使用PHP找到URL的一部分

Question

把這個域名：

http://www.?.co.uk/elderly-care-advocacy/mental-capacity-act-advance-medical-directive.html

我如何使用PHP來找到第一和第二斜線之間的一切，無論它是否改變或沒有？

即，老年保健倡導

任何helo將不勝感激。

Answer 1

//strip the "http://" part. Note: Doesn't work for HTTPS! 
$url = substr("http://www.example.com/elderly-care-advocacy/mental-capacity-act-advance-medical-directive.html", 7); 

// split the URL in parts 
$parts = explode("/", $url); 

// The second part (offset 1) is the part we look for 
if (count($parts) > 1) { 
    $segment = $parts[1]; 
} else { 
    throw new Exception("Full URLs please!"); 
} 

Answer 2

我認爲Regular Expression應該沒問題。

嘗試使用例如：/[^/]+/應該給你/elderly-care-advocacy/作爲你的例子中數組的第二個索引。

（第一個字符串是/www.?.com/）

Answer 3

了我的頭頂部：

$url = http://www.example.co.uk/elderly-care-advocacy/mental-capacity-act-advance-medical-directive.html 
$urlParts = parse_url($url); // An array 
$target_string = $urlParts[1] // 'elderly-care-advocacy' 

乾杯

Answer 4

explode('/', $a);

Answer 5

所有你應該這樣做，首先解析url，然後爆炸字符串並獲得第一部分。隨着一些理智的檢查，將樂像以下：

$url = 'http://www.?.co.uk/elderly-care-advocacy/mental-capacity-act-advance-medical-directive.html'; 
$url_parts = parse_url($url); 
if (isset($url_parts['path'])) { 
    $path_components = explode('/', $ul_parts['path']); 
    if (count($path_components) > 1) { 
     // All is OK. Path's first component is in $path_components[0] 
    } else { 
     // Throw an error, since there is no directory specified in path 
     // Or you could assume, that $path_components[0] is the actual path 
    } 
} else { 
    // Throw an error, since there is no path component was found 
} 

Answer 6

$url = "http://www.example.co.uk/elderly-care-advocacy/mental-capacity-act-advance-medical-directive.html"; 
$parts = parse_url($url); 
$host = $parts['host']; 
$path = $parts['path']; 

$items = preg_split('/\//',$path,null,PREG_SPLIT_NO_EMPTY); 

$firstPart = $items[0]; 

Answer 7

Parse_URL是您最佳的選擇。它將URL字符串分解爲組件，您可以選擇查詢。

此功能可用於：

function extract_domain($url){ 

    if ($url_parts = parse_url($url), $prefix = 'www.', $suffix = '.co.uk') { 
     $host = $url_parts['host']; 
     $host = str_replace($prefix,'',$host); 
     $host = str_replace($suffix,'',$host); 
     return $host; 
    } 
    return false; 
} 

$host_component = extract_domain($_SERVER['REQUEST_URI']); 

Answer 8

我很驚訝過，但這個工程。

$url='http://www.?.co.uk/elderly-care-advocacy/...' 
$result=explode('/',$url)[3];