Asked  7 Months ago    Answers:  5   Viewed   35 times

Users can input URLs using a HTML form on my website, so they might enter something like this:, it can be anything. I need to extract the value of a certain query parameter, in this case 'test' (the value 123). Is there a way to do this?



You can use parse_url and parse_str like this:

$query = parse_url('', PHP_URL_QUERY);
parse_str($query, $params);
$test = $params['test'];

parse_url allows to split an URL in different parts (scheme, host, path, query, etc); here we use it to get only the query (test=123&random=abc). Then we can parse the query with parse_str.

Wednesday, March 31, 2021
answered 7 Months ago

This will work only for non-nested parentheses:

    $regex = <<<HERE
    /  "  ( (?:[^"\\]++|\\.)*+ ) "
     | '  ( (?:[^'\\]++|\\.)*+ ) '
     | ( ( [^)]*                  ) )
     | [s,]+

    $tags = preg_split($regex, $str, -1,
                       | PREG_SPLIT_DELIM_CAPTURE);

The ++ and *+ will consume as much as they can and give nothing back for backtracking. This technique is described in perlre(1) as the most efficient way to do this kind of matching.

Wednesday, March 31, 2021
answered 7 Months ago

The standard disclaimer applies: Parsing HTML with regular expressions is not ideal. Success depends on the well-formedness of the input on a character-by-character level. If you cannot guarantee this, the regex will fail to do the Right Thing at some point.

Having said that:

<ab[^>]*>(.*?)</a>   // match group one will contain the link text
Saturday, May 29, 2021
answered 5 Months ago

According to the docs of route object, you have access to a $route object from your components, which exposes what you need. In this case

//from your component
console.log(this.$route.query.test) // outputs 'yay'
Sunday, June 27, 2021
answered 4 Months ago

See urllib.urlparse

Example code:

from urlparse import urlparse
o = urlparse('')

url_without_query_string = o.scheme + "://" + o.netloc + o.path

Example output:

Python 2.6.1 (r261:67515, Jun 24 2010, 21:47:49) 
[GCC 4.2.1 (Apple Inc. build 5646)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> from urlparse import urlparse
>>> o = urlparse('')
>>> url_without_query_string = o.scheme + "://" + o.netloc + o.path
>>> print url_without_query_string
Sunday, September 19, 2021
Micah Alcorn
answered 1 Month ago
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