sev*_*ers 7 php recursion pass-by-reference
我有一个以下结构类别的树:
[6] => Array
(
[id] => 6
[name] => computers
[productCount] => 0
[children] => Array
(
[91] => Array
(
[id] => 91
[name] => notebook
[productCount] => 5
[children] => Array
(
)
)
[86] => Array
(
[id] => 86
[name] => desktop
[productCount] => 0
[children] => Array
(
)
)
)
)
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除了子类别,每个类别可能包含产品(如文件夹可能包含子文件夹和文件).
我正在尝试编写一个递归函数,我希望将此数组作为参考,并使用[productCount] = 0和包含此类空节点的所有父类别剥离两个叶类别.换句话说,在处理之后,我想只有那些在任何子级上持有产品的类别.
我已经写了一些代码,现在调试它并且它不会剥离空节点.可能是我没有正确使用引用.如果可能的话,请帮我修理一下.
function pruneTree( & $node) {
if ( ! $node['children'] && ! $node['productCount']) {
unset($node);
}
if ( ! empty($node['children'])) {
foreach ($node['children'] as $key => $child) {
pruneTree($node['children'][$key]);
}
}
return;
}
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unset 仅删除引用但不删除引用的变量:
如果PASSED BY REFERENCE的变量
unset()位于函数内部,则仅销毁局部变量.调用环境中的变量将保留与之前unset()调用的值相同的值.
所以你需要传递父数组和密钥来删除该变量:
function pruneTree(&$parent, $key) {
$node = &$parent[$key];
if (!$node['children'] && !$node['productCount']) {
unset($parent[$key]);
}
if (!empty($node['children'])) {
foreach ($node['children'] as $key => &$child) {
pruneTree($node['children'], $key);
}
}
}
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您还可以更改函数中的参数以采用节点数组而不是单个节点。这稍微改变了递归,并防止需要传递密钥:
function pruneTree(&$nodes) {
foreach ($nodes as $key => $node) {
if (!$node['children'] && !$node['productCount']) {
unset($nodes[$key]);
} elseif (!empty($node['children'])) {
pruneTree($nodes[$key]['children']);
// This line checks if all the children have been pruned away:
if (empty($nodes[$key]['children'])) {
unset($nodes[$key]);
}
}
}
}
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此外,添加了一项检查,以确保如果所有子节点都被修剪,则父(现在是叶)节点也会被修剪。
希望这可以帮助!
测试数据:
$data = array(
6 => array(
'id' => 6,
'name' => 'computers',
'productCount' => 0,
'children' => array(
91 => array(
'id' => 91,
'name' => 'notebook',
'productCount' => 5,
'children' => array()
),
86 => array(
'id' => 86,
'name' => 'desktop',
'productCount' => 0,
'children' => array()
)
)
)
);
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电话:
pruneTree($data);
echo '<pre>';
print_r($data);
echo '</pre>';
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