I know this has been asked before as ive just been reading those answers but still cant get this to work (properly).
Im very new to regex and am trying to do something that sounds pretty simple:
The string would be:
http://www.something.com/section/filter/colour/red-#998682/size/small/
What i would like to do is a preg_replace to remove the -#?????? so the url looks like:
http://www.something.com/section/filter/colour/red/size/small/
So i tried:
$string = $theURL;
$pattern = '/-\#(.*)\//i';
$replacement = '/';
$newURL = preg_replace($pattern, $replacement, $string);
That sort of works but it doesnt stop. If I have anything after the -#?????? it also removes that as well. But I thought having the / on the end would stop it doing that?
Hoping someone can help and thanks for reading
PCRE is greedy by default, meaning that .* will match as big a chunk as possible. Make it ungreedy by adding the U flag (for the entire pattern) or use .*? (for just that wildcard part):
/-\#(.*)\//iU
or
/-\#(.*?)\//i
You need to use non-greedy quantifier.
$pattern = '/-\#(.*?)\//i';
Your regex is greedy, which means that (.*)\/ looks for the last slash, not the first one.
demo
(.*) pattern is gready, which means it'll match as many characters as possible. To match everything to the first slash use (.*?):
$pattern = '/-\#(.*?)\//i';
Related
I am looking for a way to get a valid url out of a string like:
$string = 'http://somesite.com/directory//sites/9/my_forms/3-895a3e/somefilename.jpg|:||:||:||:|19845';
My original solution was:
preg_match('#^[^:|]*#', str_replace('//', '/', $string), $modifiedPath);
But obviously its going to remove a slash from the http:// instead of the one in the middle of the string.
My expected output that I want from the original is:
http://somesite.com/directory/sites/9/my_forms/3-895a3e/somefilename.jpg
I could always break off the http part of the string first but would like a more elegant solution in the form of regex if possible. Thanks.
This will do exactly what you are asking:
<?php
$string = 'http://somesite.com/directory//sites/9/my_forms/3-895a3e/somefilename.jpg|:||:||:||:|19845';
preg_match('/^([^|]+)/', $string, $m); // get everything up to and NOT including the first pipe (|)
$string = $m[1];
$string = preg_replace('/(?<!:)\/\//', '/' ,$string); // replace all occurrences of // as long as they are not preceded by :
echo $string; // outputs: http://somesite.com/directory/sites/9/my_forms/3-895a3e/somefilename.jpg
exit;
?>
EDIT:
(?<!X) in regular expressions is the syntax for what is called a lookbehind. The X is replaced with the character(s) we are testing for.
The following expression would match every instance of double slashes (/):
\/\/
But we need to make sure that the match we are looking for is NOT preceded by the : character so we need to 'lookbehind' our match to see if the : character is there. If it is then we don't want it to be counted as a match:
(?<!:)\/\/
The ! is what says NOT to match in our lookbehind. If we changed it to (?=:)\/\/ then it would only match the double slashes that did have the : preceding them.
Here is a Quick tutorial that can explain it all better than I can lookahead and lookbehind tutorial
Assuming all your strings are in the form given, you don't need any but the simplest of regexes to do this; if you want an elegant solution, then a regex is definitely not what you need. Also, double slashes are legal in a URL, just like in a Unix path, and mean the same thing a single slash does, so you don't really need to get rid of them at all.
Why not just
$url = array_shift(preg_split('/\|/', $string));
?
If you really, really care about getting rid of the double slashes in the URL, then you can follow this with
$url = preg_replace('/([^:])\/\//', '$1/', $url);
or even combine them into
$url = preg_replace('/([^:])\/\//', '$1/', array_shift(preg_split('/\|/', $string)));
although that last form gets a little bit hairy.
Since this is a quite strictly defined situation, I'd consider just one preg to be the most elegant solution.
From the top of my head:
$sanitizedURL = preg_replace('~((?<!:)/(?=/)|\\|.+)~', '', $rawURL);
Basically, what this does is look for any forward slash that IS NOT preceded by a colon (:), and IS followed bij another forward slash. It also searches for any pipe character and any character following it.
Anything found is removed from the result.
I can explain the RegEx in more detail if you like.
I'am trying to get all {{product.smth}} with preg_match_all, but if i have few of this in one line i get wrong result.
Example:
$smth = '<name>{{product.name}}</name><getname>{{product.getName()}}</getname>';
$pattern = '/\{\{product\.(.*)\}\}/';
preg_match_all($pattern, $smth, $matches);
//returns '{{product.name}}</name><getname>{{product.getName()}}'
//instad of '{{product.name}}' and '{{product.getName()}}'
What iam doing wrong? Please help.
