Object ES6 中没有变量声明的解构赋值

Question

Is it possible to have a deconstructing assignment expression in ES6 without the need of declaring variables?

In other words, why is the following code syntactically incorrect?

(I know there are a lot of ways this code could be rewritten.)

'use strict';

let var1 = 'defaultVal1', var2 = 'defaultVal2';
const obj = { group: { var1: 'newVal1', var2: 'newVal2' } }
if (obj) {
  { group: { var1, var2 } } = obj; // No 'let'/'const' keyword, i.e. no redeclaration, but invalid expression
}
console.log(var1);
console.log(var2);

Answer 1

The brackets are interpreted as block statement , but you need an expression. This can be archived by wrapping it with parentheses as an assignment without declaration .

 'use strict'; let var1 = 'defaultVal1', var2 = 'defaultVal2'; const obj = { group: { var1: 'newVal1', var2: 'newVal2' } } if (obj) { ({ group: { var1, var2 } } = obj); // No 'let'/'const' keyword, ie no redeclaration, but invalid expression } console.log(var1); console.log(var2);

Answer 2

Instead of declaring the variables, and the conditionally destructure, you can combine them, and use short-circuit evaluation to use a fallback ( || {} ), and then use defaults as the values:

 'use strict'; const obj = { group: { var1: 'newVal1', var2: 'newVal2' } }; const { group: { var1 = 'defaultVal1', var2 = 'defaultVal2' } = {}} = obj || {}; console.log(var1); console.log(var2);