I'm trying to capitalize "words" that have at least one number, letter, and special character such as a period or dash.
Things like: 3370.01b, 6510.01.b, m-5510.30, and drm-2013-c-004914.
I don't want it to match things like: hello, sk8, and mixed-up
I'm trying to use lookaheads, as suggested, but I can't get it to match anything.
$output = preg_replace_callback('/\b(?=.*[0-9]+)(?=.*[a-z]+)(?=.*[\.-]+)\b/i', function($matches){return strtoupper($matches[0]);}, $input);
You can use this regex to match the strings you want,
(?=\S*[a-z])(?=\S*\d)[a-z\d]+(?:[.-][a-z\d]+)+
Explanation:
(?=\S*[a-z]) - This look ahead ensures that there is at least an alphabet character in the incoming word
(?=\S*\d) - This look ahead ensures that there is at least a digit in the incoming word
[a-z\d]+(?:[.-][a-z\d]+)+ - This part captures a word contain alphanumeric word containing at least one special character . or -
Online Demo
Here is the PHP code demo modifying your code,
$input = '3370.01b, 6510.01.b, m-5510.30, and drm-2013-c-004914 hello, sk8, and mixed-up';
$output = preg_replace_callback('/(?=\S*[a-z])(?=\S*\d)[a-z\d]+(?:[.-][a-z\d]+)+/i', function($matches){return strtoupper($matches[0]);}, $input);
echo $output;
Prints,
3370.01B, 6510.01.B, M-5510.30, and DRM-2013-C-004914 hello, sk8, and mixed-up
Regular expression:
https://regex101.com/r/sdmlL8/1
(?=.*\d)(.*)([-.])(.*)
PHP code:
https://ideone.com/qEBZQc
$input = '3370.01b';
$output = preg_replace_callback('/(?=.*\d)(.*)([-.])(.*)/i', function($matches){return strtoupper($matches[0]);}, $input);
I don't think you never captured anything to put into matches...
$input = '3370.01b foo';
$output = preg_replace_callback('/(?=.*[0-9])(?=.*[a-z])(\w+(?:[-.]\w+)+)/i', function($matches){return strtoupper($matches[0]);}, $input);
echo $output;
Output
3370.01B foo
Sandbox
https://regex101.com/r/syJWMN/1
Related
Quick note: I know markdown parsers don't care about this issue. It's for the sake of visual consistency in the md file and also experimentation.
Sample:
# this
##that
###or this other
Goal: read each line and,if a markdown header does not have a space after the pound/hashtag sign, add one so that it would look like:
# this
## that
### or this other
My non-regex attempt:
function inelegantFunction (string $string){
$array = explode('#',$string);
$num = count($array);
$text = end($array);
return str_repeat('#', $num-1)." ".$text;
}
echo inelegantFunction("###or this other");
// returns ### or this other
This works, but it has no mechanism to match the unlikely case of seven '#'.
Regardless of efficacy, I would like to figure out how to do this with regex in php (and perhaps javascript if that matters).
Try to match (?m)^#++\K\S which matches lines starting with one or more number signs then replace it with $0 in your function:
return preg_replace('~(?m)^#++\K\S~', ' $0', $string);
See live demo here
To limit the number of #s to six use:
(?m)^(?!#{7})#++\K\S
I'm guessing that a simple expression with a right char-list boundary might be working here, maybe:
(#)([a-z])
If we might be having more chars, we can simply add it to [a-z].
Demo
Test
$re = '/(#)([a-z])/m';
$str = '#this
##that
###that
### or this other';
$subst = '$1 $2';
$result = preg_replace($re, $subst, $str);
echo "The result of the substitution is ".$result;
i'm not very firm with regular Expressions, so i have to ask you:
How to find out with PHP if a string contains a word starting with # ??
e.g. i have a string like "This is for #codeworxx" ???
I'm so sorry, but i have NO starting point for that :(
Hope you can help.
Thanks,
Sascha
okay thanks for the results - but i did a mistake - how to implement in eregi_replace ???
$text = eregi_replace('/\B#[^\B]+/','\\1', $text);
does not work??!?
why? do i not have to enter the same expression as pattern?
