让我们说，我有以下数组：
$array = array(
array(
"id" => 1,
"name" => "Europe",
"path" => "/"
),
array(
"id" => 2,
"name" => "Germany",
"path" => "/1/"
),
array(
"id" => 3,
"name" => "France",
"path" => "/1/"
),
array(
"id" => 4,
"name" => "Berlin",
"path" => "/1/2/"
),
array(
"id" => 5,
"name" => "Munich",
"path" => "/1/2/"
)
);
正如你所看到的，它是一个多维数组，在earch第二级数组中有3个属性：id，name和path。路径是基于其父级的父标识的路径结构。例如，德国（id = 2）属于欧洲，因此路径为＆＃34; / 1 /＆＃34; （身份证1 =欧洲）德国的柏林有路径＆＃34; / 1/2 /＆＃34;这意味着＆＃34; /欧洲/德国/＆＃34;
现在，我正在尝试创建一个树形数组，它应该看起来像：
$result = array(
1 => array(
"id" => 1,
"name" => "Europe",
"path" => "/",
"childs" => array(
2 => array(
"id" => 2,
"name" => "Germany",
"path" => "/1/",
"childs" => array(
4 => array(
"id" => 4,
"name" => "Berlin",
"path" => "/1/2/"
),
5 => array(
"id" => 5,
"name" => "Munich",
"path" => "/1/2/"
)
)
),
3 => array(
"id" => 3,
"name" => "France",
"path" => "/1/"
)
)
)
);
我已经尝试使用内部引用创建一个函数，但这对我没有用：
public static function pathToTree($items) {
$array = array();
$result = array();
foreach($items AS $res) {
$ids = explode("/", ltrim($res["path"] . $res["id"], "/"));
$count = count($ids);
$tmp = &$result;
foreach( $ids AS $id) {
if($count == 1) {
$tmp = $res;
$tmp["childs"] = array();
$tmp = &$tmp["childs"];
}
else {
$tmp[$id] = array(
"childs" => array()
);
$tmp = &$tmp[$id]["childs"];
}
$count--;
}
}
return $array;
}
答案 0 :(得分：0)
正如我在评论中所说，你需要至少三个循环来实现你的目标。这是功能：
function pathToTree($array){
$tree = Array();
for($i=0; $i < count($array); $i++){
if(substr_count($array[$i]["path"], '/') == 1)
$tree[$array[$i]["id"]] = $array[$i];
}
for($i=0; $i < count($array); $i++){
if(substr_count($array[$i]["path"], '/') == 2){
$num = (int)str_replace("/","",$array[$i]["path"]);
$tree[$num]["childs"][$array[$i]["id"]] = $array[$i];
}
}
for($i=0; $i < count($array); $i++){
if(substr_count($array[$i]["path"], '/') == 3){
$num = explode("/", $array[$i]["path"]);
$tree[$num[1]]["childs"][$num[2]]["childs"][$array[$i]["id"]] = $array[$i];
}
}
return $tree;
}
示例：
考虑这个数组：
$array = array(
array(
"id" => 1,
"name" => "Europe",
"path" => "/"
),
array(
"id" => 2,
"name" => "Germany",
"path" => "/1/"
),
array(
"id" => 3,
"name" => "France",
"path" => "/1/"
),
array(
"id" => 4,
"name" => "Berlin",
"path" => "/1/2/"
),
array(
"id" => 5,
"name" => "Munich",
"path" => "/1/2/"
),
array(
"id" => 6,
"name" => "Asia",
"path" => "/"
),
array(
"id" => 7,
"name" => "China",
"path" => "/6/"
),
array(
"id" => 8,
"name" => "Bangladesh",
"path" => "/6/"
),
array(
"id" => 9,
"name" => "Beijing",
"path" => "/6/7/"
),
array(
"id" => 10,
"name" => "Dhaka",
"path" => "/6/8/"
)
);
如果我运行此代码：
print_r(pathToTree($array));
输出将是：
Array
(
[1] => Array
(
[id] => 1
[name] => Europe
[path] => /
[childs] => Array
(
[2] => Array
(
[id] => 2
[name] => Germany
[path] => /1/
[childs] => Array
(
[4] => Array
(
[id] => 4
[name] => Berlin
[path] => /1/2/
)
[5] => Array
(
[id] => 5
[name] => Munich
[path] => /1/2/
)
)
)
[3] => Array
