Related
I'm not very good at regular expressions at all.
I've been using a lot of framework code to date, but I'm unable to find one that is able to match a URL like http://www.example.com/etcetc, but it is also is able to catch something like www.example.com/etcetc and example.com/etcetc.
For matching all kinds of URLs, the following code should work:
<?php
$regex = "((https?|ftp)://)?"; // SCHEME
$regex .= "([a-z0-9+!*(),;?&=$_.-]+(:[a-z0-9+!*(),;?&=$_.-]+)?#)?"; // User and Pass
$regex .= "([a-z0-9\-\.]*)\.(([a-z]{2,4})|([0-9]{1,3}\.([0-9]{1,3})\.([0-9]{1,3})))"; // Host or IP address
$regex .= "(:[0-9]{2,5})?"; // Port
$regex .= "(/([a-z0-9+$_%-]\.?)+)*/?"; // Path
$regex .= "(\?[a-z+&\$_.-][a-z0-9;:#&%=+/$_.-]*)?"; // GET Query
$regex .= "(#[a-z_.-][a-z0-9+$%_.-]*)?"; // Anchor
?>
Then, the correct way to check against the regex is as follows:
<?php
if(preg_match("~^$regex$~i", 'www.example.com/etcetc', $m))
var_dump($m);
if(preg_match("~^$regex$~i", 'http://www.example.com/etcetc', $m))
var_dump($m);
?>
Courtesy: Comments made by splattermania in the PHP manual: preg_match
RegEx Demo in regex101
This worked for me in all cases I had tested:
$url_pattern = '/((http|https)\:\/\/)?[a-zA-Z0-9\.\/\?\:#\-_=#]+\.([a-zA-Z0-9\&\.\/\?\:#\-_=#])*/';
Tests:
http://test.test-75.1474.stackoverflow.com/
https://www.stackoverflow.com
https://www.stackoverflow.com/
http://wwww.stackoverflow.com/
http://wwww.stackoverflow.com
http://test.test-75.1474.stackoverflow.com/
http://www.stackoverflow.com
http://www.stackoverflow.com/
stackoverflow.com/
stackoverflow.com
http://www.example.com/etcetc
www.example.com/etcetc
example.com/etcetc
user:pass#example.com/etcetc
example.com/etcetc?query=aasd
example.com/etcetc?query=aasd&dest=asds
http://stackoverflow.com/questions/6427530/regular-expression-pattern-to-match-url-with-or-without-http-www
http://stackoverflow.com/questions/6427530/regular-expression-pattern-to-match-url-with-or-without-http-www/
Every valid Internet URL has at least one dot, so the above pattern will simply try to find any at least two strings chained by a dot and has valid characters that URL may have.
Try this:
/^http:\/\/|(www\.)?[a-z0-9]+([\-\.]{1}[a-z0-9]+)*\.[a-z]{2,5}(:[0-9]{1,5})?(\/.*)?$/
It works exactly like the people want.
It takes with or with out http://, https://, and www.
You can use a question mark after a regular expression to make it conditional so you would want to use:
http:\/\/(www\.)?
That will match anything that has either http://www. or http:// (with no www.)
You could just use a replace method to remove the above, thus getting you the domain. It depends on what you need the domain for.
Try something like this:
.*([\w-]+\.)+[a-z]{2,5}(/[\w-]+)*
Use:
/(https?://)?((?:(\w+-)*\w+)\.)+(?:[a-z]{2})(\/?\w?-?=?_?\??&?)+[\.]?([a-z0-9\?=&_\-%#])?/g
It matches something.com, http(s):// or www. It does not match other [something]:// URLs though, but for my purpose that's not necessary.
