I got this code:
if(preg_match("/^(?:https?:\/\/)?(?:www\.)?youtu(.be\/|be\.com\/watch\?v=)(\w{11})$/", $url)){
preg_match("/^(?:https?:\/\/)?(?:www\.)?youtu(.be\/|be\.com\/watch\?v=)(\w{11})$/", $url, $matches);
$vid = str_replace(' ', '', $matches[0]);
}
That pretty much checks if the URL is a Youtube video. How do I assign the last 11 characters from the URL to the $vid variable?
Example:
URL: https://www.youtube.com/watch?v=ASDASDASDAS
$vid = ASDASDASDAS
First, you don't want to check for the last 11 characters, because the video ID might be changed any time to a different length. What you should always do is make something flexible enough. So in this case, you should check for ?.*v=([^&]+), since that will get the match until the next &. So your code would look like this:
//make this a variable, since you're using it multiple times.
$re = "/^(?:https?:\/\/)?(?:www\.)?youtu(.be|be\.com)\/watch\?.*?v=([^&#]+).*$/";
if (preg_match($re, $url)){
preg_match($re, $url, $matches);
$vid = $matches[2]; //the 2nd group in the match, so the 2nd set of ()s
}
In that code, video urls that contain other URL parameters will also still work, and it's much more flexible with the length of the video ID.
Demo - look to the right where it says 2., it will mention the video ID there (which is what's put inside $vid).
The parse_str and parse_url functions will do the job ( no need for regular expressions ):
$link = 'https://www.youtube.com/watch?v=ASDASDASDAS';
parse_str( parse_url( $link, PHP_URL_QUERY ), $query );
print_r( $query );
/*
Array
(
[v] => ASDASDASDAS
)
*/
Perhaps something like:
$vid = substr($url,strlen($url)-11);
The straightforward (correct) answers have been made, but I would encourage you to take a look at oembed standard and library for php like https://code.google.com/p/php-oembed/. This would help you get info from almost all image/video services out there, generate embed code and more.
Related
I have a url from where i am fetching value using GET method and i want to replace that value with the 4 digit random number in the string like
This is my main URL:
http://localhost/ab/index.php?id=3345
These are the strings in my table (fetching from database):
http://anyurl/index.php?id=4876&abc=any
http://anyurl/index.php?id=8726&abc=any
http://anyurl/index.php?id=9026&abc=any
So whenever i open the main url the id's of the table should be replaced according to the main url
you can get the id parameter using global GET variable
$id = $_GET["id"]
then you can change the urls in the table according to it
$url = "http://anyurl/index.php?id=".$id."&abc=any"
Hope this will help you
If you want to replace the id with preg_replace in string then you can do like below:
<?php
$string = 'http://anyurl/index.php?id=4876&abc=any';
$new_string = preg_replace('/[0-9]+/', $_GET["id"], $string);
echo $new_string;
// Will display http://anyurl/index.php?id=3345&abc=any
?>
I know that it was asked years ago, but, probably, someone will find my solution helpful
So, I also offer to use preg_replace(), as Amit Gupta, but improve it for cases when you could have other numbers before ID value:
$url = 'http://anyurl/index.php?foo=0713&id=4876&abc=any';
$new_id = $_GET['id'];
// regex: catch 1 or more digits after 'id='
$new_url = preg_replace( '/id=(\d+)/', $new_id, $url );
If $_GET['id'] is 4920, for example, $new_url will be equal to http://anyurl/index.php?foo=0713&id=4920&abc=any
My url like this:
http://mywebsite.com/movies/937-lan-kwai-fong-2?file=Rae-Ingram&q=
http://mywebsite.com/movies/937-big-daddy?file=something&q=
I want to get "lan-kwai-fong-2" and "big-daddy", so I use this code but it doesn't work. Please help me fix it ! If you can shorten it, it is so great !
