如何从URL中获取所有可能的目录组合

Question

我很难找出算法来检测URL列表中的重复目录模式，任何人都可以为此提供一种方法吗？林相当肯定它将需要一个递归调用，但我不能决定如何为每个可能的模式保存记录。

注意：这是在PHP中。

Lests说你有一些网址：

1. http://www.goodfood.com/recipes/special_occasion/desserts/pie/chocolate-pie.html 
2. http://www.goodfood.com/recipes/special_occasion/desserts/pie/cherry-pie.html 
3. http://www.goodfood.com/recipes/special_occasion/apps/chex-mix.html 
4. http://www.goodfood.com/recipes/special_occasion/soup/tomato.html 
5. http://www.goodfood.com/special/special_occasion/soup/beef-stew.html 
6. http://www.goodfood.com/special/special_occasion/soup/vegetable.html 

我想找到一种方法来确定一个以上的网址有目录的所有可能的模式。因此，其结果将是这个样子：

'recipes/special_occasion' is found in urls 1, 2, 3 and 4. 
'recipes/special_occasion/desserts' is found in urls 1, and 2. 
'recipes/special_occasion/desserts/pie' is found in urls 1, and 2. 
'special_occasion/desserts/pie' is found in urls 1, and 2. 
'desserts/pie' is found in urls 1, and 2. 
'special_occasion/desserts' is found in urls 1, and 2. 
'special_occasion/desserts/pie' is found in urls 1, and 2. 
'special/special_occasion' is found in urls 5, and 6. 
'special/special_occasion/soup' is found in urls 5, and 6. 
'special_occasion/soup' is found in urls 5, and 6. 

我的想法是要经过的每个网址，并拉出每一个可能的新格局，并将其存储在数组中。到目前为止，我有： $ commonDomains = array（）;

 foreach($query AS $row) { 


     $urlPath = parse_url($row['href'], PHP_URL_PATH); 
     echo "$urlPath<br/>"; 

     $urlChunks = explode('/', $urlPath); 
     //var_dump($urlChunks); 

     foreach($urlChunks AS $domain) { 
      if(strlen($domain) > 0) { 
       $thisDomain = $domain.'/'; 
       $commonDomains[$thisDomain][] = $row['id']; 
      } 
     } 
     var_dump($commonDomains); 
    } 

有没有人跑过这个呢？它尖叫着我的模式，但我无法在网上找到答案。我想到的一切都非常复杂。请帮忙，谢谢。

---

我有什么即时通讯工作的一个例子：http://phpfiddle.org/main/code/kn4-zyh

我的继承人的结果为止

/recipes/special_occasion/desserts/pie/grandmas-chocolate-pie.html 
array(5) { [0]=> string(7) "recipes" [1]=> string(16) "special_occasion" [2]=> string(8) "desserts" [3]=> string(3) "pie" [4]=> string(27) "grandmas-chocolate-pie.html" } 

0 : 4 : recipes/special_occasion/desserts/pie/grandmas-chocolate-pie.html 
0 : 3 : recipes/special_occasion/desserts/pie 
0 : 2 : recipes/special_occasion/desserts 
0 : 1 : recipes/special_occasion 

1 : 4 : special_occasion/desserts/pie/grandmas-chocolate-pie.html 
2 : 4 : desserts/pie/grandmas-chocolate-pie.html 
3 : 4 : pie/grandmas-chocolate-pie.html 

0 : 4 : recipes/special_occasion/desserts/pie/grandmas-chocolate-pie.html 
1 : 3 : special_occasion/desserts/pie 


**Im missing: 
2 : 3 : special_occasion/desserts 
1 : 2 : recipes/special_occasion 

**

Answer 1

实例搜索一个目录：

$links = array(
    'http://www.goodfood.com/recipes/special_occasion/desserts/pie/chocolate-pie.html', 
    'http://www.goodfood.com/recipes/special_occasion/desserts/pie/cherry-pie.html', 
    'http://www.goodfood.com/recipes/special_occasion/apps/chex-mix.html', 
    'http://www.goodfood.com/recipes/special_occasion/soup/tomato.html', 
    'http://www.goodfood.com/special/special_occasion/soup/beef-stew.html', 
    'http://www.goodfood.com/special/special_occasion/soup/vegetable.html', 
); 

$dirs = array(); 
foreach ($links as $key => $link) { 
    $urlPath = parse_url($link, PHP_URL_PATH); 
    $arrayUrlPath = explode('/', $urlPath); 
    $dirs[$key] = array(); 
    $counter = 0; 
    foreach ($arrayUrlPath as $dir) { 
     if (empty($dir) || in_array(substr($dir, -5), array('.html'))) { 
      continue; 
     } 
     $dirs[$key][$counter++] = $dir; 
    } 
} 

$searchDirs = $dirs; 

foreach ($searchDirs as $key => $dir) { 
    foreach ($dir as $name) { 
     echo 'dir: ' . $name . ', found in: ' . search($name, $key, $dirs) . "\n"; 
    } 
} 

function search($name, $excludeKey, $dirs) 
{ 
    $return = array(); 
    foreach ($dirs as $key => $dir) { 
     if ($key === $excludeKey) { 
      continue; 
     } 
     if (in_array($name, $dir)) { 
      $return[] = (int)$key + 1; 
     } 
    } 
    return join(', ', $return); 
} 

如果你想搜索长字符串重建search功能，添加explode为$name和比较的研究$key，如果dir是aaa/bbb/ccc，检查index 0是“AAA”和index 1是bbb和index 2是ccc，除非移动指针index+1并再次检查。

我希望我能帮上忙。