javascript - 陷入递归返回值

Question

我写了一个小的递归函数来查找 menuItem 的级别。这是函数:

const getSelectedMenuItemLevel = (menuItems, path, level = 0) => menuItems.forEach((menuItem) => {
  console.log(menuItem.path, path, menuItem.path === path, level);
  if (menuItem.path === path) {
    return level;
  }
  return getSelectedMenuItemLevel(menuItem.children, path, level + 1);
});

我是这样调用它的:

console.log(getSelectedMenuItemLevel(menuItems, 'two/three'));

这是 menuItems 数组:

[
  {
    path: 'one',
    name: 'One',
    children: [
      { path: 'one/one', name: 'One/One', children: [] },
      { path: 'one/two', name: 'One/Two', children: [] },
    ],
  },
  {
    path: 'two',
    name: 'Two',
    children: [
      { path: 'two/one', name: 'Two/One', children: [] },
      { path: 'two/two', name: 'Two/Two', children: [] },
      { path: 'two/three', name: 'Two/Three', children: [] },
    ],
  },
{
    path: 'three',
    name: 'Three',
    children: [],
  }
]

这个递归函数总是返回undefined。我希望它返回 level

Answer 1

您需要函数中的一个变量，并通过使用 Array#some 检查找到的级别的迭代来存储它, 因为如果找到有效级别，此方法会使用短路。

基本上，您需要将所需的返回值扩展为 undefined 或数字值。

const getSelectedMenuItemLevel = (menuItems, path, level = 0) => {
    let value;
    menuItems.some((menuItem) => {
        if (menuItem.path === path) {
            value = level;
            return true;
        }
        const temp = getSelectedMenuItemLevel(menuItem.children, path, level + 1);
        if (temp) {
            value = temp;
            return true;
        }
      return false;
    });
    return value;
};


var menuItems = [
  {
    path: 'one',
    name: 'One',
    children: [
      { path: 'one/one', name: 'One/One', children: [] },
      { path: 'one/two', name: 'One/Two', children: [] },
    ],
  },
  {
    path: 'two',
    name: 'Two',
    children: [
      { path: 'two/one', name: 'Two/One', children: [] },
      { path: 'two/two', name: 'Two/Two', children: [] },
      { path: 'two/three', name: 'Two/Three', children: [] },
    ],
  },
{
    path: 'three',
    name: 'Three',
    children: [],
  }
];

console.log(getSelectedMenuItemLevel(menuItems, 'two/three'));