Javascript 如何将PHP数组转换为对象的JS嵌套数组

用于呈现树视图的特定jQuery插件需要数据字符串“嵌套对象数组”。我的树视图数据（在相同的结构中）在PHP数组中可用。我需要以jQuery插件可以读取数据的方式回显PHP数组

我已经尝试了json\u编码PHP数组，但得到的结果与jQuery插件预期的完全不同。可在此处查看数据的示例/要求格式： 以及以下各项：

var data = [
    {
        name: 'node1',
        children: [
            { name: 'child1' },
            { name: 'child2' }
        ]
    },
    {
        name: 'node2',
        children: [
            { name: 'child3' }
        ]
    }
];

我需要将PHP数组转换为上述JavaScript格式（不过，这只是一个示例）：

---

编辑： 以下是生成阵列的方式：

$objectTempRoles = $this->roleRepository->findAll();
    $aTempRoles = [];
    foreach($objectTempRoles as $oRole){
        if($oRole->getIsroot() == 1){
            $aTempRoles[$oRole->getUid()] = [];
            $aTempRoles[$oRole->getUid()]['name'] = $oRole->getTitle();
            $aTempRoles[$oRole->getUid()]['id'] = $oRole->getUid();
            $aTempRoles[$oRole->getUid()]['children'] = $this->functionGetChildren($oRole);
        }
}


public function functionGetChildren($oRole){
    $aChildrenToReturn = [];
    if($oRole->getChildren() != null && $oRole->getChildren() != false){
        foreach($oRole->getChildren() as $oChild){
            $aChildrenToReturn[$oChild->getUid()] = [];
            $aChildrenToReturn[$oChild->getUid()]['name'] = $oChild->getTitle();
            $aChildrenToReturn[$oChild->getUid()]['id'] = $oChild->getUid();
            $aChildrenToReturn[$oChild->getUid()]['children'] = $this->functionGetChildren($oChild);
        }
    }
    return $aChildrenToReturn;
}

array (
  0 => 
  array (
    'name' => 'CEO',
    'id' => 1,
    'children' => 
    array (
      3 => 
      array (
        'name' => 'Director 1',
        'id' => 3,
        'children' => 
        array (
          4 => 
          array (
            'name' => 'Senior Manager 1',
            'id' => 4,
            'children' => 
            array (
              5 => 
              array (
                'name' => 'Manager 1',
                'id' => 5,
                'children' => 
                array (
                ),
              ),
            ),
          ),
        ),
      ),
      6 => 
      array (
        'name' => 'Director 2',
        'id' => 6,
        'children' => 
        array (
          7 => 
          array (
            'name' => 'Senior Manager 2',
            'id' => 7,
            'children' => 
            array (
            ),
          ),
        ),
      ),
    ),
  ),
)

=====

编辑： 这是我的阵列的var_转储：

$objectTempRoles = $this->roleRepository->findAll();
    $aTempRoles = [];
    foreach($objectTempRoles as $oRole){
        if($oRole->getIsroot() == 1){
            $aTempRoles[$oRole->getUid()] = [];
            $aTempRoles[$oRole->getUid()]['name'] = $oRole->getTitle();
            $aTempRoles[$oRole->getUid()]['id'] = $oRole->getUid();
            $aTempRoles[$oRole->getUid()]['children'] = $this->functionGetChildren($oRole);
        }
}


public function functionGetChildren($oRole){
    $aChildrenToReturn = [];
    if($oRole->getChildren() != null && $oRole->getChildren() != false){
        foreach($oRole->getChildren() as $oChild){
            $aChildrenToReturn[$oChild->getUid()] = [];
            $aChildrenToReturn[$oChild->getUid()]['name'] = $oChild->getTitle();
            $aChildrenToReturn[$oChild->getUid()]['id'] = $oChild->getUid();
            $aChildrenToReturn[$oChild->getUid()]['children'] = $this->functionGetChildren($oChild);
        }
    }
    return $aChildrenToReturn;
}

array (
  0 => 
  array (
    'name' => 'CEO',
    'id' => 1,
    'children' => 
    array (
      3 => 
      array (
        'name' => 'Director 1',
        'id' => 3,
        'children' => 
        array (
          4 => 
          array (
            'name' => 'Senior Manager 1',
            'id' => 4,
            'children' => 
            array (
              5 => 
              array (
                'name' => 'Manager 1',
                'id' => 5,
                'children' => 
                array (
                ),
              ),
            ),
          ),
        ),
      ),
      6 => 
      array (
        'name' => 'Director 2',
        'id' => 6,
        'children' => 
        array (
          7 => 
          array (
            'name' => 'Senior Manager 2',
            'id' => 7,
            'children' => 
            array (
            ),
          ),
        ),
      ),
    ),
  ),
)

=====

编辑： json_encode参数我用过far:json_FORCE_OBJECT

=====

编辑： 我现在已经成功地生成了所需的数据结构。为此，我使用以下函数：

public function getRoleChildrenJson($aParentObject){
        $json = "";
        $i = 1;
        foreach($aParentObject['children'] as $aObject){
            $tmbObjectStr = "{name: \"".$aObject['name']."\",id: ".$aObject['id'];
            if(!empty($aObject['children'])){
                $tmbObjectStr .= ",children: [";
                $tmbObjectStr .= $this->getRoleChildrenJson($aObject);
                if($i < count($aParentObject['children'])){
                    $tmbObjectStr .= "]},";
                }
            }
            else{
                $tmbObjectStr .= "}]}";
            }
            $json .= $tmbObjectStr;
            $i++;
        }
        return $json;
    }

但是，以下不起作用：

                        var data = [
                            {
                                name: 'node1', id: 1,
                                children: [
                                    { name: 'child1', id: 2 },
                                    { name: 'child2', id: 3 }
                                ]
                            },
                            {
                                name: 'node2', id: 4,
                                children: [
                                    { name: 'child3', id: 5 }
                                ]
                            }
                        ];
                        $('#rolestree').tree({
                            data: data
                        });

                ajaxRequest = $.ajax({
                    url: "/index.php" + $getData,
                    type: "POST",
                    data: "",
                    success: function (jsondata, textStatus, jqXHR) {
                        $('#rolestree').tree({
                            data: jsondata
                        });
                    }
                });

---

尽管相同的json字符串在AJAX上得到了完美的加载（我使用console进行了检查）。我是否首先需要评估或解析通过AJAX加载的数据？

如果要从PHP返回json作为字符串，则必须在AJAX响应中解析它

在ajax响应中添加内容类型

$.ajax({
          url: "/index.php" + $getData,
          type: "POST",                   data: "",
          success: function (jsondata, textStatus, jqXHR) {
                   console.log(jsondata);
                   var json = $.parseJSON(jsondata);

                   console.log(json);
                   $('#rolestree').tree({
                        data: json
                   });
          }
      });

---

问候

请向我们展示您的尝试。我们确实希望帮助您解决现有代码中的问题，但我们不是来为您完成所有工作的。请添加一个示例PHP数组，我们可以使用它来测试这一点（与其var_dump表示相反）您需要在php中构建正确的数据结构，然后在javascript中使用

json\u encode（）

。和

array\u values（）

将帮助您将php数组转换为“实”/“零”基数值数组。了解如何创建此php数组可能会很有用。我相信这可以通过一种符合您要求的方式来完成，而不是试图将此数组捏造成真正的数组required@ChrisG我想您要求的是

var\u export（）

，仅供将来参考