修改 JavaScript 对象的副本会导致原始对象发生更改

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If, however, you want to preserve both of the original objects, you can do so by passing an empty object as the target:

var object = $.extend({}, object1, object2);

var tempMyObj = $.extend({}, myObj);

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最佳答案

It is clear that you have some misconceptions of what the statement var tempMyObj = myObj; does.

In JavaScript objects are passed and assigned by reference (more accurately the value of a reference), so tempMyObj and myObj are both references to the same object.

Here is a simplified illustration that may help you visualize what is happening

// [Object1]<--------- myObj


var tempMyObj = myObj;


// [Object1]<--------- myObj
//         ^
//         |
//         ----------- tempMyObj

As you can see after the assignment, both references are pointing to the same object.

You need to create a copy if you need to modify one and not the other.

// [Object1]<--------- myObj


const tempMyObj = Object.assign({}, myObj);


// [Object1]<--------- myObj
// [Object2]<--------- tempMyObj

Old Answer:

Here are a couple of other ways of creating a copy of an object

Since you are already using jQuery:

var newObject = jQuery.extend(true, {}, myObj);

With vanilla JavaScript

function clone(obj) {
if (null == obj || "object" != typeof obj) return obj;
var copy = obj.constructor();
for (var attr in obj) {
if (obj.hasOwnProperty(attr)) copy[attr] = obj[attr];
}
return copy;
}


var newObject = clone(myObj);

See here and here

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Try using the create() method like as mentioned below.

var tempMyObj = Object.create(myObj);

This will solve the issue.

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This might be very tricky, let me try to put this in a simple way. When you "copy" one variable to another variable in javascript, you are not actually copying its value from one to another, you are assigning to the copied variable, a reference to the original object. To actually make a copy, you need to create a new object use

The tricky part is because there's a difference between assigning a new value to the copied variable and modify its value. When you assign a new value to the copy variable, you are getting rid of the reference and assigning the new value to the copy, however, if you only modify the value of the copy (without assigning a new value), you are modifying the copy and the original.

Hope the example helps!

let original = "Apple";
let copy1 = copy2 = original;
copy1 = "Banana";
copy2 = "John";


console.log("ASSIGNING a new value to a copied variable only changes the copy. The ogirinal variable doesn't change");
console.log(original); // Apple
console.log(copy1); // Banana
console.log(copy2); // John


//----------------------------


original = { "fruit" : "Apple" };
copy1 = copy2 = original;
copy1 = {"animal" : "Dog"};
copy2 = "John";


console.log("\n ASSIGNING a new value to a copied variable only changes the copy. The ogirinal variable doesn't change");
console.log(original); //{ fruit: 'Apple' }
console.log(copy1); // { animal: 'Dog' }
console.log(copy2); // John */


//----------------------------
// HERE'S THE TRICK!!!!!!!


original = { "fruit" : "Apple" };
let real_copy = {};
Object.assign(real_copy, original);
copy1 = copy2 = original;
copy1["fruit"] = "Banana"; // we're not assiging a new value to the variable, we're only MODIFYING it, so it changes the copy and the original!!!!
copy2 = "John";




console.log("\n MODIFY the variable without assigning a new value to it, also changes the original variable")
console.log(original); //{ fruit: 'Banana' } <====== Ops!!!!!!
console.log(copy1); // { fruit: 'Banana' }
console.log(copy2); // John
console.log(real_copy); // { fruit: 'Apple' } <======== real copy!

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deep clone object with JSON.parse() and JSON.stringify

// Deep Clone
obj = { a: 0 , b: { c: 0}};
let deepClone = JSON.parse(JSON.stringify(obj));

refrence: this article

Better reference: this article

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If you have the same problem with arrays then here is the solution

let sectionlist = [{"name":"xyz"},{"name":"abc"}];
let mainsectionlist = [];
for (let i = 0; i < sectionlist.length; i++) {
mainsectionlist[i] = Object.assign({}, sectionlist[i]);
}

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use three dots to spread object in the new variable

const a = {b: 1, c: 0};
let d = {...a};

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Serialize the original object into JSON and Deserialize to another object variable of same type. This will give you copy of object with all property values. And any modification to original object will not impact the copied object.

string s = Serialize(object); //Serialize to JSON
//Deserialize to original object type
tempSearchRequest = JsonConvert.DeserializeObject<OriginalObjectType>(s);

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To sum it all up, and for clarification, there's four ways of copying a JS object.

A normal copy. When you change the original object's properties, the copied object's properties will change too (and vice versa).

const a = { x: 0}
const b = a;
b.x = 1; // also updates a.x

A shallow copy. Top level properties will be unique for the original and the copied object. Nested properties will be shared across both objects though. Use the spread operator ...{} or Object.assign().

const a = { x: 0, y: { z: 0 } };
const b = {...a}; // or const b = Object.assign({}, a);


b.x = 1; // doesn't update a.x
b.y.z = 1; // also updates a.y.z

A deep copy. All properties are unique for the original and the copied object, even nested properties. For a deep copy, serialize the object to JSON and parse it back to a JS object.

const a = { x: 0, y: { z: 0 } };
const b = JSON.parse(JSON.stringify(a));


b.y.z = 1; // doesn't update a.y.z

A full deep copy. With the above technique, property values that are not valid in JSON (like functions) will be discarded. If you need a deep copy and keep nested properties that contain functions, you might want to look into a utility library like lodash.

import { cloneDeep } from "lodash";
const a = { x: 0, y: { z: (a, b) => a + b } };
const b = cloneDeep(a);


console.log(b.y.z(1, 2)); // returns 3

Using Object.create() does create a new object. The properties are shared between objects (changing one also changes the other). The difference with a normal copy, is that properties are added under the new object's prototype __proto__. When you never change the original object, this could also work as a shallow copy, but I would suggest using one of the methods above, unless you specifically need this behaviour.

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As I couldn't find this code anywhere around suggested answers for shallow copy/cloning cases, I'll leave this here:

// shortcuts
const {
create,
getOwnPropertyDescriptors,
getPrototypeOf
} = Object;


// utility
const shallowClone = source => create(
getPrototypeOf(source),
getOwnPropertyDescriptors(source)
);


// ... everyday code ...


const first = {
_counts: 0,
get count() {
return ++this._counts;
}
};


first.count;  // 1


const second = shallowClone(first);


// all accessors are preserved
second.count; // 2
second.count; // 3
second.count; // 4


// but `first` is still where it was
first.count;  // just 2

The main difference compared to Object.assign or {...spread} operations, is that this utility will preserve all accessors, symbols, and so on, in the process, including the inheritance.

Every other solution in this space seems to miss the fact cloning, or even copying, is not just about properties values as retrieved once, but accessors and inheritance might be more than welcome in daily cases.

For everything else, use native structuredClone method or its polyfill 👋

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In Javascript objects are passed as reference and they using shallow comparison so when we change any instance of the object the same changes is also referenced to the main object.

To ignore this replication we can stringify the JSON object.

example :-

let obj = {
key: "value"
}
function convertObj(obj){
let newObj = JSON.parse(obj);
console.log(newObj)
}
convertObj(JSON.stringify(obj));

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The following would copy objA to objB without referencing objA

let objA = { prop: 1 },
let objB = Object.assign( {}, objA )
objB.prop = 2;


console.log( objA , objB )