在json中的數據下獲取mysqli行顯示

Question

我有一個看起來像這樣的mysqli表：

在此表之外，我正在創建一個Json-File以便以后在其他Files中使用它。

這是我的代碼：

<?php

$categories = Category::getTopCategories();
$categories = array_values($categories);
echo json_encode($categories);

class Category
{
    /**
     * The information stored in the database for each category
     */
    public $id;
    public $parent;
    public $name;
    public $vlink;

    // The child categories
    public $children;

    public function __construct()
    {
        // Get the child categories when we get this category
        $this->getChildCategories();
    }

    /**
     * Get the child categories
     * @return array
     */
    public function getChildCategories()
    {
        if ($this->children) {
            return $this->children;
        }
        return $this->children = self::getCategories("parent = {$this->id}");
    }

    ////////////////////////////////////////////////////////////////////////////

    /**
     * The top-level categories (i.e. no parent)
     * @return array
     */
    public static function getTopCategories()
    {
        return self::getCategories('parent = 0');
    }

    /**
     * Get categories from the database.
     * @param string $where Conditions for the returned rows to meet
     * @return array
     */
    public static function getCategories($where = '')
    {
        if ($where) $where = " WHERE $where";
        $conn=mysqli_connect('localhost', 'root', '','praktikum');
        mysqli_set_charset($conn,"UTF8"); 

        $sql = "SELECT id,name,parent,vlink FROM mdl_praktikum$where";

        $result = mysqli_query($conn,$sql);

        $categories = array();
        while ($category = mysqli_fetch_object($result, 'Category'))
            $categories[] = $category;

        mysqli_free_result($result);
        return $categories;
    }
}



?>

這是我的結果的摘要：

[{
    "id": "1"
    , "parent": "0"
    , "name": "root"
    , "vlink": "0"          //this is wrong
    , "children": [{
        "id": "2"
        , "parent": "1"
        , "name": "Verschiedenes"
        , "vlink": "0"
        , "children": []
    }, {
        "id": "3"
        , "parent": "1"
        , "name": "EBSN"
        , "vlink": "0"
        , "children": [{
            "id": "6"
            , "parent": "3"
            , "name": "EBSN 1415"
            , "vlink": "0"
            , "children": [{
                "id": "23"
                , "parent": "6"
                , "name": "General Information"
                , "vlink": "0"
                , "children": [{
                    "id": "208"
                    , "parent": "23"
                    , "name": "03-05 SocialMediaMining3"
                    , "vlink": "198"
                    , "children": []
                }, {
                    "id": "209"
                    , "parent": "23"
                    , "name": "06-08 Business Model Blocks - Relationship"
                    , "vlink": "221"
                    , "children": []
                }, {........

現在，我希望將“ vlink”行顯示為json文件如下所示的數據對象：

[{
    "id": "1"
    , "parent": "0"
    , "name": "root"
    , "data": {
    , "data": {"vlink": "0"}              // This should be standing here
    , "children": [{
        "id": "2"
        , "parent": "1"
        , "name": "Verschiedenes"
        , "data": { "vlink": "0"}
        , "children": []
    }, {
        "id": "3"
        , "parent": "1"
        , "name": "EBSN"
        , "data": { "vlink": "0"}
        , "children": [{
            "id": "6"
            , "parent": "3"
            , "name": "EBSN 1415"
            , "data": { "vlink": "0"}
            , "children": [{
                "id": "23"
                , "parent": "6"
                , "name": "General Information"
                , "data": { "vlink": "0"}
                , "children": [{
                    "id": "208"
                    , "parent": "23"
                    , "name": "03-05 SocialMediaMining3"
                    , "data": { "vlink": "198"}
                    , "children": []
                }, {
                    "id": "209"
                    , "parent": "23"
                    , "name": "06-08 Business Model Blocks - Relationship"
                    , "data": { "vlink": "221"}
                    , "children": []
                }, {......

我無法回答這個問題，我想在代碼中更改什么，或者是否必須更改表中的內容才能正確顯示...。

希望你理解我在做什么

問候阿明

Answer 1

也許這就是您想要的：

    while ($category = mysqli_fetch_object($result, 'Category')) {
        $vlink = array('vlink' => $category->vlink);
        unset($category->vlink);
        $category->data = $vlink;
        $categories[] = $category;
    }

這將從行對象中刪除vlink屬性，將其替換為包含vlink作為對象的data屬性。 它實際上是一個PHP關聯數組，但是關聯數組和對象在編碼為JSON時都會變成對象。

Answer 2

請不要從靜態函數中調用構造函數。 僅在創建實際對象實例時運行。 同樣，一個人只需要一個遞歸循環來拾取遞歸下降的子代。

編輯：啊，然后要將該vlink移動到數據元素中，請復制並取消設置它。

$categories = Category::getCategories();
echo '<pre>'.json_encode($categories);

class Category
{
    /**
     * The information stored in the database for each category
     */
    public $id;
    public $parent;
    public $name;
    public $vlink;

    // The child categories
    public $children;

    /**
     * Get categories from the database.
     * @param string $where Conditions for the returned rows to meet
     * @return array
     */
    public static function getCategories($parent_id='0')
    {

        $parent_id = (integer) $parent_id;
        $conn=mysqli_connect('localhost', 'root', '','praktikum');
        mysqli_set_charset($conn,"UTF8");

        $sql = "SELECT id, name, parent, vlink FROM mdl_praktikum WHERE parent = $parent_id";

        $result = mysqli_query($conn,$sql);

        $categories = array();
        while ($category = mysqli_fetch_object($result, 'Category')) {
            $category->children = Self::getCategories($category->id);
            $category->data = array('vlink'=>$category->vlink);
            unset($category->vlink);
            $categories[] = $category;
        }
        mysqli_free_result($result);

        return $categories;
    }
}