(
[id] => 3
[name] => France
[path] => /1/
)
)
)
[6] => Array
(
[id] => 6
[name] => Asia
[path] => /
[childs] => Array
(
[7] => Array
(
[id] => 7
[name] => China
[path] => /6/
[childs] => Array
(
[9] => Array
(
[id] => 9
[name] => Beijing
[path] => /6/7/
)
)
)
[8] => Array
(
[id] => 8
[name] => Bangladesh
[path] => /6/
[childs] => Array
(
[10] => Array
(
[id] => 10
[name] => Dhaka
[path] => /6/8/
)
)
)
)
)
)
以下是phpfiddle link，以防您自己尝试。
答案 1 :(得分：0)
好的，我想我刚刚找到了解决方案：
function pathToTree($array){
$tree = array();
foreach($array AS $item) {
$pathIds = explode("/", ltrim($item["path"], "/") . $item["id"]);
$current = &$tree;
foreach($pathIds AS $id) {
if(!isset($current["childs"][$id])) $current["childs"][$id] = array();
$current = &$current["childs"][$id];
if($id == $item["id"]) {
$current = $item;
}
}
}
return $tree["childs"];
}
这是1-n深度的动力解决方案。看看我在http://ideone.com/gn0XLp的例子。在这里，我测试了一些水平：
答案 2 :(得分：0)
我创建了一个递归函数：
//数组
$array = array(
array(
"id" => 1,
"name" => "Europe",
"path" => "/"
),
array(
"id" => 2,
"name" => "Germany",
"path" => "/1/"
),
array(
"id" => 3,
"name" => "France",
"path" => "/1/"
),
array(
"id" => 4,
"name" => "Berlin",
"path" => "/1/2/"
),
array(
"id" => 5,
"name" => "Munich",
"path" => "/1/2/"
),
array(
"id" => 6,
"name" => "Asia",
"path" => "/"
),
array(
"id" => 7,
"name" => "India",
"path" => "/6/"
),
array(
"id" => 7,
"name" => "Mumbai",
"path" => "/6/7"
),
array(
"id" => 8,
"name" => "Delhi",
"path" => "/6/7"
),
);
//递归函数
function createTree($input, &$result = array(), $key = null) {
if ($key == "id") {
$result["temp"]["id"] = $input;
}
if ($key == "name") {
$result["temp"]["name"] = $input;
}
if ($key == "path") {
$result["temp"]["path"] = $input;
$levels = is_string($input) ? array_values(array_filter(explode('/', $input))) : null;
if ($input == "/") {
$result[$result["temp"]["id"]] = $result["temp"];
}
if (count($levels) == 1) {
$result[$levels[0]]["childs"][$result["temp"]["id"]] = $result["temp"];
}
if (count($levels) == 2) {
$result[$levels[0]]["childs"][$levels[1]]["childs"][$result["temp"]["id"]] = $result["temp"];
}
unset($result["temp"]);
}
if (is_array($input)) {
foreach($input as $key => $value) {
createTree($value, $result, $key);
}
}
return $result;
}
//结果
array (
1 =>
array (
'id' => 1,
'name' => 'Europe',
'path' => '/',
'childs' =>
array (
2 =>
array (
'id' => 2,
'name' => 'Germany',
'path' => '/1/',
'childs' =>
array (
4 =>
array (
'id' => 4,
'name' => 'Berlin',
'path' => '/1/2/',
),
5 =>
array (
'id' => 5,
'name' => 'Munich',
'path' => '/1/2/',
),
),
),
3 =>
array (
'id' => 3,
'name' => 'France',
'path' => '/1/',
),
),
),
6 =>
array (
'id' => 6,
'name' => 'Asia',
'path' => '/',
'childs' =>
array (
7 =>
array (
'id' => 7,
'name' => 'India',
'path' => '/6/',
'childs' =>
array (
7 =>
array (
'id' => 7,
'name' => 'Mumbai',
'path' => '/6/7',
),
8 =>
array (
'id' => 8,
'name' => 'Delhi',
'path' => '/6/7',
),
),
),
),
),
)