The regex matches e.g.:
http://foo.co.uk/
www.regex.com/foo.html?q=bar$some=thi-ng,regex
regex.foo.com/blog
You can try this:
r"(http[s]:\/\/)?([\w-]+\.)+([a-z]{2,5})(\/+\w+)? "
Selection:
may be start with http:// or https:// (optional)
anything (word) end with dot (.)
followed by 2 to 5 character [a-z]
followed by "/[anything]" (optional)
followed by space
Try this
$url_reg = /(ftp|https?):\/\/(\w+:?\w*#)?(\S+)(:[0-9]+)?(\/([\w#!:.?+=&%#!\/-])?)?/;
I have been using the following, which works for all my test cases, as well as fixes any issues where it would trigger at the end of a sentence preceded by a full-stop (end.), or where there were single character initials, such as 'C.C. Plumbing'.
The following regex contains multiple {2,}s, which means two or more matches of the previous pattern.
((http|https)\:\/\/)?[a-zA-Z0-9\.\/\?\:#\-_=#]{2,}\.([a-zA-Z0-9\&\.\/\?\:#\-_=#]){2,}
Matches URLs such as, but not limited to:
https://example.com
http://example.com
example.com
example.com/test
example.com?value=test
Does not match non-URLs such as, but not limited to:
C.C Plumber
A full-stop at the end of a sentence.
Single characters such as a.b or x.y
Please note: Due to the above, this will not match any single character URLs, such as: a.co, but it will match if it is preceded by a URL scheme, such as: http://a.co.
I was getting so many issues getting the answer from anubhava to work due to recent PHP allowing $ in strings and the preg match wasn't working.
Here is what I used:
// Regular expression
$re = '/((https?|ftp):\/\/)?([a-z0-9+!*(),;?&=.-]+(:[a-z0-9+!*(),;?&=.-]+)?#)?([a-z0-9\-\.]*)\.(([a-z]{2,4})|([0-9]{1,3}\.([0-9]{1,3})\.([0-9]{1,3})))(:[0-9]{2,5})?(\/([a-z0-9+%-]\.?)+)*\/?(\?[a-z+&$_.-][a-z0-9;:#&%=+\/.-]*)?(#[a-z_.-][a-z0-9+$%_.-]*)?/i';
// Match all
preg_match_all($re, $blob, $matches, PREG_SET_ORDER, 0);
// Print the entire match result
var_dump($matches);
// The first element of the array is the full match
This PHP Composer package URL highlight is doing a good job in PHP:
<?php
use VStelmakh\UrlHighlight\UrlHighlight;
$urlHighlight = new UrlHighlight();
$matches = $urlHighlight->getUrls($string);
?>
If it does not have to be regex, you could always use the validate filters that are in PHP.
filter_var('http://example.com', FILTER_VALIDATE_URL);
filter_var (mixed $variable [, int $filter = FILTER_DEFAULT [, mixed $options ]]);
Types of Filters
Validate Filters
Regex if you want to ensure a URL starts with HTTP/HTTPS:
https?:\/\/(www\.)?[-a-zA-Z0-9#:%._\+~#=]{1,256}\.[a-zA-Z0-9()]{1,6}\b([-a-zA-Z0-9()#:%_\+.~#?&//=]*)
If you do not require the HTTP protocol:
[-a-zA-Z0-9#:%._\+~#=]{1,256}\.[a-zA-Z0-9()]{1,6}\b([-a-zA-Z0-9()#:%_\+.~#?&//=]*)
I have a url that looks some what like this
for-sale/stuff/state/used-bla-bla2-bla3-bla4-(bla5)---f10-85934.html
i'm trying to validate the format, in my function using this regex.
if (preg_match('/(?:^|(?:\-))(\w+)/g', $pathInfo, $matches)) {
echo $digit = $matches[0];
}
$pathInfo is the url given above.
Basically i want to match
make sure the directory is for-sale/stuff/
used-bla-bla2-bla3-bla4-(bla5)---f10-85934.html file must start with either used/new and end with a integer.html
no spaces are allowed.
After i validate, i want to get the ID. which in this case is 85934
Seems like you want something like this,
'~^for-sale/stuff/\S+/(?:used|new)\S*?(\d+)\.html$~'
DEMO
I'd suggest this sample piece of code and the following regex:
$re = "~\\bfor\\-sale\\/stuff\\/[^<> ]*?\\/(?:used|new)[^/ ]*?\\-(\\d+)\\.html\\b~";
$str = "\n";
preg_match_all($re, $str, $matches);
Regex: \bfor\-sale\/stuff\/[^<> ]*?\/(?:used|new)[^/ ]*?\-(\d+)\.html\b
I assume you have several URLs to validate in a variable string of text, thus I sugget using \b, and that the URL is inside some tag, so I'd use [^<> ]*? in order to limit capturing to just inside a tag.
