Dav*_*ill 74 php arrays recursion function
我正在寻找一个函数,它接受一个页面/类别数组(来自平面数据库结果),并根据父ID生成一个嵌套页面/类别项的数组.我想以递归方式执行此操作,以便可以完成任何级别的嵌套.
例如:我在一个查询中获取所有页面,这就是数据库表的样子
+-------+---------------+---------------------------+
| id | parent_id | title |
+-------+---------------+---------------------------+
| 1 | 0 | Parent Page |
| 2 | 1 | Sub Page |
| 3 | 2 | Sub Sub Page |
| 4 | 0 | Another Parent Page |
+-------+---------------+---------------------------+
这是我想在我的视图文件中处理的数组:
Array
(
[0] => Array
(
[id] => 1
[parent_id] => 0
[title] => Parent Page
[children] => Array
(
[0] => Array
(
[id] => 2
[parent_id] => 1
[title] => Sub Page
[children] => Array
(
[0] => Array
(
[id] => 3
[parent_id] => 1
[title] => Sub Sub Page
)
)
)
)
)
[1] => Array
(
[id] => 4
[parent_id] => 0
[title] => Another Parent Page
)
)
我已经查看并尝试了几乎我遇到过的所有解决方案(Stack Overflow上有很多这样的解决方案,但是没有运气得到足够通用的东西,可以用于页面和类别.
这是我得到的最接近的,但它不起作用,因为我将孩子分配给第一级父母.
function page_walk($array, $parent_id = FALSE)
{
$organized_pages = array();
$children = array();
foreach($array as $index => $page)
{
if ( $page['parent_id'] == 0) // No, just spit it out and you're done
{
$organized_pages[$index] = $page;
}
else // If it does,
{
$organized_pages[$parent_id]['children'][$page['id']] = $this->page_walk($page, $parent_id);
}
}
return $organized_pages;
}
function page_list($array)
{
$fakepages = array();
$fakepages[0] = array('id' => 1, 'parent_id' => 0, 'title' => 'Parent Page');
$fakepages[1] = array('id' => 2, 'parent_id' => 1, 'title' => 'Sub Page');
$fakepages[2] = array('id' => 3, 'parent_id' => 2, 'title' => 'Sub Sub Page');
$fakepages[3] = array('id' => 4, 'parent_id' => 3, 'title' => 'Another Parent Page');
$pages = $this->page_walk($fakepages, 0);
print_r($pages);
}
dec*_*eze 208
一些非常简单的通用树构建:
function buildTree(array $elements, $parentId = 0) {
$branch = array();
foreach ($elements as $element) {
if ($element['parent_id'] == $parentId) {
$children = buildTree($elements, $element['id']);
if ($children) {
$element['children'] = $children;
}
$branch[] = $element;
}
}
return $branch;
}
$tree = buildTree($rows);
算法非常简单:
0/无/ null/无论如何).parent_id元素的元素与您在1中得到的当前父ID匹配,则该元素是父元素的子元素.将它放在当前孩子的列表中(这里:) $branch.children元素.换句话说,该函数的一次执行返回一个元素列表,这些元素是给定父ID的子元素.调用它buildTree($myArray, 1),它将返回一个父id为1的元素列表.最初调用父函数为0的函数,因此返回没有父id的元素,它们是根节点.该函数以递归方式调用自身以查找子项的子项.
Ale*_*s G 12
我知道这个问题已经过时了,但我遇到了一个非常类似的问题 - 除非有大量的数据.经过一番努力,我设法在结果集的一次传递中构建树 - 使用引用.这段代码并不漂亮,但它的工作原理非常快.它是非递归的 - 也就是说,结果集上只有一个传递,最后一个传递array_filter:
$dbh = new PDO(CONNECT_STRING, USERNAME, PASSWORD);
$dbs = $dbh->query("SELECT n_id, n_parent_id from test_table order by n_parent_id, n_id");
$elems = array();
while(($row = $dbs->fetch(PDO::FETCH_ASSOC)) !== FALSE) {
$row['children'] = array();
$vn = "row" . $row['n_id'];
${$vn} = $row;
if(!is_null($row['n_parent_id'])) {
$vp = "parent" . $row['n_parent_id'];
if(isset($data[$row['n_parent_id']])) {
${$vp} = $data[$row['n_parent_id']];
}
else {
${$vp} = array('n_id' => $row['n_parent_id'], 'n_parent_id' => null, 'children' => array());
$data[$row['n_parent_id']] = &${$vp};
}
${$vp}['children'][] = &${$vn};
$data[$row['n_parent_id']] = ${$vp};
}
$data[$row['n_id']] = &${$vn};
}
$dbs->closeCursor();
$result = array_filter($data, function($elem) { return is_null($elem['n_parent_id']); });
print_r($result);
在对此数据执行时:
mysql> select * from test_table;
+------+-------------+
| n_id | n_parent_id |
+------+-------------+
| 1 | NULL |
| 2 | NULL |
| 3 | 1 |
| 4 | 1 |
| 5 | 2 |
| 6 | 2 |
| 7 | 5 |
| 8 | 5 |
+------+-------------+
最后一个print_r产生这个输出:
Array
(
[1] => Array
(
[n_id] => 1
[n_parent_id] =>
[children] => Array
(
[3] => Array
(
[n_id] => 3
[n_parent_id] => 1
[children] => Array
(
)
)
[4] => Array
(
[n_id] => 4
[n_parent_id] => 1
[children] => Array
(
)
)
)
)
[2] => Array
(
[n_id] => 2
[n_parent_id] =>
[children] => Array
(
[5] => Array
(
[n_id] => 5
[n_parent_id] => 2
[children] => Array
(
[7] => Array
(
[n_id] => 7
[n_parent_id] => 5
[children] => Array
(
)
)
[8] => Array
(
[n_id] => 8
[n_parent_id] => 5
[children] => Array
(
)
)
)
)
[6] => Array
(
[n_id] => 6
[n_parent_id] => 2
[children] => Array
(
)
)
)
)
)
这正是我想要的.