Match anything with has some whitespace in front of a # followed by something else than whitespace:
$ cat 1812901.php
<?php
echo preg_match("/\B#[^\B]+/", "This should #match it");
echo preg_match("/\B#[^\B]+/", "This should not# match");
echo preg_match("/\B#[^\B]+/", "This should match nothing and return 0");
echo "\n";
?>
$ php 1812901.php
100
break your string up like this:
$string = 'simple sentence with five words';
$words = explode(' ', $string );
Then you can loop trough the array and check if the first character of each word equals "#":
if ($stringInTheArray[0] == "#")
Assuming you define a word a sequence of letters with no white spaces between them, then this should be a good starting point for you:
$subject = "This is for #codeworxx";
$pattern = '/\s*#(.+?)\s/';
preg_match($pattern, $subject, $matches);
print_r($matches);
Explanation:
\s*#(.+?)\s - look for anything starting with #, group all the following letters, numbers, and anything which is not a whitespace (space, tab, newline), till the closest whitespace.
See the output of the $matches array for accessing the inner groups and the regex results.
#OP, no need regex. Just PHP string methods
$mystr='This is for #codeworxx';
$str = explode(" ",$mystr);
foreach($str as $k=>$word){
if(substr($word,0,1)=="#"){
print $word;
}
}
Just incase this is helpful to someone in the future
/((?<!\S)#\w+(?!\S))/
This will match any word containing alphanumeric characters, starting with "#." It will not match words with "#" anywhere but the start of the word.
Matching cases:
#username
foo #username bar
foo #username1 bar #username2
Failing cases:
foo#username
#username$
##username
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 am trying to make a regex that will look behind .txt and then behind the "-" and get the first digit .... in the example, it would be a 1.
$record_pattern = '/.txt.+/';
preg_match($record_pattern, $decklist, $record);
print_r($record);
.txt?n=chihoi%20%283-1%29
I want to write this as one expression but can only seem to do it as two. This is the first time working with regex's.
You can use this:
$record_pattern = '/\.txt.+-(\d)/';
Now, the first group contains what you want.
Your regex would be,
\.txt[^-]*-\K\d
You don't need for any groups. It just matches from the .txt and upto the literal -. Because of \K in our regex, it discards the previously matched characters. In our case it discards .txt?n=chihoi%20%283- string. Then it starts matching again the first digit which was just after to -
DEMO
Your PHP code would be,
<?php
$mystring = ".txt?n=chihoi%20%283-1%29";
$regex = '~\.txt[^-]*-\K\d~';
if (preg_match($regex, $mystring, $m)) {
$yourmatch = $m[0];
echo $yourmatch;
}
?> //=> 1
i'm not very firm with regular Expressions, so i have to ask you:
How to find out with PHP if a string contains a word starting with # ??
e.g. i have a string like "This is for #codeworxx" ???
I'm so sorry, but i have NO starting point for that :(
Hope you can help.
Thanks,
Sascha
okay thanks for the results - but i did a mistake - how to implement in eregi_replace ???
$text = eregi_replace('/\B#[^\B]+/','\\1', $text);
does not work??!?
why? do i not have to enter the same expression as pattern?
Match anything with has some whitespace in front of a # followed by something else than whitespace:
$ cat 1812901.php
<?php
echo preg_match("/\B#[^\B]+/", "This should #match it");
echo preg_match("/\B#[^\B]+/", "This should not# match");
echo preg_match("/\B#[^\B]+/", "This should match nothing and return 0");
echo "\n";
?>
$ php 1812901.php
100
break your string up like this:
$string = 'simple sentence with five words';
$words = explode(' ', $string );
Then you can loop trough the array and check if the first character of each word equals "#":
if ($stringInTheArray[0] == "#")
Assuming you define a word a sequence of letters with no white spaces between them, then this should be a good starting point for you:
$subject = "This is for #codeworxx";
$pattern = '/\s*#(.+?)\s/';
preg_match($pattern, $subject, $matches);
print_r($matches);
Explanation:
\s*#(.+?)\s - look for anything starting with #, group all the following letters, numbers, and anything which is not a whitespace (space, tab, newline), till the closest whitespace.
See the output of the $matches array for accessing the inner groups and the regex results.
#OP, no need regex. Just PHP string methods
$mystr='This is for #codeworxx';
$str = explode(" ",$mystr);
foreach($str as $k=>$word){
if(substr($word,0,1)=="#"){
print $word;
}
}
Just incase this is helpful to someone in the future
/((?<!\S)#\w+(?!\S))/
This will match any word containing alphanumeric characters, starting with "#." It will not match words with "#" anywhere but the start of the word.
Matching cases:
#username
foo #username bar
foo #username1 bar #username2
Failing cases:
foo#username
#username$
##username