我想让每个类别下的所有孩子和子孩子都得到。我有一个数组，但我需要在其对应的子类别的列表中得到它。
这些是我想要的列表。
-Category a
---category child b
-------category child c
---category d
-category b
这是我得到的数组：
Array
(
[0] => Array
(
[StudentCategory] => Array
(
[id] => 1
[name] => Category A
[parent_id] => 0
[lft] => 1
[rght] => 8
[user_id] => 1
[description] => Category A
[is_active] => 1
[is_deleted] => 0
[created] => 2014-04-16 19:43:01
[updated] => 2014-04-17 02:27:28
)
[children] => Array
(
[0] => Array
(
[StudentCategory] => Array
(
[id] => 2
[name] => Category A
[parent_id] => 1
[lft] => 2
[rght] => 5
[user_id] => 1
[description] => Category A
[is_active] => 1
[is_deleted] => 0
[created] => 2014-04-16 19:44:43
[updated] => 2014-04-17 01:15:39
)
[children] => Array
(
[0] => Array
(
[StudentCategory] => Array
(
[id] => 3
[name] => Category A
[parent_id] => 2
[lft] => 3
[rght] => 4
[user_id] => 1
[description] => Category A
[is_active] => 1
[is_deleted] => 0
[created] => 2014-04-16 19:45:39
[updated] => 2014-04-16 19:45:39
)
[children] => Array
(
)
)
)
)
[1] => Array
(
[StudentCategory] => Array
(
[id] => 4
[name] => category 2
[parent_id] => 1
[lft] => 6
[rght] => 7
[user_id] => 1
[description] => category 2
[is_active] => 0
[is_deleted] => 0
[created] => 2014-04-16 20:57:28
[updated] => 2014-04-16 20:57:28
)
[children] => Array
(
)
)
)
)
[1] => Array
(
[StudentCategory] => Array
(
[id] => 5
[name] => category 21
[parent_id] => 0
[lft] => 9
[rght] => 10
[user_id] => 1
[description] => category 21
[is_active] => 1
[is_deleted] => 0
[created] => 2014-04-16 21:00:33
[updated] => 2014-04-16 21:00:33
)
[children] => Array
(
)
)
)
我正在使用此解决方案来尝试完成此操作：
$level = 0;
$catresult = array();
$ch = $this->Common->make_category_children_array($v['children'],$level,$catresult);
pr($ch);
function make_category_children_array($childarray,&$level,&$catresult){
foreach($childarray as $item){
$catresult[$level] = $item['StudentCategory'];
$childarray = $item['children'];
if (is_array($childarray) && $childarray) { $level++;
$this->make_category_children_array($childarray,$level,$catresult);
}
}
return $catresult;
}
答案 0 :(得分：1)
或许可以选择以下内容：
function ex(&$ary, $prefix)
{
$retval = '';
foreach(array_keys($ary) as $i) {
...
$retval .= $prefix + $ary[$i]['StudentCategory']['name'];
if (array_key_exists($ary[$i], 'children'))
$retval .= ex($ary[$i]['children'], '---' . $prefix);
}
}
答案 1 :(得分：0)
递归解决方案可能看起来有点像：
/**
* An initial function to set up the recursive loop
*/
function explore(array $structure)
{
exploreLevel($structure, 0);
}
/**
* The actual recursion happens in here
*/
function exploreLevel(array $subStructure, $level)
{
foreach ($subStructure as $item)
{
// Let's render the category here. The repeated spaces give us
// some indentation depending where we are in the hierarchy
$category = $item['StudentCategory'];
echo str_repeat(' ', $level) . $category . "\n";
// Only recurse if the key is an array and it is not empty
$children = $item['children'];
if (is_array($children) && $children)
{
exploreLevel($children, $level + 1);
}
}
}
尽管可以使用非递归算法探索和修改嵌套集算法，但这种数据结构具体是分层的，并且可能是递归构建的。因此，我们需要一个递归函数来探索它。
递归编程中一种流行的方法是传递一个级别编号，最初为零，并在调用自身时递增1。因此，当一系列调用变得更深时，它会探索级别0，级别0 + 1，级别0 + 1 + 1，依此类推。只要有更多关卡探索，这个系列就会继续。如果我们希望增加探索深度的限制，$level计数器也很有用。