Question

我希望我的类别列表在返回用户时格式很好。我从数据库得到的是：

[
{
    "id": 1,
    "name": "pet",
    "parent_id": null
},
{
    "id": 2,
    "name": "page",
    "parent_id": null
},
{
    "id": 3,
    "name": "dog",
    "parent_id": 1
},
{
    "id": 4,
    "name": "cat",
    "parent_id": 1
},
{
    "id": 5,
    "name": "rodent",
    "parent_id": 1
},...

我希望它保持树形结构，如：

{
    "id": 1,
    "name": "pet",
    "parent_id": null,
    "children": [
        {
            "id": 3,
            "name": "dog",
            "parent_id": 1
        },
        {
            "id": 4,
            "name": "cat",
            "parent_id": 1
        },...

等

有没有一种简单的方法可以做到这一点，或者我必须遍历数据库结果并创建新的有组织的数组才能返回？这样做的最佳方法是什么？问题是子类别也可能有子类别。或者我应该保留从数据库中获取的结构并将子ID作为数组添加（因为我可以参考它们）？

我很感激你的帮助。

谢谢。

Answer 1

<强>解决方案：

function normalize_db_animals(){

    $values[] = ["id" => 1, "name" => "pet", "parent_id" => null];
    $values[] = ["id" => 2, "name" => "dog", "parent_id" => 1];
    $values[] = ["id" => 3, "name" => "cat", "parent_id" => 1];
    $values[] = ["id" => 4, "name" => "rodent", "parent_id" => 1];

    $values[] = ["id" => 5, "name" => "wild", "parent_id" => null];
    $values[] = ["id" => 6, "name" => "tiger", "parent_id" => 5];
    $values[] = ["id" => 7, "name" => "rhino", "parent_id" => 5];

    $normalize = function () use ($values) {
        $tree = [];
        $i = 0;
        do {
            $pet = $values[$i];
            if ($pet['parent_id']) {
                if (array_key_exists($pet['parent_id'], $tree)) {
                    $tree[$pet['parent_id']]['children'][] = $pet;
                }
            } else {
                $tree[$pet['id']] = $pet;
            }

            $i++;
        } while ($i < count($values));

        return $tree;
    };

    $tree = $normalize();
    echo json_encode($tree);
}

<强>结果：

   {"1":{"id":1,"name":"pet","parent_id":null,"children":[{"id":2,"name":"dog","parent_id":1},{"id":3,"name":"cat","parent_id":1},{"id":4,"name":"rodent","parent_id":1}]},"5":{"id":5,"name":"wild","parent_id":null,"children":[{"id":6,"name":"tiger","parent_id":5},{"id":7,"name":"rhino","parent_id":5}]}}

Answer 2

试试这个

$a = json_decode('[{
      "id": 1,
      "name": "pet",
      "parent_id": null
    },
    {
      "id": 2,
      "name": "page",
      "parent_id": null
    },
    {
      "id": 3,
      "name": "dog",
      "parent_id": 1
    },
    {
      "id": 4,
      "name": "cat",
      "parent_id": 1
    },
    {
      "id": 5,
      "name": "rodent",
      "parent_id": 4
    },
    {
      "id": 6,
      "name": "rodent",
      "parent_id": 2
    }]');

    $a = collect($a);

    $filtered = $a;

    foreach ($filtered as $key => $value) {
      $children = $a->where('parent_id', $value->id);
      if(!$children->isEmpty()){
        $value->children = $children;
       $filtered->forget(array_values(array_keys($children->toArray())));

      }

    }
    dd($filtered);

PHP使用修改后的数据返回JSON

2 个答案: