-- 2019-05-16 01:00：
Deep Clone.png
var _deepCloneRecursion = (obj, cloned)=>{
// deep first approach, loop each prop
// until the prop has no key/ not an obj
var keys = Object.keys(obj);
keys.forEach(key=>{
var value = obj[key];
if(value instanceof Object){
// save parent node key
cloned[key] = {};
// pass the parent node to the next recursion
this._deepCloneRecursion(value, cloned[key]);
}else{
cloned[key] = value;
}
})
return cloned;
}
var objOrigin = {
"a":{
"b":"c",
"d":"e",
"f":{
"g":"h"
}
},
"A":{
"B":"C"
}
};
var clone = (obj)=>{
var cloned = {};
_deepCloneRecursion(obj, cloned);
return res;
}
var objCloned = clone(objOrigin);
objOrigin.a.f.g = "x";
objCloned.a.f.g = "y";
console.log(objOrigin, objCloned);
-- 2018-07-29
Object.assign 可以实现一层的copy，于是我想用这个方法来迭代复制每一层的，实现deep clone。
目前实现了两层的clone，当object层数超过三层后，不会被复制到新的obj中
Idea，用entries来遍历Object，每个entry保存了obj的key 和 value。
· 当value不是object的时候，就可以去用assign去copy了；
· 否则的话，把key和value（is an object）传入到loop函数里面
· 当存在key时，需要指定key作为节点进行assign，目前是两层，所以只有一个key的情况
· 是否可以通过迭代这个写好的twoLayerClone方法来实现deepClone，因为在key的管理上很棘手。
"use strict";
// Deep clone
/* const a = {a:1,b:{c:1,d:{e:1,f:1}}, c:1}; */
const a = { a: 1, b: { c: {e: 1}, d:3}, c: 1 };
// Two layer clone
const twoLayerClone = function(obj) {
const resObj = {};
// [[a,1],[b,{c:1,d:1}]]
//let keyArr = [];
const _loopObj = (obj, key) => {
const entries = Object.entries(obj);
for (let entry of entries) {
const _obj = {};
if (typeof entry[1] !== "object") {
if (!key) {
_obj[entry[0]] = entry[1];
Object.assign(resObj, _obj);
} else {
if (typeof _obj[key] === "undefined") {
_obj[key] = {};
}
_obj[key][entry[0]] = entry[1];
if (typeof resObj[key] === "undefined") {
resObj[key] = {};
}
Object.assign(resObj[key], _obj[key]);
//keyArr.pop();
}
} else {
//keyArr.push(entry[0]);
_loopObj(entry[1], entry[0]);
}
}
};
_loopObj(obj);
return resObj;
};
const deepClone = function(obj) {
};
const aCloned = twoLayerClone(a);
a.a = 2;
a.b.d = 2;
console.log("a", a);
console.log("twoLayerClone", aCloned);
还用Object.keys(), Object.values()尝试得到key之间的关系的规律 ，目前有点mess。
Idea，
· 用allKeys来记录每次迭代时的keys，如果是新的keys，就push进去。当key的value不是object时，需要将allKeys中，这个key删除掉。
· key node表示这个key下面还有object。所有的key node应该在all keys里面出现才对。当key node不在当前allKeys中时，把当前的allKeys pop一层出来，同时pop preKeys，进行迭代查找和更新，找到的时候应该是这个key正确的prekeys了。。
"use strict";
// Objet loop
const aObj = {
a: 1,
b: 2,
c: { d: 1, e: { f: 1, g: { n: 1 } } },
d: { d: 1, e: { f: 1, g: function() {} } }
};
let preKeys = [];
// key has chance to be a key node
let allKeys = [];
const getValues = aObj => {
let values = Object.values(aObj);
let keys = Object.keys(aObj);
if (allKeys.length === 0 || allKeys[allKeys.length - 1] !== keys) {
allKeys.push(keys);
}
console.log("allKeys", allKeys);
console.log("preKeys",preKeys);
for (let key of keys) {
if (typeof aObj[key] === "object") {
console.log("key node", key);
findKey(key);
getValues(aObj[key]);
console.log("---");
} else {
// Delete the latest track
allKeys[allKeys.length-1].splice(0,1);
console.log("keys", keys);
console.log("currKey", key, "preKeys", preKeys);
console.log("---");
/* console.log('value',aObj) */
}
}
};
const findKey = key => {
if (!allKeys[allKeys.length - 1].includes(key)) {
allKeys.pop();
preKeys.pop();
findKey(key);
} else {
preKeys.push(key);
}
};
getValues(aObj);
