Question

我正在创建一个树状结构菜单，就像wordpress一样。我能够像wordpress一样做。我的UI看起来像这样：

在这里，用户可以添加他们需要的子菜单数量。下面的JSON代码格式：

[
   {
      "layout":0,
      "layouttype":1,
      "deleted":0,
      "new":0,
      "deleter":0,
      "slug":"Home",
      "name":"Home",
      "id":"12122017033717-1"
   },
   {
      "layout":0,
      "layouttype":4,
      "deleted":0,
      "new":0,
      "deleter":0,
      "slug":"About",
      "name":"About",
      "id":"12122017033717-2"
   },
   {
      "layout":0,
      "layouttype":1,
      "deleted":0,
      "new":0,
      "deleter":0,
      "slug":"Services",
      "name":"Servicessss",
      "id":"12122017033717-3",
      "children":[
         {
            "layout":0,
            "layouttype":6,
            "deleted":0,
            "new":0,
            "deleter":1,
            "slug":"Services-1",
            "name":"Services 1",
            "id":"12122017033717-4"
         },
         {
            "layout":0,
            "layouttype":1,
            "deleted":0,
            "new":0,
            "deleter":1,
            "slug":"ss",
            "name":"ss",
            "id":"12122017033920-1",
            "children":[
               {
                  "layout":0,
                  "layouttype":1,
                  "deleted":0,
                  "new":0,
                  "deleter":1,
                  "slug":"fddzfsdz",
                  "name":"fddzfsdz",
                  "id":"12122017040829-1"
               },
               {
                  "layout":0,
                  "layouttype":3,
                  "deleted":0,
                  "new":0,
                  "deleter":1,
                  "slug":"fddzfsdzfcg",
                  "name":"Sarath",
                  "id":"12122017040829-2"
               },
               {
                  "layout":0,
                  "layouttype":1,
                  "deleted":0,
                  "new":0,
                  "deleter":1,
                  "slug":"asassssss",
                  "name":"asassssss",
                  "id":"12122017041219-1"
               }
            ]
         }
      ]
   }
]

获取和更新一个子菜单的当前代码来了。

$arr = json_decode($arr);
foreach($arr as $item) {
    $menuid = $item - > id;
    if ($menuid == $id) {
        $item - > layouttype = $page_layout;
    }
    if (property_exists($item, 'children')) {
        foreach($item - > children as $innerRow) {
            $submenuid = $innerRow - > id;
            if ($submenuid == $id) {
                $innerRow - > layouttype = $page_layout;
            }
        }
    }
}

我需要编辑数据＆＃39; layouttype＆＃39;动态。我尝试使用foreach()并且它可以正常运行子菜单的一个或两个子菜单，但如果用户添加了更多子菜单，则代码将无法容纳。

那么有人可以告诉我这个foreach循环可以修改的最佳方式，以便它根据子菜单子的数量来运作吗？

Answer 1

如果您想要无限期/无限嵌套，则需要使用循环函数。

E.g。（伪代码）

function recurrentMenuProcess(branch) {
    foreach (item in branch) {
        //processing here
        if (item.hasChildren) {
            recurrentMenuProcess(item);
        }
    }
}
recurrentMenuProcess(json);

更新动态添加的子数组

1 个答案: