Question

我有一个对象数组，每个对象都有一个父对象，而那个父对象有孩子。

我想有一个仅包含父母的名字和ID以及他的孩子的名字和ID的数组。以下示例显示了一个对象当前的外观。

@id: "/course_categories/1"
@type: "CourseCategory"
children: Array(5)
   0: {@id: "/course_categories/2", @type: "CourseCategory", id: 2, name: "PHP voor beginners", left: 2, …}
   1: {@id: "/course_categories/3", @type: "CourseCategory", id: 3, name: "PHP advanced", left: 40, …}
   2: {@id: "/course_categories/4", @type: "CourseCategory", id: 4, name: "PHP voor dummies", left: 42, …}
   3: {@id: "/course_categories/5", @type: "CourseCategory", id: 5, name: "PHP voor experts", left: 44, …}
   4: {@id: "/course_categories/6", @type: "CourseCategory", id: 6, name: "PHP design patterns", left: 46, …}
id: 1
key: "0-1"
left: 1
level: 0
name: "PHP"
parent: null
right: 48

我想要的样子：

data: [{
  title: 'PHP', id: 1, children: [
    {
      title: 'PHP voor beginners', id: 2, children: []
    },
    {
      title: 'PHP advanced', id: 3, children: []
    },
    {
      // PHP voor dummies, PHP voor experts, etc...
    }
  ]
},
{
  title: 'REACT', id: 4, children: [
    {
      title: 'React voor beginners', id: 5, children: [
        {
          title: 'React voor beginners 2', id: 6, children: []
        }
      ]
    },
    {
      // ....
    }
  ]
}]

孩子也有可能拥有孩子。因此，孩子的孩子应该具有相同的格式。（上面的示例）我知道可以通过使用递归函数来实现，但是我想知道过滤当前对象的最佳方法是什么。

Answer 1

您可以使用函数通过过滤键来删除对象的属性，然后使用reduce从剩下的键中获取对象。您可以递归调用此函数，以对子项执行相同的操作。

然后您可以编写一个单独的函数来将name属性转换为title：

const data = [{
  "@id": "/course_categories/1",
  "@type": "CourseCategory",
  children: [{
      "@id": "/course_categories/2",
      "@type": "CourseCategory",
      id: 2,
      name: "PHP voor beginners",
      left: 2,
    },
    {
      "@id": "/course_categories/3",
      "@type": "CourseCategory",
      id: 3,
      name: "PHP advanced",
      left: 40,
    },
    {
      "@id": "/course_categories/4",
      "@type": "CourseCategory",
      id: 4,
      name: "PHP voor dummies",
      left: 42,
    },
    {
      "@id": "/course_categories/5",
      "@type": "CourseCategory",
      id: 5,
      name: "PHP voor experts",
      left: 44,
    },
    {
      "@id": "/course_categories/6",
      "@type": "CourseCategory",
      id: 6,
      name: "PHP design patterns",
      left: 46,
    }
  ],
  id: 1,
  key: "0-1",
  left: 1,
  level: 0,
  name: "PHP",
  parent: null,
  right: 48,
}]

const keepProperties = (object, properties) => {
  const objectWithOnlyRequiredProperties = Object.keys(object)
    .filter(key => properties.indexOf(key) > -1)
    .reduce((prev, curr) => ({
      ...prev,
      [curr]: object[curr],
    }), {});

  const children = object.hasOwnProperty('children') ?
    object.children.map(child => keepProperties(child, properties)) : [];

  return {
    ...objectWithOnlyRequiredProperties,
    children,
  };
}

const transformNameToTitle = object => {
  const children = object.hasOwnProperty('children') ? {
    children: object.children.map(transformNameToTitle)
  } : false;

  const objectKeysWithoutName = Object
    .keys(object)
    .filter(key => key !== 'name')

  const objectWithoutName = keepProperties(
    object,
    objectKeysWithoutName,
  );

  return {
    ...objectWithoutName,
    title: object.name,
    ...children,
  }
}

const test = transformNameToTitle(
  keepProperties(data[0], ['name', 'id'])
);

console.dir(test)

如何保留嵌套对象数组中的特定对象属性

1 个答案: