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
Related
I'm trying to retrieve the followed by count on my instagram page. I can't seem to get the Regex right and would very much appreciate some help.
Here's what I'm looking for:
y":{"count":
That's the beginning of the string, and I want the 4 numbers after that.
$string = preg_replace("{y"\"count":([0-9]+)\}","",$code);
Someone suggested this ^ but I can't get the formatting right...
You haven't posted your strings so it is a guess to what the regex should be... so I'll answer on why your codes fail.
preg_replace('"followed_by":{"count":\d')
This is very far from the correct preg_replace usage. You need to give it the replacement string and the string to search on. See http://php.net/manual/en/function.preg-replace.php
Your second usage:
$string = preg_replace(/^y":{"count[0-9]/","",$code);
Is closer but preg_replace is global so this is searching your whole file (or it would if not for the anchor) and will replace the found value with nothing. What your really want (I think) is to use preg_match.
$string = preg_match('/y":\{"count(\d{4})/"', $code, $match);
$counted = $match[1];
This presumes your regex was kind of correct already.
Per your update:
Demo: https://regex101.com/r/aR2iU2/1
$code = 'y":{"count:1234';
$string = preg_match('/y":\{"count:(\d{4})/', $code, $match);
$counted = $match[1];
echo $counted;
PHP Demo: https://eval.in/489436
I removed the ^ which requires the regex starts at the start of your string, escaped the { and made the\d be 4 characters long. The () is a capture group and stores whatever is found inside of it, in this case the 4 numbers.
Also if this isn't just for learning you should be prepared for this to stop working at some point as the service provider may change the format. The API is a safer route to go.
This regexp should capture value you're looking for in the first group:
\{"count":([0-9]+)\}
Use it with preg_match_all function to easily capture what you want into array (you're using preg_replace which isn't for retrieving data but for... well replacing it).
Your regexp isn't working because you didn't escaped curly brackets. And also you didn't put count quantifier (plus sign in my example) so it would only capture first digit anyway.
I found a regex pattern for PHP that does the exact OPPOSITE of what I'm needing, and I'm wondering how I can reverse it?
Let's say I have the following text: Item_154 ($12)
This pattern /\((.*?)\)/ gets what's inside the parenthesis, but I need to get "Item_154" and cut out what's in parenthesis and the space before the parenthesis.
Anybody know how I can do that?
Regex is above my head apparently...
/^([^( ]*)/
Match everything from the start of the string until the first space or (.
If the item you need to match can have spaces in it, and you only want to get rid of whitespace immediately before the parenthetical, then you can use this instead:
/^([^(]*?)\s*\(/
The following will match anything that looks like text (...) but returns just the text part in the match.
\w+(?=\s*\([^)]*\))
Explanation:
The \w includes alphanumeric and underscore, with + saying match one or more.
The (?= ) group is positive lookahead, saying "confirm this exists but don't match it".
Then we have \s for whitespace, and * saying zero or more.
The \( and \) matches literal ( and ) characters (since its normally a special chat).
The [^)] is anything non-) character, and again * is zero or more.
Hopefully all makes sense?
/(.*)\(.*\)/
What is not in () will now be your 1st match :)
One site that really helped me was http://gskinner.com/RegExr/
It'll let you build a regex and then paste in some sample targets/text to test it against, highlighting matches. All of the possible regex components are listed on the right with (essentially) a tooltip describing the function.
<?php
$string = 'Item_154 ($12)';
$pattern = '/(.*)\(.*?\)/';
preg_match($pattern, $string, $matches);
var_dump($matches[1]);
?>
Should get you Item_154
The following regex works for your string as a replacement if that helps? :-
\s*\(.*?\)
Here's an explanation of what's it doing...
Whitespace, any number of repetitions - \s*
Literal - \(
Any character, any number of repetitions, as few as possible - .*?
Literal - \)
I've found Expresso (http://www.ultrapico.com/) is the best way of learning/working out regular expressions.
HTH
Here is a one-shot to do the whole thing
$text = 'Item_154 ($12)';
$text = preg_replace('/([^\s]*)\s(\()[^)]*(\))/', $1$2$3, $text);
var_dump($text);
//Outputs: Item_154()
Keep in mind that using any PCRE functions involves a fair amount of overhead, so if you are using something like this in a long loop and the text is simple, you could probably do something like this with substr/strpos and then concat the parens on to the end since you know that they should be empty anyway.
That said, if you are looking to learn REGEXs and be productive with them, I would suggest checking out: http://rexv.org
I've found the PCRE tool there to very useful, though it can be quirky in certain ways. In particular, any examples that you work with there should only use single quotes if possible, as it doesn't work with double quotes correctly.
Also, to really get a grip on how to use regexs, I would check out Mastering Regular Expressions by Jeffrey Friedl ISBN-13:978-0596528126
Since you are using PHP, I would try to get the 3rd Edition since it has a section specifically on PHP PCRE. Just make sure to read the first 6 chapters first since they give you the foundation needed to work with the material in that particular chapter. If you see the 2nd Edition on the cheap somewhere, that pretty much the same core material, so it would be a good buy as well.
I'm trying to extract only certain elements of a string using regular expressions and I want to end up with only the captured groups.
For example, I'd like to run something like (is|a) on a string like "This is a test" and be able to return only "is is a". The only way I can partially do it now is if I find the entire beginning and end of the string but don't capture it:
.*?(is|a).*? replaced with $1
However, when I do this, only the characters preceding the final found/captured group are eliminated--everything after the last found group remains.
is is a test.
