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:

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如何跟踪浏览器支持

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。

或

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)

ECMAScript 6 添加了另一个关键字来宣布变量,而不是“放下”。

“Let”和“Let”在“var”上引入的主要目标是,而不是传统的词汇,区块点点点点点点点点点点点点点点点点点点点点点点点点。

相反, var 可以像下面那样粘贴。 { console.log(cc); // undefined. 由于粘贴 var cc = 23; } { console.log(bb); // ReferenceError: bb 没有定义 let bb = 23; } 事实上, Per @Bergi, 两者都是粘贴。

ES6 引入了两个新的关键字(let 和 const) 替代到 var。

当你需要一个区块水平下降时,你可以用Let和Const而不是VAR去。

下面的表总结了 var, let 和 const 之间的差异

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