subject);
I have string like below,
$string = "test coontevt [gallery include=\"12,24\"] first [gallery include=\"12,24\"] second";
i need to remove the string starts with [gallery to first ocuurance of it's ].
i already use this one,
$string12 = preg_replace('/[gallery.+?)+(/])/i', '', $string);
but i get empty string only.
Finally i want result for the above string is,
$string ="test coontevt first second".
How can i do this using regular expression?.
plz help me?
The character [ is a regex meta-character. TO match a literal [ you need to escape it.
$string12 = preg_replace('/\[gallery.+?\]/i', '', $string);
or
$string12 = preg_replace('/\[gallery[^\]]+\]/i', '', $string);
You need to escape the square brackets
$string12 = preg_replace('/\[gallery.+?\]/i', '', $string);
The round brackets are unnecessary so I removed them, also the quantifier between those brackets and the forward slash before the last square bracket.
To avoid multiple space in the result, I would match also the surrounding spaces and replace with 1 space.
\s+\[gallery.+?\]\s+ and replace with one space
$string12 = preg_replace('/\s+\[gallery.+?\]\s+/i', ' ', $string);
See this expression here online on Regexr
Try it like this:
$string12 = preg_replace('/\[gallery[^\]]+\]/i', '', $string);
[^\]]+ means that there can be one or more character that is not ]. And there is no need for any ( and ) if you don't want to use the backreferences.