如何根據點符號和“型別”屬性將平面陣列轉換為樹陣列？

我想轉換下面的平面陣列：

$array = [
   [
    "hierarchy" => "1",
    "title" => "Fruits",
    "type" => "category_label"
   ],
   [
    "hierarchy" => "1.1",
    "title" => "Citruses",
    "type" => "category_label"
   ],
   [
    "hierarchy" => "1.1.1",
    "title" => "Orange",
    "type" => "item"
   ],
   [
    "hierarchy" => "1.1",
    "title" => "Mango",
    "type" => "item"
   ],
   [
    "hierarchy" => "1.2",
    "title" => "Grape",
    "type" => "item"
   ]
];

從上面可以看出，它有兩個分層的點符號字串“1.1”，因此我使用“type”屬性來區分它們。

我期待的結果是什么：

/*
1. Fruits
-  1.1. Citruses
--- 1.1.1. Orange
-- 1.1. Mango
-- 1.2. Grape
*/

//Desired result
[
    "hierarchy" => "1",
    "title" => "Fruits",
    "type" => "category_label",
    "children" => [
        [
            "hierarchy" => "1.1",
            "title" => "Citruses",
            "type" => "category_label"
            "children" => [
                [
                    "hierarchy" => "1.1.1",
                    "title" => "Orange",
                    "type" => "item"
                ]
            ]
        ],
        [
            "hierarchy" => "1.1",
            "title" => "Mango",
            "type" => "item"
        ],
        [
            "hierarchy" => "1.2",
            "title" => "Grape",
            "type" => "item"
        ]
    ]
];

使用How to build a tree from a concatenated string in PHP? 中描述的方法。我無法達到我想要的結果。我現在能做什么？

uj5u.com熱心網友回復：

您可以首先在新的關聯陣列中按層次結構鍵入所有專案。然后再次迭代這些項，并洗掉 的最后一部分hierarchy以獲取父項的鍵，將其定位在新的關聯陣列中，并將其添加到其子項中。如果它沒有父級，則將其添加到結果陣列中。

由于您的hierarchy分隔符是一個點，您可以使用該pathinfo函式從中洗掉最后一個“擴展名”：

事實證明，您的層次結構編號不是唯一的（例如“1.1”）。如果打算父節點可以與“item”型別的兄弟節點具有相同的編號，但相同型別的兩個節點永遠不會具有相同的層次結構，則通過型別和層次結構的組合來鍵入節點：

foreach ($array as $item) $keyed[$item["type"] . $item["hierarchy"]] = $item;

foreach ($keyed as &$item) {
    $parent = pathinfo($item["hierarchy"], PATHINFO_FILENAME);
    if ($parent == $item["hierarchy"]) $result[] =& $item;
    else $keyed["category_label$parent"]["children"][] =& $item;
}

運行后，$result將具有所需的結構。

但是，我建議使層次結構代碼唯一，這樣它們就不會在型別之間發生沖突。那么上面的代碼可以改成：

foreach ($array as $item) $keyed[$item["hierarchy"]] = $item;

foreach ($keyed as &$item) {
    $parent = pathinfo($item["hierarchy"], PATHINFO_FILENAME);
    if ($parent == $item["hierarchy"]) $result[] =& $item;
    else $keyed[$parent]["children"][] =& $item;
}

uj5u.com熱心網友回復：

看看這個邏輯是否有幫助，我會嘗試在 1 天內編輯答案并解釋。

$tree_array = [];
foreach($array as $key => $value){
    if(count(explode('.',$value['hierarchy']) == 1){
        array_push($array_tree, $value)
    }
}

foreach($array as $value){
    if(count(explode('.',$value['hierarchy'])) == 2){
        if($value['type'] == 'category_label'){
            foreach($array_tree as $key =>  $tree){
                if($tree['hierarchy'] == explode('.',$value['hierarchy'])[0]){
                    array_push($array_tree['children'],$value);
                }
            }
        } else {
            foreach($array_tree as $key =>  $tree){
                if($tree['hierarchy'] == explode('.',$value['hierarchy'])[0]){
                    array_push($array_tree[$key],$value);
                }
            }
        }
    }
}

foreach($array as $value){
    if(count(explode('.',$value['hierarchy'])) == 3){
        if($value['type'] == 'item'){
            foreach($array_tree as $key => $tree){
                foreach($tree as $item){
                    if($item['hierarchy'] == explode('.',$value['hierarchy'],2)[1]){
                        array_push($array_tree[$key]['children'],$value);
                    }
                }
            }
        }
    }
}

標籤：php 数组 多维数组

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