实现 lodash 的_.get

上期回顾

一堆整数，分三份，尽量和相等

function a1(arr, count) {
  arr.sort((a, b) => b - a);
  const arg = arr.reduce((a, total) => a + total) / count;
  const resArr = [];
  let current = 0;
  for (let i = 0; i < count - 1; i++) {
    if (current + arr[arr.length - 1] / 2 < arg && i) {
      arr.pop();
      resArr[i - 1].push(arr[arr.length - 1]);
    }
    current = 0;
    resArr[i] = [];
    arr.forEach((item, index) => {
      current += item;
      arr.splice(index, 1);
      resArr[i].push(item);
      if (current > arg) {
        arr.splice(index, 0, item);
        resArr[i].pop();
        current -= item;
      }
    });
  }
  resArr[count - 1] = arr;
  return resArr;
}

今日解题

_.get函数是为了解决像a.b.c.d.e这样嵌套复杂的情况，提高容错性

function get(object, path, defaulltValue = undefined) {
  /// a[3].b -> a.3.b -> [a,3,b]
  // path中也可能是数组的路径，全部转化成 . 运算符并组成数组
  const paths = path.replace(/\[(\d+)\]/g, ".$1").split(".");
  let result = object;
  for (const p of paths) {
    // null 与 undefined 取属性会报错，所以使用Object包装一下
    result = Object(result)[p];
    if (result == undefined) {
      return defaultValue;
    }
  }
  return result;
}

console.log(get({ a: null }, "a.b.c", 3)); /// 3
console.log(get({ a: undefined }, "a", 3)); /// 3
console.log(get({ a: null }, "a", 3)); /// 3
console.log(get({ a: [{ b: 1 }] }, "a[0].b", 3)); /// 1

如果不考虑数组的情况

function get(object, paths, defaultValue = undefined) {
  return (
    paths.split(".").reduce((total, now) => (total || {})[now], object) ||
    defaultValue
  );
}