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in Technique[技术] by (71.8m points)

为什么then(console.log(myVar))在函数体内的第一个awiat执行后被callback了

let myVar = ""

let asyncAccessURL = async function () {
    myVar = "landmark1"
    const f1 = await fetch('https://www.baidu.com');
    myVar = "landmark2"
    const f2 = await fetch('https://www.baidu.com');
    myVar = "landmark3"
    console.log(f1);
    console.log(f2);
};

asyncAccessURL().then(console.log(myVar))

image.png


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1 Answer

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by (71.8m points)

then接收的是一个回调函数作为参数,
then(res=>{})//是个未执行的函数。
你这么写then(console.log())//相当于把console.log()的返回结果(undefined)传给了then;
而console.log(myVar)在第一时间执行并输出在控制台了。

    asyncAccessURL().then(undefined);

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