The ID will be in the first capturing group (captured by \d+).
Spaces are also disallowed: [^<> ]*?, [^/ ]*?.
I'm not very good at regular expressions at all.
I've been using a lot of framework code to date, but I'm unable to find one that is able to match a URL like http://www.example.com/etcetc, but it is also is able to catch something like www.example.com/etcetc and example.com/etcetc.
For matching all kinds of URLs, the following code should work:
<?php
$regex = "((https?|ftp)://)?"; // SCHEME
$regex .= "([a-z0-9+!*(),;?&=$_.-]+(:[a-z0-9+!*(),;?&=$_.-]+)?#)?"; // User and Pass
$regex .= "([a-z0-9\-\.]*)\.(([a-z]{2,4})|([0-9]{1,3}\.([0-9]{1,3})\.([0-9]{1,3})))"; // Host or IP address
$regex .= "(:[0-9]{2,5})?"; // Port
$regex .= "(/([a-z0-9+$_%-]\.?)+)*/?"; // Path
$regex .= "(\?[a-z+&\$_.-][a-z0-9;:#&%=+/$_.-]*)?"; // GET Query
$regex .= "(#[a-z_.-][a-z0-9+$%_.-]*)?"; // Anchor
?>
Then, the correct way to check against the regex is as follows:
<?php
if(preg_match("~^$regex$~i", 'www.example.com/etcetc', $m))
var_dump($m);
if(preg_match("~^$regex$~i", 'http://www.example.com/etcetc', $m))
var_dump($m);
?>
Courtesy: Comments made by splattermania in the PHP manual: preg_match
RegEx Demo in regex101
This worked for me in all cases I had tested:
$url_pattern = '/((http|https)\:\/\/)?[a-zA-Z0-9\.\/\?\:#\-_=#]+\.([a-zA-Z0-9\&\.\/\?\:#\-_=#])*/';
Tests:
http://test.test-75.1474.stackoverflow.com/
https://www.stackoverflow.com
https://www.stackoverflow.com/
http://wwww.stackoverflow.com/
http://wwww.stackoverflow.com
http://test.test-75.1474.stackoverflow.com/
http://www.stackoverflow.com
http://www.stackoverflow.com/
stackoverflow.com/
stackoverflow.com
http://www.example.com/etcetc
www.example.com/etcetc
example.com/etcetc
user:pass#example.com/etcetc
example.com/etcetc?query=aasd
example.com/etcetc?query=aasd&dest=asds
http://stackoverflow.com/questions/6427530/regular-expression-pattern-to-match-url-with-or-without-http-www
http://stackoverflow.com/questions/6427530/regular-expression-pattern-to-match-url-with-or-without-http-www/
Every valid Internet URL has at least one dot, so the above pattern will simply try to find any at least two strings chained by a dot and has valid characters that URL may have.
Try this:
/^http:\/\/|(www\.)?[a-z0-9]+([\-\.]{1}[a-z0-9]+)*\.[a-z]{2,5}(:[0-9]{1,5})?(\/.*)?$/
It works exactly like the people want.
It takes with or with out http://, https://, and www.
You can use a question mark after a regular expression to make it conditional so you would want to use:
http:\/\/(www\.)?
That will match anything that has either http://www. or http:// (with no www.)
You could just use a replace method to remove the above, thus getting you the domain. It depends on what you need the domain for.
Try something like this:
.*([\w-]+\.)+[a-z]{2,5}(/[\w-]+)*
Use:
/(https?://)?((?:(\w+-)*\w+)\.)+(?:[a-z]{2})(\/?\w?-?=?_?\??&?)+[\.]?([a-z0-9\?=&_\-%#])?/g
It matches something.com, http(s):// or www. It does not match other [something]:// URLs though, but for my purpose that's not necessary.
