JavaScript中的简单异步功能

Question

我正在使用async / await和以下代码，在5秒钟后响应const中的setTimeout完成之前，响应不应该为空或为null吗？并且响应不应该返回xyz而不是1吗？

async function test() {
  try {
    const response = await setTimeout(
      function() { 
        const obj = {};
        obj.text = "xyz";
        console.log('should console log after 5 seconds')
        return obj;
      }, 
    5000)

    if (response) {
      console.log(`response is ${response}`)
      console.log(`response.text is ${response.text}`)
    }
  } 
  catch(err) {
    console.log(err)
  }
}

test();

Answer 1

您必须做出一个承诺来等待setTimeout（）。

async function test() {
try {
    const response = await new Promise((resolve) => {
        setTimeout(
            function() {
                const obj = {};
                obj.text = "xyz";
                console.log('should console log after 5 seconds')
                return resolve(obj);
            },
            5000)
    });

    if (response) {
        console.log(`response is ${response}`)
        console.log(`response.text is ${response.text}`)
    }
}
catch(err) {
    console.log(err)
}
}

test();

Answer 2

为使代码按预期运行，您需要将设置的超时包装在Promise中。检查代码段。无需将其包装在promise中，setTimeout立即返回值，它是计时器的ID值。

async function test() {
  try {
    const response = await new Promise(function(resolve, reject) {
      setTimeout(
        function() {
          const obj = {};
          obj.text = "xyz";
          console.log('should console log after 5 seconds')
          resolve(obj);
        },
        5000)
    })

    if (response) {
      console.log(`response is ${response}`)
      console.log(`response.text is ${response.text}`)
    }
  } catch (err) {
    console.log(err)
  }
}

test();