如何获取URL的查询字符串（问号后面的部分）？

Question

How can I get only the content from a query string staring from ?:

asegment?param1=value1&param2=value2

NOTE: I only need all the content after the "?", also I want to ommit that character.

Answer 1

Here is the Regex \?(.+), and a Rubular to prove it. And to use it:

preg_match("/\?(.+)/", $inputString, $matches);
if (!$matches) {
    // no matches
}

$result = $matches[1];