使用不是有效变量名称的属性名进行对象解构

灰太狼 2023-07-01 08:51 89阅读 0赞

使用不是有效变量名称的属性名进行对象解构(Object destructuring with property names that are not valid variable names)  
JavaScript IT屋 2017/8/3 19:48:36

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问 题

Does anyone know if you can use object destructuring with spaces in the property name? Maybe this cannot be done and I realize the JavaScript notation is incorrect but I cannot change the server json response.

var obj1 = \{name: ‘Mr Smith’, age: 21\};  
//destructure  
var \{name, age\} = obj1;  
//name=‘Mr Smith’ and age=21  
This works as expected.

But when I have the following object structure can I use object destructuring or not?

var obj2 = \{“my name”: “Mr Jones”, age: 22\};  
var \{‘my name’, age\} = obj2;  
If this is not possible It would be nice if I could assign the variable with some sort of syntax like ‘as’…

var \{‘my name’ as name, age\} = obj2; //name=‘Mr Jones’;  
Thanks

解决方案  
You can assign it a valid variable name using this syntax:

var \{“my name”: myName, age\} = obj2;

// use myName here  
本文地址：IT屋 » Object destructuring with property names that are not valid variable names

问 题  
有没有人知道您是否可以使用属性名称中的空格对象解构？也许这不能完成，我意识到JavaScript符号是不正确的，但我不能更改服务器json响应。

var obj1 = \{名字：史密斯先生，年龄：21岁;  
// destructure  
var \{name，age\} = obj1;  
// name ='史密斯先生和年龄= 21

这样可以预期。

但是当我有以下对象结构可以使用对象解构吗？

code> var obj2 = \{我的名字”：琼斯先生”，年龄：22\};  
var \{‘my name’，age\} = obj2;

如果这是不可能的，如果我可以使用某种类型的语法’as’…

var \{'my name’as name，age\} = obj2; // name =‘琼斯先生’;

谢谢