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('~((?<!:)/(?=/)|\\|.+)~', '', $rawURL);
Basically, what this does is look for any forward slash that IS NOT preceded by a colon (:), and IS followed bij another forward slash. It also searches for any pipe character and any character following it.
Anything found is removed from the result.
I can explain the RegEx in more detail if you like.
Related
Sorry if the title is confusing. All I'm trying to do is some simple regex:
The text: /thing/images/info.gif
And what I want is: info
My regex (not fully working): ([^\/]+$)(.*?)(?=\.gif)
(Note: [^\/]+$ returns info.gif)
Thanks for any help!
I'd say you don't need to match all the string, so you can be much more generic. If you know your string always contains a path you can just use:
preg_match( '/([^\/]+)\.\w+$/', "/thing/images/info.gif", $matches) ;
print_r( $matches );
and it will be valid for any filename, even names that contains dots like my_file.name.jpg or spaces like /thing/images/my image.gif
Demo here.
The structure is (from the end of the regex moving to the left):
Match before the end of the string
any number of characters preceded by a dot
any character that is not a slash (your filename, if there is a slash, there starts the directories)
Not sure how much more complex the string is but this seems to work on the test string:
preg_match('![^/.]+(?=\.gif)!', '/thing/images/info.gif', $m);
Matching NOT / NOT . followed by .gif.
In editors (Sublime):
Find:^(.*)(\/)(.*)(\.)(.*)$
Replace it with:\3
In PHP:
<?php
preg_match('/^(.*)(\/)(.*)(\.)(.*)$/', '/thing/images/info.gif', $match);
echo $match[3];
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'm working in PHP with friendly URL paths in the form of:
/2011/09/here-is-the-title
/2011/09/here-is-the-title/2
I need to standardize these URL paths to remove anything after the 4 slash including the slash itself. The value after the 4th slash is sometimes a number, but can also be any parameter.
Any thoughts on how I could do this? I imagine regex could handle it, but I'm terrible with it. I also thought a combination of strpos and substr might be able to handle it, but cannot quite figure it out.
You can use explode() function:
$parts = explode('/', '/2011/09/here-is-the-title/2');
$output = implode('/', array_slice($parts, 0, 4));
Replace
%^((/[^/]*){3}).*%g
with $1.
see http://regexr.com?2vlr8 for a live example
If your regex implementation support arbitrary length look-behind assertions you could replace
(?<=^[^/]*(/[^/]*){3})/.*$
with an empty string.
If it does not, you can replace
^([^/]*(?:/[^/]*){3})/.*$
with the contents of the first capturing group. A PHP example for the second one can be found at ideone.com.
you could also use a loop:
result="";
for char c in URL:
if(c is a slash) count++;
if(count<4) result=result+c;
else break;
I'm trying to strip part of a string (which happens to be a url) with Regex. I'm getting better out regex but can't figure out how to tell it that content before or after the string is optional. Here is what I have
$string='http://www.example.com/username?refid=22';
$new_string= preg_replace('/[/?refid=0-9]+/', '', $string);
echo $new_string;
I'm trying to remove the ?refid=22 part to get http://www.example.com/username
Ideas?
EDIT
I think I need to use Regex instead of explode becuase sometimes the url looks like http://example.com/profile.php?id=9999&refid=22 In this case I also want to remove the refid but not get id=9999
parse_url() is good for parsing URLs :)
$string = 'http://www.example.com/username?refid=22';
$url = parse_url($string);
// Ditch the query.
unset($url['query']);
echo array_shift($url) . '://' . implode($url);
CodePad.
Output
http://www.example.com/username
If you only wanted to remove that specific GET param, do this...
parse_str($url['query'], $get);
unset($get['refid']);
$url['query'] = http_build_query($get);
CodePad.
Output
http://example.com/profile.php?id=9999
If you have the extension, you can rebuild the URL with http_build_url().
Otherwise you can make assumptions about username/password/port and build it yourself.
Update
Just for fun, here is the correction for your regular expression.
preg_replace('/\?refid=\d+\z/', '', $string);
[] is a character class. You were trying to put a specific order of characters in there.
\ is the escape character, not /.
\d is a short version of the character class [0-9].
I put the last character anchor (\z) there because it appears it will always be at the end of your string. If not, remove it.
Dont use regexs if you dont have to
echo current( explode( '?', $string ) );
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