I am trying to add a span tag to the last word of a string. It works if the string has no special characters. I can't figure out the correct regex for it.
$string = "Onun Mesajı";
echo preg_replace("~\W\w+\s*\S?$~", ' <span>' . '\\0' . '</span>', $string);
Here is the Turkish character set : ÇŞĞÜÖİçşğüöı
You need to use /u modifier to allow processing Unicode characters in the pattern and input string.
preg_replace('~\w+\s*$~u', '<span>$0</span>', $string);
^
Full PHP demo:
$string = "Onun Mesajı";
echo preg_replace("~\w+\s*$~u", '<span>$0</span>', $string);
Also, the regex you need is just \w+\s*$:
\w+ - 1 or more alphanumerics
\s* - 0 or more whitespace (trailing)
$ - end of string
Since I removed the \W from the regex, there is no need to "hardcode" the leading space in the replacement string (removed, too).
You should use the u modifier for regular expressions to set the engine into unicode mode:
<?php
$subject = "Onun äöüß Mesajı";
$pattern = '/\w+\s*?$/u';
echo preg_replace($pattern, '<span>\\0</span>', $subject);
The output is:
Onun äöüß <span>Mesajı</span>
This regex will do the trick for you, and is a lot shorter then the other solutions:
[ ](.*?$)
Here is an example of it:
$string = "Onun Mes*ÇŞĞÜÖİçşğüöıajı";
echo preg_replace('~[ ](.*?$)~', ' <span>' .'${1}'. '</span>', $string);
Will echo out:
Onun <span>Mes*ÇŞĞÜÖİçşğüöıajı</span>
The way this regex works is that we look for any characters without space in lazy mode [ ].*?.
then we add the $ identifier, so it matches from the end of the string instead.
Related
I have a string as
This is a sample text. This text will be used as a dummy for "various" RegEx "operations" using PHP.
I want to select and replace all the first alphabet of each word (in the example : T,i,a,s,t,T,t,w,b,u,a,d,f,",R,",u,P). How do I do it?
I tried /\b.{1}\w+\b/. I read the expression as "select any character that has length of 1 followed by word of any length" but didn't work.
You may try this regex as well:
(?<=\s|^)([a-zA-Z"])
Demo
Your regex - /\b.{1}\w+\b/ - matches any string that is not enclosed in word characters, starts with any symbol that is in a position after a word boundary (thus, it can even be whitespace if there is a letter/digit/underscore in front of it), followed with 1 or more alphanumeric symbols (\w) up to the word boundary.
That \b. is the culprit here.
If you plan to match any non-whitespace preceded with a whitespace, you can just use
/(?<!\S)\S/
Or
/(?<=^|\s)\S/
See demo
Then, replace with any symbol you need.
You may try to use the following regex:
(.)[^\s]*\s?
Using the preg_match_all and implode the output result group 1
<?php
$string = 'This is a sample text. This text will be used as a dummy for'
. '"various" RegEx "operations" using PHP.';
$pattern = '/(.)[^\s]*\s?/';
$matches;
preg_match_all($pattern, $string, $matches);
$output = implode('', $matches[1]);
echo $output; //Output is TiastTtwbuaadf"R"uP
For replace use something like preg_replace_callback like:
$pattern = '/(.)([^\s]*\s?)/';
$output2 = preg_replace_callback($pattern,
function($match) { return '_' . $match[2]; }, $string);
//result: _his _s _ _ample _ext. _his _ext _ill _e _sed _s _ _ummy _or _various" _egEx _operations" _sing _HP.
I have a list of string like this
$16,500,000(#$2,500)
$34,000(#$11.00)
$214,000(#$18.00)
$12,684,000(#$3,800)
How can I extract all symbols and the (#$xxxx) from these strings so that they can be like
16500000
34000
214000
12684000
\(.*?\)|\$|,
Try this.Replace by empty string.See demo.
https://regex101.com/r/vD5iH9/42
$re = "/\\(.*?\\)|\\$|,/m";
$str = "\$16,500,000(#\$2,500)\n\$34,000(#\$11.00)\n\$214,000(#\$18.00)\n\$12,684,000(#\$3,800)";
$subst = "";
$result = preg_replace($re, $subst, $str);
To remove the end (#$xxxx) characters, you could use the regex:
\(\#\$.+\)
and replace it with nothing:
preg_replace("/\(\#\$.+\)/g"), "", $myStringToReplaceWith)
Make sure to use the g (global) modifier so the regex doesn't stop after it finds the first match.
Here's a breakdown of that regex:
\( matches the ( character literally
\# matches the # character literally
\$ matches the $ character literally
.+ matches any character 1 or more times
\) matches the ) character literally
Here's a live example on regex101.com
In order to remove all of these characters:
$ , ( ) # .
From a string, you could use the regex:
\$|\,|\(|\)|#|\.
Which will match all of the characters above.
The | character above is the regex or operator, effectively making it so
$ OR , OR ( OR ) OR # OR . will be matched.
Next, you could replace it with nothing using preg_replace, and with the g (global) modifier, which makes it so the regex doesn't return on the first match:
preg_replace("/\$|\,|\(|\)|#|\./g"), "", $myStringToReplaceWith)
Here's a live example on regex101.com
So in the end, your code could look like this:
$str = preg_replace("/\(\#\$.+\)/g"), "", $str)
$str = preg_replace("/\$|\,|\(|\)|#|\./g"), "", $str)
Although it isn't in one regex, it does not use any look-ahead, or look-behind (both of which are not bad, by the way).
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 '
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 have been trying to replace a portion of a string if two of more non alphanumeric characters are found.
I have it partly working but can not replace when a underscore is in there.
This is what i am trying.
$str = "-dxs_ s";
$str = preg_replace('/\W{2,}|\_{2,}/', ' ', $str);
reults in -dxs_ s should be -dxs s.
So how do you replace if two or more non alphanumeric characters are found in a string?
Simply
$str = preg_replace('/(\W|_){2,}/', ' ', $str);
What this is doing is grouping the "non-word or underscore" part and applies the 2+ quantifier to it as a whole.
See it in action.
\W also excludes _ therefore you need your own characters class :
/[^a-zA-Z0-9]{2,}/
or
$result = preg_replace('/[^a-z\d]{2,}/i', ' ', $subject);