我正在尝试将旧的HTML站点转换为新的CMS。要获取正确的菜单hierachy（深度不同），我想读取所有文件，并提取/解析菜单（嵌套的无序列表）中的关联阵列
root.html
<ul id="menu">
<li class="active">Start</li>
<ul>
<li><a href="file1.html">Sub1</a></li>
<li><a href="file2.html">Sub2</a></li>
</ul>
</ul>
file1.html
<ul id="menu">
<li><a href="root.html">Start</a></li>
<ul>
<li class="active">Sub1</li>
<ul>
<li><a href="file3.html">SubSub1</a></li>
<li><a href="file4.html">SubSub2</a></li>
<li><a href="file5.html">SubSub3</a></li>
<li><a href="file6.html">SubSub4</a></li>
</ul>
</ul>
</ul>
file3.html
<ul id="menu">
<li><a href="root.html">Start</a></li>
<ul>
<li><a href="file1.html">Sub1</a></li>
<ul>
<li class="active">SubSub1</li>
<ul>
<li><a href="file7.html">SubSubSub1</a></li>
<li><a href="file8.html">SubSubSub2</a></li>
<li><a href="file9.html">SubSubSub3</a></li>
</ul>
</ul>
</ul>
</ul>
file4.html
<ul id="menu">
<li><a href="root.html">Start</a></li>
<ul>
<li><a href="file1.html">Sub1</a></li>
<ul>
<li><a href="file3.html">SubSub1</a></li>
<li class="active">SubSub2</li>
<li><a href="file5.html">SubSub3</a></li>
<li><a href="file6.html">SubSub4</a></li>
</ul>
</ul>
</ul>
我想循环浏览所有文件，提取'id ="菜单"'，并在保留层次结构和文件信息
的同时创建这样的数组（或类似的数组）Array
[file] => root.html
[child] => Array
[Sub1] => Array
[file] => file1.html
[child] => Array
[SubSub1] => Array
[file] => file3.html
[child] => Array
[SubSubSub1] => Array
[file] => file7.html
[SubSubSub2] => Array
[file] => file8.html
[SubSubSub3] => Array
[file] => file9.html
[SubSub2] => Array
[file] => file4.html
[SubSub3] => Array
[file] => file5.html
[SubSub4] => Array
[file] => file6.html
[Sub2] => Array
[file] => file2.html
借助PHP简单的HTML DOM解析器库，我成功地读取了该文件并提取了菜单
$html = file_get_html($file);
foreach ($html->find("ul[id=menu]") as $ul) {
..
}
仅解析菜单的活动部分（省略了指向1个或更多级别的链接）我使用了
$ul->find("ul",-1)
找到外部UL内部的最后一个UL。这非常适合一个文件。
，但是我很难循环浏览所有文件/菜单并保留父/子信息，因为每个菜单都有不同的深度。
感谢所有建议，提示和帮助！
编辑：好，毕竟这不是那么容易：）
顺便说一句，这个库确实是一个很好的工具。对写它的人表示敬意。
这是一个可能的解决方案：
class menu_parse {
static $missing = array(); // list of missing files
static private $files = array(); // list of source files to process
// initiate menu parsing
static function start ($file)
{
// start with root file
self::$files[$file] = 1;
// parse all source files
for ($res=array(); current(self::$files); next(self::$files))
{
// get next file name
$file = key(self::$files);
// parse the file
if (!file_exists ($file))
{
self::$missing[$file] = 1;
continue;
}
$html = file_get_html ($file);
// get menu root (if any)
$root = $html->find("ul[id=menu]",0);
if ($root) self::menu ($root, $res);
}
// reorder missing files array
self::$missing = array_keys (self::$missing);
// that's all folks
return $res;
}
// parse a menu at a given level
static private function menu ($menu, &$res)
{
foreach ($menu->children as $elem)
{
switch ($elem->tag)
{
case "li" : // name and possibly source file of a menu
// grab menu name
$name = $elem->plaintext;
// see if we can find a link to the menu file
$link = $elem->children(0);
if ($link && $link->tag == 'a')
{
// found the link
$file = $link->href;
$res[$name]->file = $file;
// add the source file to the processing list
self::$files[$file] = 1;
}
break;
case "ul" : // go down one level to grab items of the current menu
self::menu ($elem, $res[$name]->childs);
}
}
}
}
用法：
// The result will be an array of menus indexed by item names.
//
// Each menu will be an object with 2 members
// - file -> source file of the menu
// - childs -> array of menu subtitems
//
$res = menu_parse::start ("root.html");
// parse_menu::$missing will contain all the missing files names
echo "Result : <pre>";
print_r ($res);
echo "</pre><br>missing files:<pre>";
print_r (menu_parse::$missing);
echo "</pre>";
您的测试案例：
Array
(
[Start] => stdClass Object
(
[childs] => Array
(
[Sub1] => stdClass Object
(
[file] => file1.html
[childs] => Array
(
[SubSub1] => stdClass Object
(
[file] => file3.html
[childs] => Array
(
[SubSubSub1] => stdClass Object
(
[file] => file7.html
)
[SubSubSub2] => stdClass Object
(
[file] => file8.html
)
[SubSubSub3] => stdClass Object
(
[file] => file9.html
)
)
)
[SubSub2] => stdClass Object
(
[file] => file3.html
)
[SubSub3] => stdClass Object
(
[file] => file5.html
)
[SubSub4] => stdClass Object
(
[file] => file6.html
)
)
)
[Sub2] => stdClass Object
(
[file] => file2.html
)
)
[file] => root.html
)
)
missing files: Array
(
[0] => file2.html
[1] => file5.html
[2] => file6.html
[3] => file7.html
[4] => file8.html
[5] => file9.html
)
注意：
- 代码假定所有项目名称在给定菜单中都是唯一的。
您可以修改代码以将（sub）菜单作为数字索引和名称作为属性的数组（以便两个具有相同名称的项目不会彼此覆盖），但这会使结果的结构变得复杂。
如果发生这种名称的重复，最好的解决方案是重命名其中一个项目，恕我直言。
- 代码还假定只有一个根菜单。
可以修改以处理多个，但这并没有多大意义（这意味着根菜单ID重复，这可能会给JavaScript引起麻烦，试图首先尝试处理它）。<<<<<<<<<<<<<<<</p>
这更像是具有向上链接的目录树。file1在第2级上的file3上的file1点，然后将其点回到1级上的文件1，这会导致"不同的深度"。考虑设置一个特定的菜单对象，向上和向下指向，并保留列表，而不是字符串数组。PHP中这种层次的起点可能是这样的类：
class menuItem {
protected $leftSibling = null;
protected $rightSibling = null;
protected $parents = array();
protected $childs = array();
protected properties = array();
// set property like menu name or file name
function setProp($name, $val) {
$this->properties[$name] = $val;
}
// get a propertue if set, false otherwise
function getProp($name) {
if ( isset($this->properties[$name]) )
return $this->properties[$name];
return false;
}
function getLeftSiblingsAsArray() {
$sibling = $this->getLeftSibling();
$siblings = array();
while ( $sibling != null ) {
$siblings[] = $sibling;
$sibling = $sibling->getLeftSibling();
}
return $siblings;
}
function addChild($item) {
$this->childs[] = $item;
}
function addLeftSibling($item) {
$sibling = $this->leftSibling;
while ( $sibling != null ) {
if ( $sibling->hasLeft() )
$sibling = $sibling->getLeftSibling();
else {
$sibling->addFinalLeft($item);
break;
}
}
}
function addFinalLeft(item) {
$sibling->leftSibling = $item;
}
....