Grab remaining text after last “/” in a php string

Question

So, lets say I have a $somestring thats holds the value \"main/physician/physician_view\".

I want to grab just \"physician_view\". I want it to also wo

Answer 1

You can use strrpos() to find the last occurence of one string in another:

substr($somestring, strrpos($somestring, '/') + 1)

Answer 2

$last_part = substr(strrchr($somestring, "/"), 1);

Examples:

php > $a = "main/physician/physician_view";
php > $b = "main/physician_view";
php > $c = "site/main/physician/physician_view";
php > echo substr(strrchr($a, "/"), 1);
    physician_view
php > echo substr(strrchr($b, "/"), 1);
    physician_view
php > echo substr(strrchr($c, "/"), 1);
    physician_view

Answer 3

For another one liner, you can use the explode trick and reverse the array:

current(array_reverse(explode('/',$url)));

Answer 4

Simply you can use :

$id = substr( $url, strrpos( $url, '/' )+1 );

Answer 5

There are many ways to do this. I would probably use:

array_pop(explode('/', $string));

Answer 6

The other solutions don't always work, or are inefficient. Here is a more useful general utility function that always works, and can be used with other search terms.

/**
 * Get a substring starting from the last occurrence of a character/string
 *
 * @param  string $str The subject string
 * @param  string $last Search the subject for this string, and start the substring after the last occurrence of it.
 * @return string A substring from the last occurrence of $startAfter, to the end of the subject string.  If $startAfter is not present in the subject, the subject is returned whole.
 */
function substrAfter($str, $last) {
    $startPos = strrpos($str, $last);
    if ($startPos !== false) {
        $startPos++;
        return ($startPos < strlen($str)) ? substr($str, $startPos) : '';
    }
    return $str;
}


// Examples

substrAfter('main/physician/physician_view', '/');  // 'physician_view'

substrAfter('main/physician/physician_view/', '/'); // '' (empty string)

substrAfter('main_physician_physician_view', '/');  // 'main_physician_physician_view'