5-ES6 destructuring assignment

Destructuring assignment is an extension to the assignment operator
It operates on arrays and objects
Advantages: code writing is simple and easy to read
Overview
 Destructuring assignment is an extension of the assignment operator.

It is a kind of pattern matching against arrays or objects, and then assigning values ​​to the variables.

The code is concise and easy to read, and the semantics are clearer; it also facilitates the acquisition of data fields in complex objects.

Deconstructing the model
 In deconstruction, the following two parts are involved:

The source of destructuring, the right part of the destructuring assignment expression.
The target of destructuring, the left part of the destructuring assignment expression.
Destructuring of an array model ( Array)
basic

let [a, b, c] = [1, 2, 3];
// a = 1
// b = 2
// c = 3
 Nestable

let [a, [[b], c]] = [1, [[2], 3]];
// a = 1
// b = 2
// c = 3
 Ignorable

let [a, , b] = [1, 2, 3];
// a = 1
// b = 3
 incomplete deconstruction

let [a = 1, b] = []; // a = 1, b = undefined
 remainder operator

let [a, ...b] = [1, 2, 3];
//a = 1
//b = [2, 3]
string etc.

In the destructuring of the array, if the target of the destructuring is a traversable object, the destructuring assignment can be performed. Traversable objects are implemented Iterator interface data.

let [a, b, c, d, e] = 'hello';
// a = 'h'
// b = 'e'
// c = 'l'
// d = 'l'
// e = 'o'
Destructuring default values

let [a = 2] = [undefined]; // a = 2
 When the destructuring pattern has a matching result, and the matching result is undefined When the default value is triggered as the return result.

let [a = 3, b = a] = [];     // a = 3, b = 3
let [a = 3, b = a] = [1];    // a = 1, b = 1
let [a = 3, b = a] = [1, 2]; // a = 1, b = 2
a and b The matching result is undefined ,Trigger defaults: a = 3; b = a =3
a Normal destructuring assignment, matching results: a = 1,b match result undefined ,Trigger defaults: b = a =1
a and b Normal destructuring assignment, matching results: a = 1,b = 2
 Deconstruction of the object model ( Object)
basic

let { foo, bar } = { foo: 'aaa', bar: 'bbb' };
// foo = 'aaa'
// bar = 'bbb'
 
let { baz : foo } = { baz : 'ddd' };
// foo = 'ddd'
Can be nested or ignored

let obj = {p: ['hello', {y: 'world'}] };
let {p: [x, { y }] } = obj;
// x = 'hello'
// y = 'world'
let obj = {p: ['hello', {y: 'world'}] };
let {p: [x, {  }] } = obj;
// x = 'hello'
incomplete deconstruction

let obj = {p: [{y: 'world'}] };
let {p: [{ y }, x ] } = obj;
// x = undefined
// y = 'world'
remainder operator

let {a, b, ...rest} = {a: 10, b: 20, c: 30, d: 40};
// a = 10
// b = 20
// rest = {c: 30, d: 40}
Destructuring default values

let {a = 10, b = 5} = {a: 3};
// a = 3; b = 5;
let {a: aa = 10, b: bb = 5} = {a: 3};
// aa = 3; bb = 5;

  

  

Overview

Destructuring assignment is an extension of the assignment operator.

It is a kind of pattern matching against arrays or objects, and then assigning values ​​to the variables.

The code is concise and easy to read, and the semantics are clearer; it also facilitates the acquisition of data fields in complex objects.

Deconstructing the model

In deconstruction, the following two parts are involved:

    The source of the destructuring, the right part of the destructuring assignment expression.
  • The target of destructuring, the left part of the destructuring assignment expression.

Destructuring the Array Model (Array)

basic

let[a, b, c] = [1, 2, 3]; // a = 1// b = 2// c = 3

Nestable

let[a, [[b], c]] = [1, [[2], 3]]; // a = 1// b = 2// c = 3

Ignorable

let[a, , b] = [1, 2, 3]; // a = 1// b = 3

incomplete deconstruction

let[a = 1, b] = []; // a = 1, b = undefined

remainder operator

let[a, ...b] = [1, 2, 3]; //a = 1//b = [2, 3]

string etc.

In the destructuring of the array, if the target of the destructuring is a traversable object, the destructuring assignment can be performed. Traversable objects are data that implement the Iterator interface.

let[a, b, c, d, e] = 'hello'; // a = 'h'// b = 'e'// c = 'l'// d = 'l'// e = 'o'

Destructuring default values

let[a = 2] = [undefined]; // a = 2

When the destructuring pattern has a matching result, and the matching result is undefined, the default value will be triggered as the return result.

let[a = 3, b = a] = []; // a = 3, b = 3let[a = 3, b = a] = [1]; // a = 1, b = 1let[a = 3, b = a] = [1, 2]; // a = 1, b = 2
  • a matches b with undefined , triggering the default value: a = 3; b = a =3
  • a normal destructuring assignment, match result: a = 1, b match result undefined , trigger default value: b = a =1
  • Normal destructuring assignment of a and b, matching result: a = 1, b = 2

Destructuring the Object Model (Object)

basic

let{foo, bar} = {foo: 'aaa', bar: 'bbb'}; // foo = 'aaa'// bar = 'bbb'let{baz : foo} = {baz : 'ddd'}; // foo = 'ddd'

Can be nested or ignored

letobj = {p: ['hello', {y: 'world'}]}; let{p: [x, {y}]} = obj; // x = 'hello'// y = 'world'letobj = {p: ['hello', {y: 'world'}]}; let{p: [x, {}]} = obj; // x = 'hello'

incomplete deconstruction

letobj = {p: [{y: 'world'}]}; let{p: [{y}, x]} = obj; // x = undefined// y = 'world'

remainder operator

let{a, b, ...rest} = {a: 10, b: 20, c: 30, d: 40}; // a = 10// b = 20// rest = {c: 30, d: 40}

Destructuring default values

let{a = 10, b = 5} = {a: 3}; // a = 3; b = 5;let{a: aa = 10, b: bb = 5} = {a: 3}; // aa = 3; bb = 5;

Tags: ECMAScript

Posted by Jeremias on Fri, 20 May 2022 04:38:53 +0300