Users can input URLs using a HTML form on my website, so they might enter something like this: http://www.example.com?test=123&random=abc, 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?
用户可以在我的网站上使用HTML表单输入url,因此他们可以输入如下内容:http://www.example.com?test= 123andrandom =abc,它可以是任何东西。我需要提取某个查询参数的值,在本例中是“test”(值123)。有办法吗?
3 个解决方案
#1
43
You can use parse_url
and parse_str
like this:
您可以这样使用parse_url和parse_str:
$query = parse_url('http://www.example.com?test=123&random=abc', PHP_URL_QUERY);
parse_str($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
.
parse_url允许将URL分割到不同的部分(方案、主机、路径、查询等);这里我们只使用它来获取查询(test=123&random=abc)。然后我们可以使用parse_str解析查询。
#2
2
I needed to check an url that was relative for our system so I couldn't use parse_str. For anyone who needs it:
我需要检查一个相对于系统的url,所以不能使用parse_str。对于任何需要它的人:
$urlParts = null;
preg_match_all("~[\?&]([^&]+)=([^&]+)~", $url, $urlParts);
#3
0
the hostname is optional but is required at least the question mark at the begin of parameter string:
主机名是可选的,但在参数字符串开始时至少需要问号:
$inputString = '?test=123&random=abc&usersList[]=1&usersList[]=2' ;
parse_str ( parse_url ( $inputString , PHP_URL_QUERY ) , $params );
print_r ( $params );
#1
43
You can use parse_url
and parse_str
like this:
您可以这样使用parse_url和parse_str:
$query = parse_url('http://www.example.com?test=123&random=abc', PHP_URL_QUERY);
parse_str($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
.
parse_url允许将URL分割到不同的部分(方案、主机、路径、查询等);这里我们只使用它来获取查询(test=123&random=abc)。然后我们可以使用parse_str解析查询。
#2
2
I needed to check an url that was relative for our system so I couldn't use parse_str. For anyone who needs it:
我需要检查一个相对于系统的url,所以不能使用parse_str。对于任何需要它的人:
$urlParts = null;
preg_match_all("~[\?&]([^&]+)=([^&]+)~", $url, $urlParts);
#3
0
the hostname is optional but is required at least the question mark at the begin of parameter string:
主机名是可选的,但在参数字符串开始时至少需要问号:
$inputString = '?test=123&random=abc&usersList[]=1&usersList[]=2' ;
parse_str ( parse_url ( $inputString , PHP_URL_QUERY ) , $params );
print_r ( $params );