js获取深层次属性

方式一：通过递归

// 获取js深层次属性兼容有[]的情况
function getJsonValue(obj, node) {
 if (!obj) {
   return null;
 }
 if (!node) {
   return null;
 }
 let nodes = node.split(".");
 let item = nodes[0]
 let newObj = obj[item]
 if (nodes[0].indexOf('[') > -1) {
   let itemArr = item.split("[")
   newObj = obj[itemArr[0]]
   newObj = newObj[itemArr[1].slice(0, -1)]
 }
 if (nodes.length == 1) {
   return newObj;
 }
 return getJsonValue(newObj, node.substring(item.length + 1));
}
var a = {
 aa: {
   aaa: [1, 2, 3]
 }
}

console.log(getJsonValue(a, 'aa.aaa[1]'))

参考下面文章后js获取深层次属性,一道很经典的面试题，在此基础上通过替换[]为.来兼容有[]的情况

ES7可选链式调用 console.log(data.user?.address?.street) //undefined

方式二：reduce

String.prototype.replaceAll = function (search, replacement) {
  var target = this;
  search = escapeRegExp(search)
  return target.replace(new RegExp(search, 'g'), replacement);
};
// 兼容正则中需要转义的
function escapeRegExp(str) {
  return str.replace(/[.*+?^${}()|[\]\\]/g, "\\$&"); // $& means the whole matched string
}
const safeGet = (o, path) => {
  // 只要把[]都替换了就好了嘛
  path = path.replaceAll('[', '.')
  path = path.replaceAll(']', '')
  console.log('path', path)
  try {
    return path.split('.').reduce((o, k) => o[k], o)
  } catch (e) {
    return void 666
  }
}

var a = {
  aa: {
    aaa: [1, 2, 3]
  }
}

console.log(safeGet(a, 'aa.aaa[1]'))