Here is the subject:
http://www.mysite.com/files/get/937IPiztQG/the-blah-blah-text-i-dont-need.mov
What I need using regex is only the bit before the last / (including that last / too)
The 937IPiztQG string may change; it will contain a-z A-Z 0-9 - _
Here's what I tried:
$code = strstr($url, '/http:\/\/www\.mysite\.com\/files\/get\/([A-Za-z0-9]+)./');
EDIT: I need to use regex because I don't actually know the URL. I have string like this...
a song
more text
oh and here goes some more blah blah
I need it to read that string and cut off filename part of the URLs.
You really don't need a regexp here. Here is a simple solution:
echo basename(dirname('http://www.mysite.com/files/get/937IPiztQG/the-blah-blah-text-i-dont-need.mov'));
// echoes "937IPiztQG"
Also, I'd like to quote Jamie Zawinski:
"Some people, when confronted with a problem, think 'I know, I'll use regular expressions.' Now they have two problems."
This seems far too simple to use regex. Use something similar to strrpos to look for the last occurrence of the '/' character, and then use substr to trim the string.
/http:\/\/www.mysite.com\/files\/get\/([^/]+)\/
How about something like this? Which should capture anything that's not a /, 1 or more times before a /.
The greediness of regexp will assure this works fine ^.*/
The strstr() function does not use a regular expression for any of its arguments it's the wrong function for regex replacement.
Are you thinking of preg_replace()?
But a function like basename() would be more appropriate.
Try this
$ok=preg_match('#mysite\.com/files/get/([^/]*)#i',$url,$m);
if($ok) $code=$m[1];
Then give a good read to these pages
http://www.php.net/preg_match
preg_replace
Note
the use of "#" as a delimiter to avoid getting trapped into escaping too many "/"
the "i" flag making match insensitive
(allowing more liberal spellings of the MySite.com domain name)
the $m array of captured results
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.
This is the text sample:
$text = "asd dasjfd fdsfsd http://11111.com/asdasd/?s=423%423%2F gfsdf http://22222.com/asdasd/?s=423%423%2F
asdfggasd http://3333333.com/asdasd/?s=423%423%2F";
This is my regex pattern:
preg_match_all( "#http:\/\/(.*?)[\s|\n]#is", $text, $m );
That match the first two urls, but how do I match the last one? I tried adding [\s|\n|$] but that will also only match the first two urls.
Don't try to match \n (there's no line break after all!) and instead use $ (which will match to the end of the string).
Edit:
I'd love to hear why my initial idea doesn't work, so in case you know it, let me know. I'd guess because [] tries to match one character, while end of line isn't one? :)
This one will work:
preg_match_all('#http://(\S+)#is', $text, $m);
Note that you don't have to escape the / due to them not being the delimiting character, but you'd have to escape the \ as you're using double quotes (so the string is parsed). Instead I used single quotes for this.
I'm not familar with PHP, so I don't have the exact syntax, but maybe this will give you something to try. the [] means a character class so |$ will literally look for a $. I think what you'll need is another look ahead so something like this:
#http:\/\/(.*)(?=(\s|$))
I apologize if this is way off, but maybe it will give you another angle to try.
See What is the best regular expression to check if a string is a valid URL?
It has some very long regular expressions that will match all urls.
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
I'm new to preg_replace() and I've been trying to get this to work, I couldn't so StackOverflow is my last chance.
I have a string with a few of these:
('pm_IDHERE', 'NameHere');">
I want it to be replaced with nothing, so it would require 2 wildcards for NameHere and pm_IDHERE.
But I've tried it and failed myself, so could someone give me the right code please, and thanks :)
Update:
You are almost there, you just have to make the replacement an empty string and escape the parenthesis properly, otherwise they will be treated as capture group (which you don't need btw):
$str = preg_replace("#\('pm_.+?', '.*?'\);#si", "", $str);
You probably also don't need the modifiers s and i but that is up to you.
Old answer:
Probably str_replace() is sufficient:
$str = "Some string that contains pm_IDHERE and NameHere";
$str = str_replace(array('pm_IDHERE', 'NameHere'), '', $str);
If this is not what you mean and pm_IDHERE is actually something like pm_1564 then yes, you probably need regular expressions for that. But if NameHere has no actual pattern or structure, you cannot replace it with regular expression.
And you definitely have to explain better what kind of string you have and what kind of string you have want to replace.
Since I am completely useless at regex and this has been bugging me for the past half an hour, I think I'll post this up here as it's probably quite simple.
hey.exe
hey2.dll
pomp.jpg
In PHP I need to extract what's between the <a> tags example:
hey.exe
hey2.dll
pomp.jpg
Avoid using '.*' even if you make it ungreedy, until you have some more practice with RegEx. I think a good solution for you would be:
'/<a[^>]+>([^<]+)<\/a>/i'
Note the '/' delimiters - you must use the preg suite of regex functions in PHP. It would look like this:
preg_match_all($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.