how to get id url with preg_replace.
this is the link:
http://www.DDDD.com.br/photo/5b87f8eaa7c20f79c3257eb3ec0a35e0/id how do I get the id? in the case would be: 5b87f8eaa7c20f79c3257eb3ec0a35e0
In this case I recommend not to use preg_match (preg_replace would be used to replace something.
Simply use
$array = explode('/',$_SERVER['REQUEST_URI']);
$id = $array[1];
If you must use preg_match:
$array = array();
preg_match('#^/photo/([0-9a-f]{32})/id$#',$_SERVER['REQUEST_URI'],$array);
$id = $array[1];
You can do this easily using strripos to find the last / in the URL.
$url = $_SERVER['REQUEST_URI'];
if (($pos = strripos($url, '/')) !== false) {
$id = substr($url, $pos + 1);
}
else {
trigger_error('You must supply a valid photo ID');
}
If you would like to just extract that id string, you can use:
$id_url = "http://www.DDDD.com.br/photo/5b87f8eaa7c20f79c3257eb3ec0a35e0/id";
$pattern = "/photo\/([a-zA-Z0-9]*)/";
preg_match($pattern, $id_url, $output_array);
echo $output_array[1];
Or, to make the replacement:
$id_url = "http://www.DDDD.com.br/photo/5b87f8eaa7c20f79c3257eb3ec0a35e0/id";
$pattern = "/photo\/([a-zA-Z0-9]*)/";
$replacement = "your replacement";
$replaced_url = preg_replace($pattern, $replacement, $id_url);
echo $replaced_url;
PHP Live Regex - a useful tool for testing your patterns
Related
I had a php file already using regex to extract m3u8 link from youtube, which was working fine until last week.
http://server.com/youtube.php?id=youtbueid
use to pass the youtube id like this.
$string = get_data('https://www.youtube.com/watch?v=' . $channelid);
if(preg_match('#"hlsManifestUrl.":."(.*?m3u8)#', $string, $match)) {
$var1=$match[1];
$var1=str_replace("\/", "/", $var1);
$man = get_data($var1);
//echo $man;
preg_match_all('/(https:\/.*\/95\/.*index.m3u8)/U',$man,$matches, PREG_PATTERN_ORDER);
$var2=$matches[1][0];
header("Content-type: application/vnd.apple.mpegurl");
header("Location: $var2");
}
else {
preg_match_all('#itag.":([^,]+),."url.":."(.*?).".*?qualityLabel.":."(.*?)p."#', $string, $match);
//preg_match_all('#itag.":([^,]+),."url.":."(.*?).".*?bitrate.":.([^,]+),#', $string, $match);
$filter_keys = array_filter($match[3], function($element) {
return $element <= 720;
});
//print_r($filter_keys);
$max_key = array_keys($filter_keys, max($filter_keys))[0];
//print_r($max_key);
$urls = $match[2];
foreach($urls as &$url) {
$url = str_replace('\/', '/', $url);
$url = str_replace('\\\u0026', '&', $url);
}
print_r($urls[$max_key]);
header('location: ' . $urls[$max_key]);
How do I solve this problem?
Based on this post, I'm guessing that the desired URLs might look like:
and we can write a simple expression such as:
(.+\?v=)(.+)
We can also add more boundaries to it, if it was necessary.
RegEx
If this expression wasn't desired, you can modify/change your expressions in regex101.com.
RegEx Circuit
You can also visualize your expressions in jex.im:
PHP Test
$re = '/(.+\?v=)(.+)/m';
$str = ' https://www.youtube.com/watch?v=_Gtc-GtLlTk';
$subst = '$2';
$result = preg_replace($re, $subst, $str);
echo $result;
JavaScript Demo
This snippet shows that we likely have a valid expression:
const regex = /(.+\?v=)(.+)/gm;
const str = ` https://www.youtube.com/watch?v=_Gtc-GtLlTk`;
const subst = `$2`;
// The substituted value will be contained in the result variable
const result = str.replace(regex, subst);
console.log('Substitution result: ', result);
It might seem easy to do but I have trouble extracting this string. I have a string that has # tags in it and I'm trying to pull the tags maps/place/Residences+Jardins+de+Majorelle/#33.536759,-7.613825,17z/data=!3m1!4b1!4m2!3m1!1s0xda62d6053931323:0x2f978f4d1aabb1aa
And here is what I want to extract 33.536759,-7.613825,17z :
$var = preg_match_all("/#(\w*)/",$path,$query);
Any way I can do this? Much appreciated.
Change your regex to this one: /#([\w\d\.\,-]*)/.
This will return the string beginning with #.