The regex matches e.g.:
http://foo.co.uk/
www.regex.com/foo.html?q=bar$some=thi-ng,regex
regex.foo.com/blog
You can try this:
r"(http[s]:\/\/)?([\w-]+\.)+([a-z]{2,5})(\/+\w+)? "
Selection:
may be start with http:// or https:// (optional)
anything (word) end with dot (.)
followed by 2 to 5 character [a-z]
followed by "/[anything]" (optional)
followed by space
Try this
$url_reg = /(ftp|https?):\/\/(\w+:?\w*#)?(\S+)(:[0-9]+)?(\/([\w#!:.?+=&%#!\/-])?)?/;
I have been using the following, which works for all my test cases, as well as fixes any issues where it would trigger at the end of a sentence preceded by a full-stop (end.), or where there were single character initials, such as 'C.C. Plumbing'.
The following regex contains multiple {2,}s, which means two or more matches of the previous pattern.
((http|https)\:\/\/)?[a-zA-Z0-9\.\/\?\:#\-_=#]{2,}\.([a-zA-Z0-9\&\.\/\?\:#\-_=#]){2,}
Matches URLs such as, but not limited to:
https://example.com
http://example.com
example.com
example.com/test
example.com?value=test
Does not match non-URLs such as, but not limited to:
C.C Plumber
A full-stop at the end of a sentence.
Single characters such as a.b or x.y
Please note: Due to the above, this will not match any single character URLs, such as: a.co, but it will match if it is preceded by a URL scheme, such as: http://a.co.
I was getting so many issues getting the answer from anubhava to work due to recent PHP allowing $ in strings and the preg match wasn't working.
Here is what I used:
// Regular expression
$re = '/((https?|ftp):\/\/)?([a-z0-9+!*(),;?&=.-]+(:[a-z0-9+!*(),;?&=.-]+)?#)?([a-z0-9\-\.]*)\.(([a-z]{2,4})|([0-9]{1,3}\.([0-9]{1,3})\.([0-9]{1,3})))(:[0-9]{2,5})?(\/([a-z0-9+%-]\.?)+)*\/?(\?[a-z+&$_.-][a-z0-9;:#&%=+\/.-]*)?(#[a-z_.-][a-z0-9+$%_.-]*)?/i';
// Match all
preg_match_all($re, $blob, $matches, PREG_SET_ORDER, 0);
// Print the entire match result
var_dump($matches);
// The first element of the array is the full match
This PHP Composer package URL highlight is doing a good job in PHP:
<?php
use VStelmakh\UrlHighlight\UrlHighlight;
$urlHighlight = new UrlHighlight();
$matches = $urlHighlight->getUrls($string);
?>
If it does not have to be regex, you could always use the validate filters that are in PHP.
filter_var('http://example.com', FILTER_VALIDATE_URL);
filter_var (mixed $variable [, int $filter = FILTER_DEFAULT [, mixed $options ]]);
Types of Filters
Validate Filters
Regex if you want to ensure a URL starts with HTTP/HTTPS:
https?:\/\/(www\.)?[-a-zA-Z0-9#:%._\+~#=]{1,256}\.[a-zA-Z0-9()]{1,6}\b([-a-zA-Z0-9()#:%_\+.~#?&//=]*)
If you do not require the HTTP protocol:
[-a-zA-Z0-9#:%._\+~#=]{1,256}\.[a-zA-Z0-9()]{1,6}\b([-a-zA-Z0-9()#:%_\+.~#?&//=]*)
I am trying to convert specific keywords in text, which are stored in array, to the links.
Example text:
$text='This text contains many keywords, but also formated keywords.'
So now I want to convert the word keywords to the #keywords.
I used the very simple preg_replace function
preg_replace('/keywords/i',' keywords ',$text);
but obviously it converts to link also the string already formatted as a link, so I get a messy html like:
$text='This text contains many keywords, but also formated keywords" title="keywords">keywords</a>.'
