Dot (.) Meta character is not matching text and showing zero(o) output. Please tell me what i am missing in this code.
$string = 'pakistan';
echo preg_match('/p.n/',$string);
The following regex:
$string = 'pakistan';
echo preg_match('~^p.+n$~',$string);
Anchors the string to the beginning and end and requires at least one (but unlimited times) character between p and n.
Use below code:-
preg_match("/p.*n/U", $string, $match);
echo '<pre>'; print_r($match[0]);
If you want to count how many times specific word is repeating in the string
Use preg_match_all
preg_match_all ("/p.*n/U", $string, $match);
echo '<pre>'; print_r($match[0]);
Hope it will help you :)
Related
I want to change the id partner=852501 to partner=985412 in my code, but stuck in this step. Example:
http://a-ads.com/?partner=852501
My attempt:
([1-z0-9\/]+)\/\/(a-ads\.com\/?)
How to do the right thing?
Regex: partner=\K[^&]+
Details:
\K resets the starting point of the reported match
[^&]+ Match a single character not present in the list & one and unlimited times
PHP code:
preg_replace('/partner=\K[^&]+/', '985412', $url);
preg_replace you can replace the value
$string = 'http://a-ads.com/?partner=852501';
$your = 985412;
echo preg_replace('/(\d+)/', $your, $string);
UPD: The value can be obtained with the help of preg_match
preg_match('/partner=(\d+)/', $string, $match);
echo $match[1];
Hi I'm new to php and I need a little help
I need to change the text that is between ** in php string and put it between html tag
$text = "this is an *example*";
But I really don't know how and i need help
personally I would use explode, you can then piece the sentence back together if the example appears in the middle of a sentence
<?php
$text = "this is an *example*";
$pieces = explode("*", $text);
echo $pieces[0];
?>
Edit:
Since you're looking for what basically amounts to custom BB Code use this
$text = "this is an *example*";
$find = '~[\*](.*?)[\*]~s';
$replace = '<span style="color: green">$1</span>';
echo preg_replace($find,$replace,$text);
You can add this to a function and have it parse any text that gets passed to it, you can also make the find and replace variables into arrays and add more codes to it
You really should use a DOM parser for things like this, but if you can guaratee it will always be the * character you can use some regex:
$text = "this is an *example*";
$regex = '/(?<=\*)(.*?)(?=\*)/';
$replacement = 'ostrich';
$new_text = preg_replace($regex, $replacement, $text);
echo $new_text;
Returns
this is an *ostrich*
Here is how the regex works:
Positive Lookbehind (?<=\*)
\* matches the character * literally (case sensitive)
1st Capturing Group (.*?)
.*? matches any character (except for line terminators)
*? Quantifier — Matches between zero and unlimited times, as few times as possible, expanding as needed (lazy)
Positive Lookahead (?=\*)
\* matches the character * literally (case sensitive)
This regex essentially starts and ends by looking at what is ahead of and behind the search character you specified and leaves those characters intact during the replacement with preg_replace().
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 am using php Regular expressions but i can't retrieve values that i group using ()
this is my input
<img src="http://www.example.com/image.jpg" title="title" />
i need only src value , this is my regex '"<img src=\"(.*?)\".*?\/>"'
if i can retrieve First group just like java patterns my problem is sloved
preg_match_all('"<img src=\"(.*?)\".*?\/>"', $source, $re);
print_r($re);
and it return full image tag like this <img src="http://www.example.com/image.jpg" title="title" />
To match a single string, preg_match function is enough. You don't need to go for preg_matchall function. If you want to match more number of strings then you could use preg_matchall function. And also first try to match the exact string through the pattern rather than to go for grouping. If it's impossible to match a particular string then go for grouping.
In the below, matching the exact value of src attribute is done.
You could get the value of src in two ways,
1. positive lookbehind
Regex:
(?<=src=\")[^\"]*
PHP code:(Through match_all)
<?php
$string = "<img src=\"http://www.example.com/image.jpg\" title=\"title\" />";
$regex = '~(?<=src=\")[^\"]*~';
preg_match_all($regex, $string, $matches);
print_r($matches);
?>
PHP code:(Through match)
<?php
$string = "<img src=\"http://www.example.com/image.jpg\" title=\"title\" />";
$regex = '~(?<=src=\")[^\"]*~';
if (preg_match($regex, $string, $m)) {
$yourmatch = $m[0];
echo $yourmatch;
}
?> //=> http://www.example.com/image.jpg
Explanation:
(?<=src=\") Positive look-behind is used here. So the regex engine puts the match marker just after to the src=".
[^\"]* Now it starts matching any character zero or more times but not of ". When it finds a ", it stops matching characters.
2. Using \K
Regex:
src=\"\K[^\"]*
PHP code (through match)
<?php
$string = "<img src=\"http://www.example.com/image.jpg\" title=\"title\" />";
$regex = '~src=\"\K[^\"]*~';
if (preg_match($regex, $string, $m)) {
$yourmatch = $m[0];
echo $yourmatch;
}
?> //=> http://www.example.com/image.jpg
Explanation:
\K resets the starting point of the reported match. Any previously consumed characters are no longer included in the final match.
src=\"\K So it discards the previously matched src=".
[^\"]* Matches any character zero or more times but not of "
You're using preg_match_all so that you need to pass index as well, use print_r($re[1]); to get results.
I Got It Accidently !
we can Code Like this for first Grouped
print_r($re[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