$url= $_SERVER['REQUEST_URI'];
preg_replace('/\?file.*/','',$url);
preg_match('/[a-z][\w\-]+$/',$url,$matches);
$matches= str_replace("-"," ",$matches[0]);
First there are issue with your code which im going to go over because they are general things:
preg_replace does not work by reference so you are never actually modifying the url. You need to assign the result of the replace to a variable:
// this would ovewrite the current value of url with the replaced value
$url = preg_replace('/\?file.*/','',$url);
It is possible that preg_match will not find anything so you need to test the result
// it should also be noted that sometimes you may need a more exact test here
// because it can return false (if theres an error) or 0 (if there is no match)
if (preg_match('/[a-z][\w\-]+$/',$url,$matches)) {
// do stuff
}
Now with that out of the way you are making this more difficult than it needs to be. There are specific function for working with urls parse_url and parse_str.
You can use these to easily work with the information:
$urlInfo = parse_url($_SERVER['REQUEST_URI']);
$movie = basename($urlInfo['path']); // yields 937-the-movie-title
Just replace
preg_replace('/\?file.*/','',$url);
with
$url= preg_replace('/\?file.*/','',$url);
Regex works, and parse_url is the right way to do it. But for something quick and dirty I would usually use explode. I think it's clearer.
#list($path, $query) = explode("?", $url, 2); // separate path from query
$match = array_pop(explode("/", $path)); // get last part of path
How about this:
$url = $_SERVER['REQUEST_URI'];
preg_match('/\/[^-]+-([^?]+)\?/', $url, $matches);
$str = isset($matches[1]) ? $matches[1] : false;`
match last '/'
match anything besides '-' until '-'
capture anything besides '?' until (not including) '?'
I have a server application which looks up where the stress is in Russian words. The end user writes a word жажда. The server downloads a page from another server which contains the stresses indicated with apostrophes for each case/declension like this жа'жда. I need to find that word in the downloaded page.
In Russian the stress is always written after a vowel. I've been using so far a regex that is a grouping of all possible combinations (жа'жда|жажда'). Is there a more elegant solution using just a regex pattern instead of making a PHP script which creates all these combinations?
EDIT:
I have a word жажда
The downloaded page contains the string жа'жда. (notice the
apostrophe, I do not before-hand know where the apostrophe in the
word is)
I want to match the word with apostrophe (жа'жда).
P.S.: So far I have a PHP script creating the string (жа'жда|жажда') used in regex (apostrophe is only after vowels) which matches it. My goal is to get rid of this script and use just regex in case it's possible.
If I understand your question,
have these options (d'isorder|di'sorder|dis'order|diso'rder|disor'der|disord'er|disorde'r|disorder') and one of these is in the downloaded page and I need to find out which one it is
this may suit your needs:
<pre>
<?php
$s = "d'isorder|di'sorder|dis'order|diso'rder|disor'der|disord'er|disorde'r|disorder'|disorde'";
$s = explode("|",$s);
print_r($s);
$matches = preg_grep("#[aeiou]'#", $s);
print_r($matches);
running example: https://eval.in/207282
Uhm... Is this ok with you?
<?php
function find_stresses($word, $haystack) {
$pattern = preg_replace('/[aeiou]/', '\0\'?', $word);
$pattern = "/\b$pattern\b/";
// word = 'disorder', pattern = "diso'?rde'?r"
preg_match_all($pattern, $haystack, $matches);
return $matches[0];
}
$hay = "something diso'rder somethingelse";
find_stresses('disorder', $hay);
// => array(diso'rder)
You didn't specify if there can be more than one match, but if not, you could use preg_match instead of preg_match_all (faster). For example, in Italian language we have àncora and ancòra :P
Obviously if you use preg_match, the result would be a string instead of an array.
Based, on your code, and the requirements that no function is called and disorder is excluded. I think this is what you want. I have added a test vector.
<pre>
<?php
// test code
$downloadedPage = "
there is some disorde'r
there is some disord'er in the example
there is some di'sorder in the example
there also' is some order in the example
there is some disorder in the example
there is some dso'rder in the example
";
$word = 'disorder';
preg_match_all("#".preg_replace("#[aeiou]#", "$0'?", $word)."#iu"
, $downloadedPage
, $result
);
print_r($result);
$result = preg_grep("#'#"
, $result[0]
);
print_r($result);
// the code you need
$word = 'also';
preg_match("#".preg_replace("#[aeiou]#", "$0'?", $word)."#iu"
, $downloadedPage
, $result
);
print_r($result);
$result = preg_grep("#'#"
, $result
);
print_r($result);
Working demo: https://eval.in/207312
I need to get the vine video id from the url
so the output from link like this
https://vine.co/v/bXidIgMnIPJ
be like this
bXidIgMnIPJ
I tried to use code form other question here for Vimeo (NOT VINE)
Get img thumbnails from Vimeo?
