ECMAScript 6 引入了许可声明。

我听说它被描述为一个当地变量,但我仍然不确定它是如何行为不同于 var 关键词。

什么是差异?什么时候应该被允许使用而不是 var?


当前回答

答答答答答答答答答答答答答答答答答

var 变量是全球性的,基本上可以到达到任何地方,而让变量不是全球性的,只有直到一个关闭的偏见杀死它们。

请参见下面的我的例子,并注意狮子(Let)变量如何在两个 console.logs 中以不同的方式行事;它在第二个 console.log 中变得无效。

var cat = "cat";
let dog = "dog";

var animals = () => {
    var giraffe = "giraffe";
    let lion = "lion";

    console.log(cat);  //will print 'cat'.
    console.log(dog);  //will print 'dog', because dog was declared outside this function (like var cat).

    console.log(giraffe); //will print 'giraffe'.
    console.log(lion); //will print 'lion', as lion is within scope.
}

console.log(giraffe); //will print 'giraffe', as giraffe is a global variable (var).
console.log(lion); //will print UNDEFINED, as lion is a 'let' variable and is now out of scope.

其他回答

使用 var 声明定义的变量在其定义的整个函数中已知,从函数的开始。 (*) 使用 let 声明定义的变量仅在其定义的区块中已知,从其定义的时刻开始。

// i IS NOT known here
// j IS NOT known here
// k IS known here, but undefined
// l IS NOT known here

function loop(arr) {
    // i IS known here, but undefined
    // j IS NOT known here
    // k IS known here, but has a value only the second time loop is called
    // l IS NOT known here

    for( var i = 0; i < arr.length; i++ ) {
        // i IS known here, and has a value
        // j IS NOT known here
        // k IS known here, but has a value only the second time loop is called
        // l IS NOT known here
    };

    // i IS known here, and has a value
    // j IS NOT known here
    // k IS known here, but has a value only the second time loop is called
    // l IS NOT known here

    for( let j = 0; j < arr.length; j++ ) {
        // i IS known here, and has a value
        // j IS known here, and has a value
        // k IS known here, but has a value only the second time loop is called
        // l IS NOT known here
    };

    // i IS known here, and has a value
    // j IS NOT known here
    // k IS known here, but has a value only the second time loop is called
    // l IS NOT known here
}

loop([1,2,3,4]);

for( var k = 0; k < arr.length; k++ ) {
    // i IS NOT known here
    // j IS NOT known here
    // k IS known here, and has a value
    // l IS NOT known here
};

for( let l = 0; l < arr.length; l++ ) {
    // i IS NOT known here
    // j IS NOT known here
    // k IS known here, and has a value
    // l IS known here, and has a value
};

loop([1,2,3,4]);

// i IS NOT known here
// j IS NOT known here
// k IS known here, and has a value
// l IS NOT known here


今天使用安全吗?

有些人会说,在未来,我们只会使用让陈述,而这些陈述会变得过时。JavaScript老师Kyle Simpson写了一篇非常复杂的文章,他认为为什么不会这样。

事实上,我们实际上需要问自己是否安全使用放弃声明,这个问题的答案取决于你的环境:

此分類上一篇


如何跟踪浏览器支持


函数 VS 区块范围:

与 var 和 let 的主要区别是,与 var 宣言的变量是函数分解的,而与 let 宣言的函数是区块分解的。

function testVar () {
  if(true) {
    var foo = 'foo';
  }

  console.log(foo);
}

testVar();  
// logs 'foo'


function testLet () {
  if(true) {
    let bar = 'bar';
  }

  console.log(bar);
}

testLet(); 
// reference error
// bar is scoped to the block of the if statement 

当第一个函数测试Var 被称为变量 foo 时,与 var 声明,仍然在 if 声明之外可用。

与Let的变量:

当第二个函数测试Let被称为变量栏,用Let表示,仅可在 if 声明中访问,因为用Let表示的变量是区块分开的(其中一个区块是曲线条之间的代码,例如,如果{},为{},函数{})。

请不要让变量变量:

变量与不要被抓住:

console.log(letVar);

let letVar = 10;
// referenceError, the variable doesn't get hoisted

与 var do 相容的变量:

console.log(varVar);

var varVar = 10;
// logs undefined, the variable gets hoisted

var bar = 5;
let foo  = 10;

console.log(bar); // logs 5
console.log(foo); // logs 10

console.log(window.bar);  
// logs 5, variable added to window object

console.log(window.foo);
// logs undefined, variable not added to window object

const name = 'Max'; let age = 33; var hasHobbies = true; name = 'Maximilian'; age = 34; hasHobbies = false; const summarizeUser = (userName, userAge, userHasHobby) => { return ( 'Name is'+ userName + ', age is'+ userAge +'and the user has hobbies:'+ userHasHobby ); } console.log(summarizeUser(name, age, hasHobbies));

正如你可以从上面的代码运行中看到的那样,当你尝试更改 const 变量时,你会发现一个错误:

试图超越一个“名称”,这是一个恒定的。

TypeError: Invalid assignment to const 'name'.

但是,看看放变量。

首先,我们宣布让年龄=33岁,然后将另一个值年龄=34岁,这是OK;当我们试图改变时,我们没有任何错误。

一些黑客与Let:

1。

    let statistics = [16, 170, 10];
    let [age, height, grade] = statistics;

    console.log(height)

二。

    let x = 120,
    y = 12;
    [x, y] = [y, x];
    console.log(`x: ${x} y: ${y}`);

3、

    let node = {
                   type: "Identifier",
                   name: "foo"
               };

    let { type, name, value } = node;

    console.log(type);      // "Identifier"
    console.log(name);      // "foo"
    console.log(value);     // undefined

    let node = {
        type: "Identifier"
    };

    let { type: localType, name: localName = "bar" } = node;

    console.log(localType);     // "Identifier"
    console.log(localName);     // "bar"

Getter 和 Setter 與 Let:

let jar = {
    numberOfCookies: 10,
    get cookies() {
        return this.numberOfCookies;
    },
    set cookies(value) {
        this.numberOfCookies = value;
    }
};

console.log(jar.cookies)
jar.cookies = 7;

console.log(jar.cookies)

msg = “Hello World” 函数 doWork() { // msg 将是可用的,因为它被定义在这个开幕式上! 让朋友 = 0; console.log(msg); // 与 VAR 虽然: for (var iCount2 = 0; iCount2 < 5; iCount2++) {} // iCount2 将在这个开幕式后可用! console.log(iCount2); for (let iCount1 = 0; iCount1 < 5; iCount1++) {} // iCount1 将没有