Expected result:
$text='This text contains many keywords, but also formated keywords.'
Any suggestions?
THX
EDIT
We are one step from the perfect function, but still not working well in this case:
$text='This text contains many keywords, but also formated
keywords.'
In this case it replaces also the word keywords in the href, so we again get the messy code like
keywords.com/keywords" title="keywords">keywords</a>
I'm not great with regular expressions, but maybe this one will work:
/[^#>"]keywords/i
What I think it will do is ignore any instances of #keywords, >keywords, and "keywords and find the rest.
EDIT:
After testing it out, it looks like that replaces the space before the word as well, and doesn't work if keywords is the beginning of the string. It also didn't preserve original capitalization. I have tested this one, and it works perfectly for me:
$string = "Keywords and keywords, plus some more keywords with the original keywords.";
$string = preg_replace("/(?<![#>\"])keywords/i", "$0", $string);
echo $string;
The first three are replaced, preserving the original capitalization, and the last one is left untouched. This one uses a negative lookbehind and backreferences.
EDIT 2:
OP edited question. With the new example provided, the following regex will work:
$string = 'This text contains many keywords, but also formated keywords.';
$string = preg_replace("/(?<![#>\".\/])keywords/i", "$0", $string);
echo $string;
// outputs: This text contains many keywords, but also formated keywords.
This will replace all instances of keywords that are not preceded by #, >, ", ., or /.
Here is the problem:
The keyword could be inside the href, the title, or the text of the link, and anywhere in there (like if the keyword was sanity and you already had href="insanity". Or even worse, you could have a non-keyword link that happens to contain a keyword, something like:
Click here to find more keywords and such!
In the above example, even though it fits every other possible criteria (it's got spaces before and after being the easiest one to test for), it still would result in a link within a link, which I think breaks the internet.
Because of this, you need to use lookaheads and lookbehinds to check if the keyword is wrapped in a link. But there is one catch: lookbehinds have to have a defined pattern (meaning no wild cards).
I thought I'd be the hero and show you the easy fix for your issue, which would be something to the effect of:
'/(?<!\<a.?>)[list|of|keywords](?!\<\/a>)/'
Except you can't do that because the lookbehind in this case has that wildcard. Without it, you end up with a super greedy expression.
So my proposed alternative is to use regex to find all link elements, then str_replace to swap them out with a placeholder, and then replacing them with the placeholder at the end.
Here's how I did it:
$text='This text contains many keywords, but also formated keywords.';
$keywords = array('text', 'formatted', 'keywords');
//This is just to make the regex easier
$keyword_list_pattern = '['. implode($keywords,"|") .']';
// First, get all matching keywords that are inside link elements
preg_match_all('/<a.*' . $keyword_list_pattern . '.*<\/a>/', $text, $links);
$links = array_unique($links[0]); // Cleaning up array for next step.
// Second, swap out all matches with a placeholder, and build restore array:
foreach($links as $count => $link) {
$link_key = "xxx_{$count}_xxx";
$restore_links[$link_key] = $link;
$text = str_replace($link, $link_key, $text);
}
// Third, we build a nice replacement array for the keywords:
foreach($keywords as $keyword) {
$keyword_links[$keyword] = "<a href='#$keyword'>$keyword</a>";
}
// Merge the restore links to the bottom of the keyword links for one mass replacement:
$keyword_links = array_merge($keyword_links, $restore_links);
$text = str_replace(array_keys($keyword_links), $keyword_links, $text);
echo $text;
You can change your RegEx so that it only targets keywords with a space in front. Since the formatted keywords do no contain a space. Here is an example.
$text = preg_replace('/ keywords/i',' keywords',$text);
I am trying to pull the anchor text from a link that is formatted this way:
<h3><b>File</b> : i_want_this</h3>
I want only the anchor text for the link : "i_want_this"
"variable_text" varies according to the filename so I need to ignore that.
I am using this regex:
<a href=\"\/en\/browse\/file\/variable_text\">(.*?)<\/a>
This is matching of course the complete link.
PHP uses a pretty close version to PCRE (PERL Regex). If you want to know a lot about regex, visit perlretut.org. Also, look into Regex generators like exspresso.