$string = 'maps/place/Residences+Jardins+de+Majorelle/#33.536759,-7.613825,17z/data=!3m1!4b1!4m2!3m1!1s0xda62d6053931323:0x2f978f4d1aabb1aa';
$string = explode('/',$string);
//$coordinates = substr($string[3], 1);
//print_r($coordinates);
foreach ($string as $substring) {
if (substr( $substring, 0, 1 ) === "#") {
$coordinates = $substring;
}
}
echo $coordinates;
This is working for me:
$path = "maps/place/Residences+Jardins+de+Majorelle/#33.536759,-7.613825,17z/data=!3m1!4b1!4m2!3m1!1s0xda62d6053931323:0x2f978f4d1aabb1aa";
$var = preg_match_all("/#([^\/]+)/",$path,$query);
print $query[1][0];
A regex would do.
/#(-*\d+\.\d+),(-*\d\.\d+,\d+z*)/
If there is only one # and the string ends with / you can use the following code:
//String
$string = 'maps/place/Residences+Jardins+de+Majorelle/#33.536759,-7.613825,17z/data=!3m1!4b1!4m2!3m1!1s0xda62d6053931323:0x2f978f4d1aabb1aa';
//Save string after the first #
$coordinates = strstr($string, '#');
//Remove #
$coordinates = str_replace('#', '', $coordinates);
//Separate string on every /
$coordinates = explode('/', $coordinates );
//Save first part
$coordinates = $coordinates[0];
//Do what you want
echo $coordinates;
do like this
$re = '/#((.*?),-(.*?),)/mi';
$str = 'maps/place/Residences+Jardins+de+Majorelle/#33.536759,-7.613825,17z/data=!3m1!4b1!4m2!3m1!1s0xda62d6053931323:0x2f978f4d1aabb1aa';
preg_match_all($re, $str, $matches);
echo $matches[2][0].'<br>';
echo $matches[3][0];
output
33.536759
7.613825
I have the following url : http:example.com/country/France/45.
With the pattern http:example.com/country/name/**NUMBER**(?$_GET possibly).
How can I extract the number with a regex (or something else then regex) ?
With regexp:
$str = 'http:example.com/country/France/45';
preg_match('/http:example\.com\/country\/(?P<name>\w+)\/(?P<id>\d+)/', $str, $matches);
print_r($matches); // return array("name"=>"France", "id" => 45);
$url = 'http:example.com/country/France/45';
$id = end(explode('/',trim($url,'/')));
Simple isn't ?
The usage of trim () is to remove trailing \
echo $last = substr(strrchr($url, "/"), 1 );
strrchr() will give last occurence of the / character and then substr() gives string after it.
use something like this
$url = "http://example.com/country/France/45";
$parts = explode('/', $url);
$number = $parts[count($parts) - 1];
and if you have GET variable at the end, you can explode further like this
$number = explode('?', $number);
$number = $number[0];
hope this helps :)
the get command for php is $_GET so to show to number do
<html>
<body>
<?php
echo $_GET["eg"];
?>
</body>
</html>
with a URL of http:example.com/country/name/?eg=**NUMBER**
Use explode():
$parts = explode('/', $url);
$number = $parts[count($parts)-1];
How can I extract 4 from this string?
$string = "Rank_1:1:4";
I'm trying to get pagerank from Googles server, and the last value (4) is the actual pagerank.
Try
$string = "Rank_1:1:4";
$data = explode(':',$string);
echo end($data);
EDIT
as per #MichaelHampton, if they add more fields later, then use as below
$string = "Rank_1:1:4";
$data = explode(':',$string);
echo $data[2];
PHP has so many string function you can use ...
Variables
$find = ":";
$string = "Rank_1:1:4";
Using substr
echo substr($string, strrpos($string, $find) + 1);
Using strrchr
echo ltrim(strrchr($string, $find),$find);
$pattern = '/:\d+$/';
preg_match($pattern, $string, $matches);
$rank = substr($matches[0],1);
I have url like this
/cp/foo-bar/another-testing
how to parse it with the pattern
/cp/{0}-{1}/{2}
results will be
0:foo
1:bar
2:another-testing
I need a global solution to parse all kind of url with a pattern like that. I mean using {0}, {1} flag.
if (preg_match('#/cp/([^/]+?)-([^/]+?)/([^/]+)#'), $url, $matches)) {
//look into $matches[1], $matches[2] and $matches[3]
}
Instead of using {0}, {1}, {2}, I offer a new way: using {$s[0]}, {$s[1]}, {$s[2]}:
$your_url = '/cp/foo-bar/another-testing';
$s = explode('/', $your_url);
if(!$s[0])
array_shift($s);
if($temp = array_pop($s))
$s[] = $temp;
//then
$result = "/cp/{$s[0]}-{$s[1]}/{$s[2]}";