The problem is that repetition is greedy. Either make it ungreedy by using .*? or better yet: disallow the } character for the repetition:
$pattern = '/\{\{product\.([^}]*)\}\}/';
If you do want to allow single } in that value (like {{product.some{thing}here}}), the equivalent solution uses a negative lookahead:
$pattern = '/\{\{product\.((?:(?!\}\}).)*)\}\}/';
For every single character included in .* it checks that that character doesn't mark the start of a }}.
I think it'll work if you change .* to .*? this will make it lazy instead of greedy and it will try to match as little as possible - so, till the first occurance of }} rather than the last.
I've been searching for a regex to replace plain text url's in a string (the string can contain more than 1 url), by:
url
and I found this:
http://mathiasbynens.be/demo/url-regex
I would like to use the diegoperini's regex (which according to the tests is the best):
_^(?:(?:https?|ftp)://)(?:\S+(?::\S*)?#)?(?:(?!10(?:\.\d{1,3}){3})(?!127(?:\.\d{1,3}){3})(?!169\.254(?:\.\d{1,3}){2})(?!192\.168(?:\.\d{1,3}){2})(?!172\.(?:1[6-9]|2\d|3[0-1])(?:\.\d{1,3}){2})(?:[1-9]\d?|1\d\d|2[01]\d|22[0-3])(?:\.(?:1?\d{1,2}|2[0-4]\d|25[0-5])){2}(?:\.(?:[1-9]\d?|1\d\d|2[0-4]\d|25[0-4]))|(?:(?:[a-z\x{00a1}-\x{ffff}0-9]+-?)*[a-z\x{00a1}-\x{ffff}0-9]+)(?:\.(?:[a-z\x{00a1}-\x{ffff}0-9]+-?)*[a-z\x{00a1}-\x{ffff}0-9]+)*(?:\.(?:[a-z\x{00a1}-\x{ffff}]{2,})))(?::\d{2,5})?(?:/[^\s]*)?$_iuS
But I want o make it global to replace all the url's in a string.
When I use this:
/_(?:(?:https?|ftp)://)(?:\S+(?::\S*)?#)?(?:(?!10(?:\.\d{1,3}){3})(?!127(?:\.\d{1,3}){3})(?!169\.254(?:\.\d{1,3}){2})(?!192\.168(?:\.\d{1,3}){2})(?!172\.(?:1[6-9]|2\d|3[0-1])(?:\.\d{1,3}){2})(?:[1-9]\d?|1\d\d|2[01]\d|22[0-3])(?:\.(?:1?\d{1,2}|2[0-4]\d|25[0-5])){2}(?:\.(?:[1-9]\d?|1\d\d|2[0-4]\d|25[0-4]))|(?:(?:[a-z\x{00a1}-\x{ffff}0-9]+-?)*[a-z\x{00a1}-\x{ffff}0-9]+)(?:\.(?:[a-z\x{00a1}-\x{ffff}0-9]+-?)*[a-z\x{00a1}-\x{ffff}0-9]+)*(?:\.(?:[a-z\x{00a1}-\x{ffff}]{2,})))(?::\d{2,5})?(?:/[^\s]*)?_iuS/g
It does not work, how do I make this regex global and what does the underscore at the beginning and the "_iuS", at the end, means?
I would like to use it with php so I am using:
preg_replace($regex, '$0', $examplestring);
The underscores are the regex delimiters, the i, u and S are pattern modifiers :
i (PCRE_CASELESS)
If this modifier is set, letters in the pattern match both upper and lower
case letters.
U (PCRE_UNGREEDY)
This modifier inverts the "greediness" of the quantifiers so that they are
not greedy by default, but become greedy if followed by ?. It is not compatible
with Perl. It can also be set by a (?U) modifier setting within the pattern
or by a question mark behind a quantifier (e.g. .*?).
S
When a pattern is going to be used several times, it is worth spending more
time analyzing it in order to speed up the time taken for matching. If this
modifier is set, then this extra analysis is performed. At present, studying
a pattern is useful only for non-anchored patterns that do not have a single
fixed starting character.
For more informations see http://www.php.net/manual/en/reference.pcre.pattern.modifiers.php
When you added the / ... /g , you added another regex delimiter plus the modifier g wich does not exists in PCRE, that's why it did not work.