This what I tried to use but I did not succeed
$url = 'https://vine.co/v/bXidIgMnIPJ';
preg_replace('~^https://(?:www\.)?vine\.co/(?:clip:)?(\d+)~','$1',$url)
basename maybe?
<?php
$url = 'https://vine.co/v/bXidIgMnIPJ';
var_dump(basename($url));
http://codepad.org/vZiFP27y
Assuming it will always be in that format, you can just split the url by the / delimiter. Regex is not needed for a simple url such as this.
$id = end(explode('/', $url));
Referring to as the question is asked here is a solution for preg_replace:
$s = 'https://vine.co/v/bXidIgMnIPJ';
$new_s = preg_replace('/^.*\//','',$s);
echo $new_s;
// => bXidIgMnIPJ
or if you need to validate that an input string is indeed a link to vine.co :
$new_s = preg_replace('/^(https?:\/\/)?(www\.)?vine\.co.*\//','',$s);
I don't know if that /v/ part is always present or is it always v... if it is then it may also be added to regex for stricter validation:
$new_s = preg_replace('/^(https?:\/\/)?(www\.)?vine\.co\/v\//','',$s);
Here's what I am using:
function getVineId($url) {
preg_match("#(?<=vine.co/v/)[0-9A-Za-z]+#", $url, $matches);
if (isset($matches[0])) {
return $matches[0];
}
return false;
}
I used a look-behind to ensure "vine.co/v/" always precedes the ID, while ignoring if the url is HTTP or HTTPS (or if it lacks a protocol altogether). It assumes the ID is alphanumeric, of any length. It will ignore any characters or parameters after the id (like Google campaign tracking parameters, etc).
I used the "#" delimiter so I wouldn't have to escape the forward slashes (/), for a cleaner look.
explode the string with '/' and the last string is what you are looking for :) Code:
$vars = explode("/",$url);
echo $vars[count($vars)-1];
$url = 'https://vine.co/v/b2PFre2auF5';
$regex = '/^http(?:s?):\/\/(?:www\.)?vine\.co\/v\/([a-zA-Z0-9]{1,13})$/';
preg_match($regex,$url,$m);
print_r($m);
1. b2PFre2auF5
if I have this url: node/95/pdf/1. How will I able to get the numeric/value 1? Tried the parse_url but gave me the wrong output.
PS: the value 1 is just an example, the id is dynamic depends on what the user click.
I would use sscanf
Untested example:
list($node_id, $pdf_id) = sscanf($url, "node/%d/pdf/%d");
$node_id contains the node id, $pdf_id contains the pdf id. According to your comment: Yes, you can output it with e.g. echo $pdf_id;.
If you need them both in an array, you can remove the list() method, doing it like this:
$ids = sscanf($url, "node/%d/pdf/%d");.
This returns an array with both node and pdf id in $ids.
Finally, if you just need the pdf id, you could do
$id = sscanf($url, "node/95/pdf/%d");.
I just showed how to fetch both because I assumed you may need both numbers from your url.
Edit
seeing all the other answers after posting my solution, I am wondering why everyone is solving this with multiple functions when there is a function available that does exactly what he needs: parsing a string according to a format. This also leads to less sql-injection prone code IMHO. And it doesn't break something when the url gets extended or query strings are appended.
Edit 2
list($node_id, $sub, $sub_id) = sscanf($url, "node/%d/%[^/]/%d"); will get you the "pdf" and it's id separate instead of "node/%d/%s/%d". This is because char / is also matched by %s. Using %[^/] matches everything except the forward slash.
You can do this:
$id = end(explode('/', 'node/95/pdf/1'));
Example:
$arr = explode('/', 'node/95/pdf/1');
$id = end($arr);
echo $id; // 1
$url = "node/95/pdf/1";
// Find the last occurence of a slash, get everything after that.
$id = substr($url, strrpos($url, "/") + 1 );
Try with:
$input = 'node/95/pdf/1';
$parts = explode('/', $input);
$output = (int) $parts[3];