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?
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
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。
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
在 MDN 中查看此链接
let x = 1;
if (x === 1) {
let x = 2;
console.log(x);
// expected output: 2
}
console.log(x);
// expected output: 1
答案缺少一点:
{
let a = 123;
};
console.log(a); // ReferenceError: a is not defined
此分類上一篇: <script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script> <p> 點擊每個數字將登錄到主機:</p> <div id="div1">1</div> <div id="div2">2</div> <div id="div3">3</div> <div id="div4">4</div> <div id="div5">5</div>
每个单点处理器都会提到相同的对象,因为只有一个对象对象持有6个,所以你每次点击获得6个。
一个通用工作是将此插入一个匿名功能,并将它作为一个论点。 这种问题现在也可以通过使用,而不是在下面的代码中显示的变化来避免。
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。