PHP Cut URL from some variable query

Question

if I have some URL like these:

http://localhost/myfile.php?start=2018&end=2019&page=1&status=

Did anynone know how to replace any words from that link that contain page= ?

The result that I want is :

http://localhost/myfile.php?start=2018&end=2019&status=

Even if the link is like this:

http://localhost/myfile.php?start=2018&end=2019&status=&page=3

I still want the result will be without page variable and value, so it might be: http://localhost/myfile.php?start=2018&end=2019&status=

any idea how to do this? Especially without regex (regex is my last option)

Without regex ? It's very complicate and can't be sure to 100% — Inazo, Aug 31 '18 at 13:29
Is it a real URL or you are getting it as a string? agreed with @Inazo — Alive to die - Anant, Aug 31 '18 at 13:30
Without regex I suppose you could tokenize the string, read it character-by-character and parse it out into a data structure based on those tokens, remove what you don't want, then serialize that data structure back into a URL. — David, Aug 31 '18 at 13:31
PHP already has functions for [parsing urls](http://php.net/manual/en/function.parse-url.php) and [building query strings](http://php.net/manual/en/function.http-build-query.php). Use them. — iainn, Aug 31 '18 at 13:33
sorry, its a URL.. regex is last option, if it cant use only some basename function — Cross Vander, Aug 31 '18 at 13:35
@Blackbam oh sorry... can't find that before... that post is what I need! thank you so much — Cross Vander, Aug 31 '18 at 13:39

score 5 · Accepted Answer · answered Aug 31 '18 at 13:37

You can use parse_url(), parse_str() and http_build_query():

// Your URL
$str = 'http://localhost/myfile.php?start=2018&end=2019&page=1&status=';

// Get parts
$parts = parse_url($str);

// Get array of arguments
parse_str($parts['query'], $args);

// Remove unwanted index
unset($args['page']);

// Rebuild your URL
echo $parts['scheme'] . '://' . $parts['host'] . $parts['path'] . '?' . http_build_query($args);

// Output: http://localhost/myfile.php?start=2018&end=2019&status=

I encouraged to read documentation or to print_r() vars of this sample code for a better understanding.

PHP Cut URL from some variable query

1 Answers1