Im trying to figure out what is the best way to achieve a reconstructed "destructured" shallow copy of an object. I mean, how can I get a referenced object with only a subset of keys, but who are connected to the original object? From let original={ a=1, b=2, c=3} to a reference connected let reconstructed= { a=1, c=3}... values of keys are strings or numbers. I tried workarounds but cannot get a comfortable solution for this. Help would be appreciated.
You could use an array containing the props you want, and then hacky nonsense with getters and setters to build an object that actually reads from/writes to a different object.
Note: I've never needed to do this. I think you might have better luck with a different general approach, like "Use the same object, but only display the properties that are appropriate instead of all of them".
let original={ a:1, b:2, c:3};
let propsInNewObject = ["a", "c"];
const reconstructed = propsInNewObject.reduce((obj, el) => {
Object.defineProperty(obj, el, {
get: () => original[el],
set: (value) => { original[el] = value },
enumerable: true
});
return obj;
}, {});
console.log(reconstructed, original);
reconstructed.a = 100;
console.log(reconstructed, original);
This might be the wrong answer because your question is unclear, especially regards which bit is the 'shallow' copy.
Your reconstructed object is a new, independent object, and not connected to the source object in any way.
let reconstructed = { a, b }
// ^ ^
// denotes a new object
The question is, what types might a and b, be? If the properties deconstructed from the source object are themselves references (other objects), then modifying the deconstructed value, either in its independent state, or when assigned to the reconstructed object, will change the source value too.
let foo = { a: 1, b: 2 }
let { a, b } = foo
a = 999
console.log(foo, a, b)
let bar = { baz: { msg: 'Hello World' } }
let { baz } = bar
let bar1 = { baz }
baz.msg = 'Hello Sailor'
console.log(bar, baz, bar1)
Related
I am copying objA to objB
const objA = { prop: 1 },
const objB = objA;
objB.prop = 2;
console.log(objA.prop); // logs 2 instead of 1
same problem for Arrays
const arrA = [1, 2, 3],
const arrB = arrA;
arrB.push(4);
console.log(arrA.length); // `arrA` has 4 elements instead of 3.
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
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
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.
Try using the create() method like as mentioned below.
var tempMyObj = Object.create(myObj);
This will solve the issue.
Try using $.extend():
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);
use three dots to spread object in the new variable
const a = {b: 1, c: 0};
let d = {...a};
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 👋
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!
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]);
}
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));
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 )
You can now use structuredClone() for deep object clones :
https://developer.mozilla.org/en-US/docs/Web/API/structuredClone
const newItem = structuredClone(oldItem);
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);
I am copying objA to objB
const objA = { prop: 1 },
const objB = objA;
objB.prop = 2;
console.log(objA.prop); // logs 2 instead of 1
same problem for Arrays
const arrA = [1, 2, 3],
const arrB = arrA;
arrB.push(4);
console.log(arrA.length); // `arrA` has 4 elements instead of 3.
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
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
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.
Try using the create() method like as mentioned below.
var tempMyObj = Object.create(myObj);
This will solve the issue.
Try using $.extend():
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);
use three dots to spread object in the new variable
const a = {b: 1, c: 0};
let d = {...a};
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 👋
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!
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]);
}
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));
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 )
You can now use structuredClone() for deep object clones :
https://developer.mozilla.org/en-US/docs/Web/API/structuredClone
const newItem = structuredClone(oldItem);
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);
This is in continuation to the answer posted for the question "Convert JavaScript dot notation object to nested object".
The code works like a charm but I'm unable to wrap my head around how!! So a few days later + a situation where my console.logs actually exceed my lines of code :P.. Here's my question:
Below code for JavaScript function:
function deepen(o) {
var oo = {}, t, parts, part;
for (var k in o) {
t = oo;
parts = k.split('.');
var key = parts.pop();
while (parts.length) {
part = parts.shift();
t = t[part] = t[part] || {};
}
t[key] = o[k]
}
return oo;
}
console.log(
deepen({ 'ab.cd.e' : 'foo', 'ab.cd.f' : 'bar', 'ab.g' : 'foo2' })
);
It deepens a JSON object from :
{
'ab.cd.e' : 'foo',
'ab.cd.f' : 'bar',
'ab.g' : 'foo2' }
Into a nested object :
{ab: {cd: {e:'foo', f:'bar'}, g:'foo2'}}
I get the part where for each key value pair, the logic pops the last element post splitting into an array by ".".
That becomes the key.
What I'm not understanding is the below.
1) The function is returning 'oo' but the operations are all on 't'. The only relationship is that t is being assigned the'empty object' "oo" at the beginning of every iteration on the flat JSON.
2) after the "while (parts.length)" loop, oo miraculously has the nested structure whereas t has one level below it. if oo is assigned to t, how is that possible?
3) I don't see the function being called recursively. How is 00 getting nested beyond the first element of the flat JSON?
