尝试将多个变量分配给同一值时，ESLint引发错误 [英] ESLint throws an error when trying to assign multiple variables to the same value

问题描述

为什么ESLint在此代码行的 a 变量声明之后抛出解析错误:意外的令牌?

const a，b，c = 1;

我的 .eslintrc.json 看起来像这样；

  {"env":{浏览器":是的，"es6":是的，"jquery":是的，"commonjs":为true}，扩展":[以"airbnb为基础"，漂亮"]，"parserOptions":{"ecmaVersion":6"sourceType":脚本"}，"plugins":["prettier"]，规则":{更漂亮/更漂亮":错误"，半":[错误"，总是"]，"quotes":[错误"，双精度"]}} 

解决方案

您无法在JS中做到这一点.

如果变量声明中列出了多个变量( var ， let 或 const )，则每个值都有其自己的初始化程序，因此在没有重复的情况下，不可能在单个声明语句中为多个变量分配单个值.

如果缺少初始化程序(不存在 = value 部分)，则变量将设置为 undefined .

因此，如果您使用 let ，则 c 将变为 1 ，而 a 和 b是 undefined :

 让a，b，c = 1;console.log(a，b，c)//未定义未定义1  

但是，在 const 声明中省略初始化程序会引发错误(将 undefined 分配给常量没有太大意义，对吧?)

因此，您的代码将失败(不仅在ESLint中，而且在正确实现ECMAScript标准的任何JS运行时中):

  const a，b，c = 1;//SyntaxError  

要将相同的值分配给多个变量，您必须:

• 重复该值(如果该值是对象文字(包括数组文字和函数表达式)或由构造函数，工厂或非纯函数返回，则将无效:

    const a = 1，b = 1，c = 1;console.log(a，b，c)//1 1 1  

• 互相分配变量(提示:将名称最短的变量放在第一位并重复该变量):

    const a = 1，b = a，c = a;console.log(a，b，c)//1 1 1  

• 使用 let :

   让a，b，c;a = b = c = 1console.log(a，b，c)//1 1 1  

• 解构一个无限迭代器(在大量变量的情况下似乎很有用):

    const [a，b，c] =(function *(v){while(true)yield v})(1);//^ ---值console.log(a，b，c)//1 1 1  

Why does ESLint throw a Parsing error: Unexpected token , after a variable declaration on this line of code?

const a, b, c = 1;

My .eslintrc.json looks like this;

{

    "env": {
        "browser": true,
        "es6": true,
        "jquery": true,
        "commonjs": true
    },
    "extends": [
        "airbnb-base",
        "prettier"
    ],
    "parserOptions": {
        "ecmaVersion": 6,
        "sourceType": "script"
    },
    "plugins": ["prettier"],
    "rules": {
        "prettier/prettier": "error",
        "semi": ["error", "always"],
        "quotes": ["error", "double"]
    }

}

解决方案

You cannot do that in JS.

If multiple variables listed in a variable declaration (either var, let or const), each value has its own initializer, so it's impossible to assign a single value to multiple variables in a single declaration statement, without repetition.

If an initializer is missing (no = value part present), the variable will be set to undefined.

So if you used let, then c would become 1, while a and b would be undefined:

let a, b, c = 1;

console.log(a, b, c) //undefined undefined 1

However, omitting the initializer in a const declaration throws an error (it doesn't make too much sense to assign undefined to a constant, right?)

Therefore, your code fails (not just in ESLint, but in any JS runtime that correctly implements the ECMAScript standard):

const a, b, c = 1; //SyntaxError

To assign the same value to multiple variables, you have to either:

• Repeat the value (this won't work if the value is an object literal (including array literals and function expressions) or returned by a constructor, a factory, or non-pure function):

  const a = 1, b = 1, c = 1;
  
  console.log(a, b, c) //1 1 1

• Assign the variables from each other (tip: put the one with the shortest name first and repeat that one):

  const a = 1, b = a, c = a;
  
  console.log(a, b, c) //1 1 1

• Use let:

  let a, b, c;
  a = b = c = 1
  
  console.log(a, b, c) //1 1 1

• Destructure an infinite iterator (that seems useful in case of a very large number of variables):

  const [a, b, c] = (function*(v){while(true) yield v})(1);
                                                     // ^--- Value
  
  console.log(a, b, c) //1 1 1

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