递归数组菜单 php

Question

如何构建一个将这段代码转换为html菜单的php函数这是我通过 ajax 对函数发布的帖子

list[0][id] 55
list[1][id] 69
list[2][children][0][id]    67
list[2][children][1][id]    68
list[2][id] 57

这是我进入函数的数组

Array
(
    [0] => Array
        (
            [id] => 55
            [children] => Array
                (
                    [0] => Array
                        (
                            [id] => 67
                            [children] => Array
                                (
                                    [0] => Array
                                        (
                                            [id] => 57
                                            [children] => Array
                                                (
                                                    [0] => Array
                                                        (
                                                            [id] => 68
                                                        )
                                                )
                                        )
                                    [1] => Array
                                        (
                                            [id] => 69
                                        )
                                )
                        )
                )
        )
)

我试过这个功能，但它不起作用

function tomenu($arr){
  $html = '<ul>'.PHP_EOL;
  foreach ($arr as $v){
    $html .= '<li>' . $v['id'];
    if (array_key_exists('children', $v)){
      $html .= $this->tomenu($v['children']);
    }
    $html .= '</li>'.PHP_EOL;
  }
  $html .= '</ul>'.PHP_EOL;
  return $html;
}

请帮助我。

Answer 1

function tomenu($arr) {
    $this->html .= '<ul>' . PHP_EOL;
    foreach ($arr as $v) {
        $this->html .= '<li>' . $v['id'];
        if (is_array($v['children'])) {
            $this->html .= $this->tomenu($v['children']);
        }
        $this->html .= '</li>' . PHP_EOL;
    }
    $this->html .= '</ul>' . PHP_EOL;
    $returnVal = $this->html;
    $this->html = '';
    return $returnVal;
}

不过我还没有测试过这段代码