I'll first redefine the function with some better names, this way explanation is a lot easier to do.
function deepen(object) {
var nestedObject = {}, cursor, nestingPath, nodeKey;
for (var dotKey in object) {
cursor = nestedObject;
nestingPath = dotKey.split('.');
var leafKey = nestingPath.pop();
while (nestingPath.length) {
nodeKey = nestingPath.shift();
cursor = cursor[nodeKey] = cursor[nodeKey] || {};
}
cursor[leafKey] = object[dotKey];
}
return nestedObject;
}
My guess is that don't entirely know how the while loop functions. Important to know is that when two variables refer to the same object both change when you change one. They are the same object, but you've chosen to have two handles.
Let me provide an example:
object = {};
cursor = object;
cursor.foo = "bar";
object; //=> {foo: "bar"}
cursor; //=> {foo: "bar"}
cursor.a = {};
object; //=> {foo: "bar", a: {}}
cursor; //=> {foo: "bar", a: {}}
cursor = cursor.a;
object; //=> {foo: "bar", a: {}}
cursor; //=> {} <- this is ^
cursor.b = "c";
object; //=> {foo: "bar", a: {b: "c"}}
cursor; //=> {b: "c"}
The while loop is mostly based upon this principal. It's not easy to explain, but I hope the above clarifies things.
Another thing that might be confusing is the line:
cursor = cursor[nodeKey] = cursor[nodeKey] || {};
// read as
cursor = (cursor[nodeKey] = (cursor[nodeKey] || {}));
This could also be written as:
if (!cursor[nodeKey]) cursor[nodeKey] = {};
cursor = cursor[nodeKey];
This assigns a new object to the dynamic nodeKey property if the property isn't there (falsy). Then cursor is assigned to the nested object within, similar to my example above cursor = cursor.a.
First, you're not working with JSON, but a JS object. Most of the time, you should see object as HashMap<String, HashMap<String, ...>> ad infinitum, if you need a Java analogy.
Your questions:
t = oo means they both refer to the same instance created at the start of the function. Why do you use a second variable?
t = t[part] You literally assign entry of t to t
I didn't test the code, but I'm pretty sure it's buggy. Test what happens with object that have multiple names in top level, eg. {'a.b':1, 'b.a':1}. You don't need recursion though, you could use stack instead.
Regarding your code:
Use descriptive names and comments, especially when asking question where other people need to understand your code
Do not define all variables at the beginning of the function. That old habit comes from the dawn of C language and needs to die
for (var k in o) is not a recommended approach to iterate over object entries. Use Object.entries
There is no need to pop from parts array, it is reset every iteration. for(const part of parts) would work just as well
so say I have this:
const mainObject = {
prop1: false,
prop2: {} or []
}
const minorObject = { name: "name", address: "address" }
const minorObject2 = {name: "name2", address: "address2" }
how do add them one at time to mainObject.prop2 without mutating mainObject with Object.assign, if it is not possible then how can it be done with ES6 spread?
When you want to keep the main reference immutable you need to work with shallow copies, similar to what happen when you work with primitives.
An example with Strings (when literally declared they are treated as primitives):
var a = 'Hello';
var b = a;
b += ' World';
console.log({a, b});
As you can see, mutating b the variable a remains immutable... this because primitives are passed by value.
Because Object is not a primitive it is passed by reference and the above example becomes:
var a = { prop: 'Hello' };
var b = a;
b.prop += ' World';
console.log({a, b});
how do add them one at time to mainObject.prop2 without mutating mainObject ...?
var a = { prop: 'Hello' }
var minorA = { prop1: ' World' };
var minorB = { prop2: ' Moon' };
var b = Object.assign({}, a); //create a shallow copy
Object.assign(b, minorA, minorB)
// because you can pass as many params you want, you can simplify the previous code in this way:
var b = Object.assign({}, a, minorA, minorB);
how can it be done with ES6 spread?
The Object spread operator, that is in esnext, you can use it with babel and Object rest spread transform
var b = {...a};
Object.assign will mutate the object passed as the first argument, so if you want to combine the properties of two objects without mutating either, simply wrap one Object.assign inside another.
Object.assign(Object.assign({},mainObject),minorObject)
How to push elements of an object into another object?
Add Javascript Object into another Javascript Object
google keyword search =
add an object to another object
plenty of more examples out there and ways of going about it.
I am copying objA to objB
const objA = { prop: 1 },
const objB = objA;
objB.prop = 2;
console.log(objA.prop); // logs 2 instead of 1
same problem for Arrays
const arrA = [1, 2, 3],
const arrB = arrA;
arrB.push(4);
console.log(arrA.length); // `arrA` has 4 elements instead of 3.
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
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
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.
Try using the create() method like as mentioned below.
var tempMyObj = Object.create(myObj);
This will solve the issue.
Try using $.extend():
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);
use three dots to spread object in the new variable
const a = {b: 1, c: 0};
let d = {...a};
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 👋
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!
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]);
}
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));
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 )
You can now use structuredClone() for deep object clones :
https://developer.mozilla.org/en-US/docs/Web/API/structuredClone
const newItem = structuredClone(oldItem);
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);