How can I isolate and replace only the captured strings (so that I end up with "is is a"), in both PHP and Perl?
Thanks!
Edit:
I see now that it's better to use m// rather than s///, but how can I apply that to PHP's preg_match? In my real regex I have several captured group, resulting in $1, $2, $3 etc -- preg_match only deals with one captured group, right?
If all you want are the matches, the there is no need for the s/// operator. You should use m//. You might want to expand on your explanation a little if the example below does not meet your needs:
#!/usr/bin/perl
use strict;
use warnings;
my $text = 'This is a test';
my #matches = ( $text =~ /(is|a)/g );
print "#matches\n";
__END__
C:\Temp> t.pl
is is a
EDIT: For PHP, you should use preg_match_all and specify an array to hold the match results as shown in the documentation.
You can't replace only captures. s/// always replaces everything included in the match. You need to either capture the additional items and include them in the replacement or use assertions to require things that aren't included in the match.
That said, I don't think that's what you're really asking. Is Sinan's answer what you're after?
You put everything into captures and then replaces only the ones you want.
(.*?)(is|a)(.*?)
demo:
$str = 'bcs >Hello >If see below!';
$repstr = preg_replace('/>[A-Z0-9].*?see below[^,\.<]*/','',$str);
echo $repstr;
What I want this tiny programme to output is "bcs >Hello ",but in fact it's only "bcs "
What's wrong with my pattern?
I think the problem is that you're misinterpreting how a non-greedy quantifier acts. Once it's in operation, yes, it stops earlier than it would otherwise. But it isn't aware of what comes before it (or potentially the text that comes later, either). It's only concerned with it's current position. Hence, the regular expression you posted will match all of:
">Hello >If see below!"
Let's see how this works:
/>[A-Z0-9].*?see below[^,\.<]*/
The regex first looks for ">" in "bcs >Hello >If see below!", and finds the first one, which is the one right before "Hello". Ok, let's check the next part of the expression:
[A-Z0-9]
The next char is a H, which matches the pattern [A-Z0-9]. Still good! Next:
.*?
Now we match all non-newline chars until we get to the first instance to match the remaining expressions of "see below[^,.<]*". If we had used just a plain greedy quantifier, we could match through multiple cases of "see below[^,.<]*" until we matched the last possible one. (So if your string had continued on, and there'd been other text match that pattern, it would have captured that as well) The non-greedy quantifier doesn't mean that your whole pattern will return the smallest possible match of all possible matches in the string. It just dictates how that particular character match functions.
You might want to try the following pattern then:
/>[A-Z0-9][^>]*?see below[^,\.<]*/
Hopefully this clears it up!
Why don't you write it like this:
$str = 'bcs >Hello >If see below!';
$repstr = preg_replace('/>If see below[^,\.<]*/','',$str);
echo $repstr;
This might be a good alternative to what you have.
The problem with your regexp is that instead of selecting what you want, you are selecting what you don't want and replacing that with an empty string.
The best approach, in my opinion, is selecting what you want, that is what the code below does. What you end up with is what is what is matched by the first sub-pattern otherwise you get your string back.
$str = 'bcs >Hello >If see below!';
$repstr = preg_replace('/^([\w]+ >[\w]+).*?see below.*?$/i', '$1', $str);
var_dump($repstr);
I hope this helps.
Hi Guys I'm very new to regex, can you help me with this.
I have a string like this "<input attribute='value' >" where attribute='value' could be anything and I want to get do a preg_replace to get just <input />
How do I specify a wildcard to replace any number of any characters in a srting?
like this? preg_replace("/<input.*>/",$replacement,$string);
Many thanks
What you have:
.*
will match "any character, and as many as possible.
what you mean is
[^>]+
which translates to "any character, thats not a ">", and there must be at least one
or altertaively,
.*?
which means
"any character, but only enough to make this rule work"
BUT DONT
Parsing HTML with regexps is Bad
use any of the existing html parsers, DOM librarys, anything, Just NOT NAïVE REGEX
For example:
<foo attr=">">
Will get grabbed wrongly by regex as
'<foo attr=" ' with following text of '">'
Which will lead you to this regex:
`<[a-zA-Z]+( [a-zA-Z]+=['"][^"']['"])*)> etc etc
at which point you'll discover this lovely gem:
<foo attr="'>\'\"">
and your head will explode.
( the syntax highlighter verifies my point, and incorrectly matches thinking i've ended the tag. )
Some people were close... but not 100%:
This:
preg_replace("<input[^>]*>", $replacement, $string);
should be this:
preg_replace("<input[^>]*?>", $replacement, $string);
You don't want that to be a greedy match.
preg_replace("<input[^>]*>", $replacement, $string);
// [^>] means "any character except the greater than symbol / right tag bracket"
This is really basic stuff, you should catch up with some reading. :-)
If I understand the question correctly, you have the code:
preg_replace("/<input.*>/",$replacement,$string);
and you want us to tell you what you should use for $replacement to delete what was matched by .*
You have to go about this the other way around. Use capturing groups to capture what you want to keep, and reinsert that into the replacement. E.g.:
preg_replace("/(<input).*(>)/","$1$2",$string);
Of course, you don't really need capturing groups here, as you're only reinserting literal text. Bet the above shows the technique, in case you want to do this in a situation where the tag can vary. This is a better solution:
preg_replace("/<input [^>]*>/","<input />",$string);
The negated character class is more specific than the dot. This regex will work if there are two HTML tags in the string. Your original regex won't.