I have this code:
$string = "123456ABcd9999";
$answer = ereg("([0-9]*)", $string, $digits);
echo $digits[0];
This outputs '123456'. I'd like it to output '1234569999' ie. all the digits. How can I achieve this. I've been trying lots of different regex things but can't figure it out.
First, don't use ereg (it's deprecated). Secondly, why not replace it out:
$answer = preg_replace('#\D#', '', $string);
Note that \D is the inverse of \d. So \d matches all decimal numeric characters (0-9), therefore \D matches anything that \d does not match...
You could use preg_replace for this, preg_replace("/[^0-9]/", "", $string) for example.
Related
i need to explode youtube url from this line:
[embed]https://www.youtube.com/watch?v=L3HQMbQAWRc[/embed]
It is possible? I need to delete [embed] & [/embed].
preg_match is what you need.
<?php
$str = "[embed]https://www.youtube.com/watch?v=L3HQMbQAWRc[/embed]";
preg_match("/\[embed\](.*)\[\/embed\]/", $str, $matches);
echo $matches[1]; //https://www.youtube.com/watch?v=L3HQMbQAWRc
$string = '[embed]https://www.youtube.com/watch?v=L3HQMbQAWRc[/embed]';
$string = str_replace(['[embed]', '[/embed]'], '', $string);
See str_replace
why not use str_replace? :) Quick & Easy
http://php.net/manual/de/function.str-replace.php
Just for good measure, you can also use positive lookbehind's and lookahead's in your regular expressions:
(?<=\[embed\])(.*)(?=\[\/embed\])
You'd use it like this:
$string = "[embed]https://www.youtube.com/watch?v=L3HQMbQAWRc[/embed]";
$pattern = '/(?<=\[embed\])(.*)(?=\[\/embed\])/';
preg_match($pattern, $string, $matches);
echo $match[1];
Here is an explanation of the regex:
(?<=\[embed\]) is a Positive Lookbehind - matches something that follows something else.
(.*) is a Capturing Group - . matches any character (except a newline) with the Quantifier: * which provides matches between zero and unlimited times, as many times as possible. This is what is matched between the groups prior to and after. This are the droids you're looking for.
(?=\[\/embed\]) is a Positive Lookahead - matches things that come before it.
I couldn't find the solution using search.
I am looking for a php solution to remove all character BEFORE the second occurance of and underscore (including the underscore)
For example:
this_is_a_test
Should output as:
a_test
I currently have this code but it will remove everything after the first occurance:
preg_replace('/^[^_]*.s*/', '$1', 'this_is_a_test');
Using a slightly different approach,
$s='this_is_a_test';
echo implode('_', array_slice( explode( '_', $s ),2 ) );
/* outputs */
a_test
preg_replace('/^.*_.*_(.*)$/U', '$1', 'this_is_a_test');
Note the U modifier which tells regex to take as less characters for .* as possible.
You can also use explode, implode along with array_splice like as
$str = "this_is_a_test";
echo implode('_',array_splice(explode('_',$str),2));//a_test
Demo
Why go the complicated way? This is a suggestion though using strrpos and substr:
<?php
$str = "this_is_a_test";
$str_pos = strrpos($str, "_");
echo substr($str, $str_pos-1);
?>
Try this one.
<?php
$string = 'this_is_a_test';
$explode = explode('_', $string, 3);
echo $explode[2];
?>
Demo
I'm still in favor of a regular expression in this case:
preg_replace('/^.*?_.*?_/', '', 'this_is_a_test');
Or (which looks more complex here but is easily adjustable to N..M underscores):
preg_replace('/^(?:.*?_){2}/', '', 'this_is_a_test');
The use of the question mark in .*? makes the match non-greedy; and the pattern has been expanded from the original post to "match up through" the second underscore.
Since the goal is to remove text the matched portion is simply replaced with an empty string - there is no need for a capture group or to use such as the replacement value.
If the input doesn't include two underscores then nothing is removed; such can be adjusted, very easily with the second regular expression, if the rules are further clarified.
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 split a word by capital letter in PHP
For example:
$string = "facebookPageUrl";
I want it like this:
$array = array("facebook", "Page", "Url");
How should I do it? I want the shortest and most efficient way.
You can use preg_split with the a look-ahead assertion:
preg_split('/(?=\p{Lu})/u', $str)
Here \p{Lu} is a character class of all Unicode uppercase letters. If you just work with US-ASCII characters, you could also use [A-Z] instead.
$string = "facebookPageUrl";
preg_match_all('((?:^|[A-Z])[^A-Z]*)', $string, $matches);
var_dump($matches);
http://ideone.com/wL9jM
I'd like to return string between two characters, # and dot (.).
I tried to use regex but cannot find it working.
(#(.*?).)
Anybody?
Your regular expression almost works, you just forgot to escape the period. Also, in PHP you need delimiters:
'/#(.*?)\./s'
The s is the DOTALL modifier.
Here's a complete example of how you could use it in PHP:
$s = 'foo#bar.baz';
$matches = array();
$t = preg_match('/#(.*?)\./s', $s, $matches);
print_r($matches[1]);
Output:
bar
Try this regular expression:
#([^.]*)\.
The expression [^.]* will match any number of any character other than the dot. And the plain dot needs to be escaped as it’s a special character.
this is the best and fast to use
function get_string_between ($str,$from,$to) {
$string = substr($str, strpos($str, $from) + strlen($from));
if (strstr ($string,$to,TRUE) != FALSE) {
$string = strstr ($string,$to,TRUE);
}
return $string;
}
If you're learning regex, you may want to analyse those too:
#\K[^.]++(?=\.)
(?<=#)[^.]++(?=\.)
Both these regular expressions use possessive quantifiers (++). Use them whenever you can, to prevent needless backtracking. Also, by using lookaround constructions (or \K), we can match the part between the # and the . in $matches[0].