For your use, know that regex is greedy. That means that when you specify that you want something, follwed by anything (any repetitions) followed by something, it will keep on going until that second something is reached.
to be more clear, what you want is this:
<a href="
any character, any number of times (regex = .* )
">
any character, any number of times (regex = .* )
</a>
beyond that, you want to capture the second group of "any character, any number of times". You can do that using what are called capture groups (capture anything inside of parenthesis as a group for reference later, also called back references).
I would also look into named subpatterns, too - with those, you can reference your choice with a human readable string rather than an array index. Syntax for those in PHP are (?P<name>pattern) where name is the name you want and pattern is the actual regex. I'll use that below.
So all that being said, here's the "lazy web" for your regex:
<?php
$str = '<h3><b>File</b> : i_want_this</h3>';
$regex = '/(<a href\=".*">)(?P<target>.*)(<\/a>)/';
preg_match($regex, $str, $matches);
print $matches['target'];
?>
//This should output "i_want_this"
Oh, and one final thought. Depending on what you are doing exactly, you may want to look into SimpleXML instead of using regex for this. This would probably require that the tags that we see are just snippits of a larger whole as SimpleXML requires well-formed XML (or XHTML).
I'm sure someone will probably have a more elegant solution, but I think this will do what you want to done.
Where:
$subject = "<h3><b>File</b> : i_want_this</h3>";
Option 1:
$pattern1 = '/(<a href=")(.*)(">)(.*)(<\/a>)/i';
preg_match($pattern1, $subject, $matches1);
print($matches1[4]);
Option 2:
$pattern2 = '()(.*)()';
ereg($pattern2, $subject, $matches2);
print($matches2[4]);
Do not use regex to parse HTML. Use a DOM parser. Specify the language you're using, too.
Since it's in a captured group and since you claim it's matching, you should be able to reference it through $1 or \1 depending on the language.
$blah = preg_match( $pattern, $subject, $matches );
print_r($matches);
The thing to remember is that regex's return everything you searched for if it matches. You need to specify that only care about the part you've surrounded in parenthesis (the anchor text). I'm not sure what language you're using the regex in, but here's an example in Ruby:
string = 'i_want_this'
data = string.match(/<a href=\"\/en\/browse\/file\/variable_text\">(.*?)<\/a>/)
puts data # => outputs 'i_want_this'
If you specify what you want in parenthesis, you can reference it:
string = 'i_want_this'
data = string.match(/<a href=\"\/en\/browse\/file\/variable_text\">(.*?)<\/a>/)[1]
puts data # => outputs 'i_want_this'
Perl will have you use $1 instead of [1] like this:
$string = 'i_want_this';
$string =~ m/<a href=\"\/en\/browse\/file\/variable_text\">(.*?)<\/a>/;
$data = $1;
print $data . "\n";
Hope that helps.
I'm not 100% sure if I understand what you want. This will match the content between the anchor tags. The URL must start with /en/browse/file/, but may end with anything.
#(.*?)#
I used # as a delimiter as it made it clearer. It'll also help if you put them in single quotes instead of double quotes so you don't have to escape anything at all.
If you want to limit to numbers instead, you can use:
#(.*?)#
If it should have just 5 numbers:
#(.*?)#
If it should have between 3 and 6 numbers:
#(.*?)#
If it should have more than 2 numbers:
#(.*?)#
This should work:
<a href="[^"]*">([^<]*)
this says that take EVERYTHING you find until you meet "
[^"]*
same! take everything with you till you meet <
[^<]*
The paratese around [^<]*
([^<]*)
group it! so you can collect that data in PHP! If you look in the PHP manual om preg_match you will se many fine examples there!
Good luck!
And for your concrete example:
<a href="/en/browse/file/variable_text">([^<]*)
I use
[^<]*
because in some examples...
.*?
can be extremely slow! Shoudln't use that if you can use
[^<]*
You should use the tool Expresso for creating regular expression... Pretty handy..
http://www.ultrapico.com/Expresso.htm