www.example.com/12-23-34
for e.g. the above written is my page. Now what I need is to get 12, 23, and 34 seperatley. How can I achieve that?
Looked for the solution, but everywhere I got the solution to get 12-23-34 together.
once you get "12-23-34", do an $myArray = explode("-",$myString), where $myArray will have 3 item : 12, 23, and 34...
You first need to get the current URI.
$uri = $_SERVER['REQUEST_URI'];
This will get you /12-23-34.
Ignore the / by getting a substring.
$sub = substr($uri, 1);
Explode the substring into parts.
$parts = explode('-', $sub);
Loop through the parts to do what you want.
foreach ($parts as $part) {
}
Here is solution for you:
$subURL = explode('/', $url); // separates url into www.example.com and 12-23-34
$numbers = explode('-', $subURL[1]); // separates 12, 23 and 34.
echo $numbers[0]; // returns 12
echo $numbers[1]; // returns 23
echo $numbers[2]; // returns 34
As mentioned remove the part you want - the path in your url and then explode it. You need to be sure your url always has a similar pattern.
$url = "www.example.com/12-23-34";
$pos = strrpos($url, "/");
$path = substr($url,$pos+1);
$my_numbers = explode("-",$path);
foreach ($my_numbers as $my_number) {
print "$my_number<br>";
}
Here is what I've used:
<?php
$value = basename(__FILE__, '.php');
echo $value;
$numbers = explode('-', $value);
echo $numbers[0];
echo $numbers[1];
echo $numbers[2];
?>
Related
I have a string $current_url that can contain 2 different values:
http://url.com/index.php&lang=en
or
http://url.com/index.php&lang=jp
in both cases I need to strip the query part so I get: http://url.com/index.php
How can I do this in php?
Thank you.
Simplest Solution
$url = 'http://url.com/index.php&lang=en';
$array = explode('&', $url);
echo $new_url =$array[0];
To only remove the lang query do this
$url = 'http://url.com/index.php&lang=en&id=1';
$array = explode('&lang=en', $url);
echo $new_url = $array[0] .''.$array[1];
//output http://url.com/index.php&id=1
So this way it only removes the lang query and keep other queries
If the value of your lang parameter is always of length 2, which should be the case for languages, you could use:
if(strpos($current_url, '&lang=') !== false){
$current_url = str_replace(substr($current_url, strpos($current_url, '&lang='), 8), '', $current_url);
}
If the substring "&lang=" is present in $current_url, it removes a substring of length 8, starting at the "&lang=" position. So it basically removes "&lang=" plus the 2 following chars.
You can Use strtok to remove the query string from url.
<?php
echo $url=strtok('http://url.com/index.php&lang=jp','&');
?>
DEMO
Answer based on comment.
You can use preg_replace
https://www.codexworld.com/how-to/remove-specific-parameter-from-url-query-string-php/
<?php
$url = 'http://url.com/index.php?page=site&lang=jp';
function remove_query_string($url_name, $key) {
$url = preg_replace('/(?:&|(\?))' . $key . '=[^&]*(?(1)&|)?/i', "$1", $url_name);
$url = rtrim($url, '?');
$url = rtrim($url, '&');
return $url;
}
echo remove_query_string($url, 'lang');
?>
DEMO
How to get string after second slash in url? URL is different every time (more slashes), but every time I need the whole text after the second slash. How to do it?
I am using this code:
<?php
$str = "$_SERVER[HTTP_HOST]$_SERVER[REQUEST_URI]";
$last = substr($str, strrpos($str, '/') - 1);
echo $last;
?>
...but it works online with some characters after slash.
Thank you very much for help.
$last = explode("/", $str, 3);
echo $last[2];
<?php
$str = "google.com/whatever/hello";
$last = GetStringAfterSecondSlashInURL($str);
echo $last;
function GetStringAfterSecondSlashInURL($the_url)
{
$parts = explode("/",$the_url,3);
if(isset($parts[2]))
return $parts[2];
}
?>
use
<?php
$str = "$_SERVER[HTTP_HOST]$_SERVER[REQUEST_URI]";
$last = explode("/",$str,3);// so at second index rest of the string will come.
echo $last[2];
see here http://www.w3schools.com/php/func_string_explode.asp
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];
This question already has answers here:
Strip off specific parameter from URL's querystring
(22 answers)
Closed 8 years ago.
I have to remove the last element in a string. I used rtrim in php but it is not working.
This is the string:
/search/listing.html?vehicle_type=&year=&make_name=&model_name=&loc_type=3&zipcode=641004&distance=100&make_order=ASC
I need to remove "&make_order=ASC"
Can anyone help me?
$s = '/search/listing.html?vehicle_type=&year=&make_name=&model_name=&loc_type=3&zipcode=641004&distance=100&make_order=ASC';
echo substr($s, 0, strrpos($s, '&'));
Edit:
$url = $base_url.trim( $_SERVER['REQUEST_URI'], "&year_order=".$arr['year_order']."" );
// ^
// |_ replace , with .
trim should work:
$string = "/search/listing.html?vehicle_type=&year=&make_name=&model_name=&loc_type=3&zipcode=641004&distance=100&make_order=ASC";
$string = trim($string, "&make_order=ASC");
There's no guarantee that make_order will be at the end of the query string - or exist at all. To remove the field properly, you'd have to use something like this:
$url = '/search/listing.html?vehicle_type=&year=&make_name=&model_name=&loc_type=3&zipcode=641004&distance=100&make_order=ASC';
// break down the URL into a path and query string
$parsed = parse_url($url);
// turn the query string into an array that we can manipulate
$qs = array();
parse_str($parsed['query'], $qs);
// remove the unwanted field
unset($qs['make_order']);
// rebuild the URL
$rebuilt = $parsed['path'];
if(!empty($qs)) {
$rebuilt .= '?' . http_build_query($qs);
}
echo $rebuilt;
$actual_link = "/search/listing.html?vehicle_type=&year=&make_name=&model_name=&loc_type=3&zipcode=641004&distance=100&make_order=ASC";
echo str_replace("&make_order=ASC","",$actual_link);
$string = "/search/listing.html?vehicle_type=&year=&make_name=&model_name=&loc_type=3&zipcode=641004&distance=100&make_order=ASC";
$args = array_pop(explode($string, "&"));
$string = implode("&", $args);
There are a bunch of ways. The easiest might be:
$i=strrpos($text,'&');
$newstring=substr($text,0,$i);
$str = "/search/listing.html?vehicle_type=&year=&make_name=&model_name=&loc_type=3&zipcode=641004&distance=100&make_order=ASC";
echo $str . "<br>";
echo trim($str,"&make_order=ASC");
if &make_order=ASC is always going to be at the end, you can use strstr() to do this
$str = '/search/listing.html?vehicle_type=&year=&make_name=&model_name=&loc_type=3&zipcode=641004&distance=100&make_order=ASC';
echo strstr($str,'&make_order=ASC',true);
Remove desired key from url.
Use:
$s = '/search/listing.html?vehicle_type=&year=&make_name=&model_name=&loc_type=3&zipcode=641004&distance=100&make_order=ASC';
echo remove_key_from_url($url, 'make_order');
Output :
/search/listing.html?vehicle_type=&year=&make_name=&model_name=&loc_type=3&zipcode=641004&distance=100
Code:
function remove_key_from_url($url, $key) {
if (strpos($url, '?') === false) return $url;
list($left, $right) = explode('?', $url, 2);
parse_str($right, $get);
if (isset($get[$key])) unset($get[$key]);
return $left . '?' . http_build_query($get);
}