I agree with #verdesmarald and used this pattern in the following function:
$string = preg_replace_callback(
"_(?:(?:https?|ftp)://)(?:\S+(?::\S*)?#)?(?:(?!10(?:\.\d{1,3}){3})(?!127(?:\.\d{1,3}){3})(?!169\.254(?:\.\d{1,3}){2})(?!192\.168(?:\.\d{1,3}){2})(?!172\.(?:1[6-9]|2\d|3[0-1])(?:\.\d{1,3}){2})(?:[1-9]\d?|1\d\d|2[01]\d|22[0-3])(?:\.(?:1?\d{1,2}|2[0-4]\d|25[0-5])){2}(?:\.(?:[1-9]\d?|1\d\d|2[0-4]\d|25[0-4]))|(?:(?:[a-z\x{00a1}-\x{ffff}0-9]+-?)*[a-z\x{00a1}-\x{ffff}0-9]+)(?:\.(?:[a-z\x{00a1}-\x{ffff}0-9]+-?)*[a-z\x{00a1}-\x{ffff}0-9]+)*(?:\.(?:[a-z\x{00a1}-\x{ffff}]{2,})))(?::\d{2,5})?(?:/[^\s]*)?_iuS",
create_function('$match','
$m = trim(strtolower($match[0]));
$m = str_replace("http://", "", $m);
$m = str_replace("https://", "", $m);
$m = str_replace("ftp://", "", $m);
$m = str_replace("www.", "", $m);
if (strlen($m) > 25)
{
$m = substr($m, 0, 25) . "...";
}
return "$m";
'), $string);
return $string;
It seem to do the trick, and resolve an issue I was having. As #verdesmarald said, removing the ^ and $ characters allowed the pattern to work even in my pre_replace_callback().
Only thing that concerns me, is how efficient is the pattern. If used in a busy/high traffic web app, could it cause a bottle neck?
UPDATE
The above regex pattern breaks if there is a trail dot at the end of the path section of a url, like so http://www.mydomain.com/page.. To solve this I modified the final part of the regex pattern by adding ^. making the final part look like so [^\s^.]. As I read it, do not match a trailing space or dot.
In my tests so far it seems to be working fine.
i'm trying to learn a bit about regex, can anyone explain to me what is going on here? And give example on a regex that would provide the expected output? Thanks!
input data = 'Sometext|even more text'
regex = '(.*)?\|?.*'
replacement = '$1'
expected output = 'Sometext'
actual output = 'Sometext|even more text'
PHP
preg_filter("(.*)?\|?.*", "$1", 'Sometext|even more text'); // returns Sometext|even more text
(.*) is greedy, so matches everything. $1 is everything then.
You are probably looking for:
/^([^|]*).*$/
Your regex is saying "all chars, followed by an optional |, followed by 0 or more chars".
Change the initial (.*) to ([^\|]*), or make the | non-optional.
* is greedy, which means it will try to match as much text as possible. In this case:
(.*)? will match all the text
\|?.* will match the "rest" (empty string)
try: regex = '\|[^|]*', replacement = ''
If you change your regex to (\w+)?\|?.*, specifically adding the + after the \w then you will get your expected answer of 'Sometext'.
The reason you were having the whole string match is that the first .* was matching the whole string. With the changes I have above, you will be matching on any word character.
Since I am completely useless at regex and this has been bugging me for the past half an hour, I think I'll post this up here as it's probably quite simple.
hey.exe
hey2.dll
pomp.jpg
In PHP I need to extract what's between the <a> tags example:
hey.exe
hey2.dll
pomp.jpg
Avoid using '.*' even if you make it ungreedy, until you have some more practice with RegEx. I think a good solution for you would be:
'/<a[^>]+>([^<]+)<\/a>/i'
Note the '/' delimiters - you must use the preg suite of regex functions in PHP. It would look like this:
preg_match_all($pattern, $string, $matches);
// matches get stored in '$matches' variable as an array
// matches in between the <a></a> tags will be in $matches[1]
print_r($matches);
This appears to work:
$pattern = '/<a.*?>(.*?)<\/a>/';
([^<]*)
I found this regular expression tester to be helpful.
Here is a very simple one:
<a.*>(.*)</a>
However, you should be careful if you have several matches in the same line, e.g.
hey.exehey2.dll
In this case, the correct regex would be:
<a.*?>(.*?)</a>
Note the '?' after the '*' quantifier. By default, quantifiers are greedy, which means they eat as much characters as they can (meaning they would return only "hey2.dll" in this example). By appending a quotation mark, you make them ungreedy, which should better fit your needs.