链接异步函数无法正常工作

Question

我正在尝试链接两个异步函数，但似乎第二个函数在第一个函数之前执行。 这是我的代码

function performAction(e) {
const ZIP = document.getElementById('zip').value;
const fellings = document.getElementById('feelings').value;
console.log(`${baseURL}${ZIP},us&appid=${key}`);
getWeather(baseURL, ZIP, key,).then((data) => {
    postData('/addweather', {temperature: data.main.temp ,date:newDate, userResponse: fellings })
}).then(
    updateUI()
)}

这是 getWeather()

const getWeather = async(baseURL, ZIP, key) => {
let URL = `${baseURL}${ZIP}&appid=${key}`;
const res = await fetch(URL)
try{
    const data = await res.json();
    return data;
}catch(error){
    console.log("error", error);
}}

这是 postData() ，它应该在 getWeather function 被执行后执行，但事实并非如此。

const postData = async ( url = '', data = {}) => {
console.log(`This is what we fetch ${data.temperature}`);
console.log(`This is what we fetch ${data.date}`);
console.log(`This is what we fetch ${data.userResponse}`);
  const response = await fetch(url, {
  method: 'POST', 
  credentials: 'same-origin',
  headers: {
      'Content-Type': 'application/json',
  },
 // Body data type must match "Content-Type" header        
  body: JSON.stringify(data), 
});
try {
    const newData = await response.json();
    console.log(`This is the new Data ${newData.temperature}`);
    return newData;
}catch(error){
  console.log("error", error);
}}

这是 updateUI()

const updateUI = async () => {
const request = await fetch('/getweather');
try{
  const allData = await request.json();
  console.log('Get request');
        document.getElementById('date').innerHTML = allData.date;
        document.getElementById('temp').innerHTML = allData.temperature;
        document.getElementById('content').innerHTML = allData.userResponse;
}catch(error){
  console.log("error", error);
}}

发生的情况是 UI 首先更新，因此它第一次获得 undefined 的值，当我重新加载页面并输入新数据时，UI 将使用上次的数据进行更新。

Answer 1

您需要返回从 postData 返回的postData ：

getWeather(baseURL, ZIP, key,).then((data) => {
   return postData('/addweather', {temperature: data.main.temp ,date:newDate, userResponse: fellings })
}).then(() => {
   return updateUI()
})

你可以这样写的另一种方式是这样的：

async function run() {
   await getWeather(baseURL, ZIP, key)
   await postData('/addweather', {temperature: data.main.temp ,date:newDate, userResponse: fellings })
   await updateUI()
}

Answer 2

您的postData()也是异步 function。 因此，您也必须等待：

getWeather(baseURL, ZIP, key,).then(async (data) => {
    await postData('/addweather', {temperature: data.main.temp ,date:newDate, userResponse: fellings })
}).then(
    updateUI()
)}

我有一段时间没有做过 javascript，但我想这样更清楚：

const performAction = async (e) => {
const ZIP = document.getElementById('zip').value;
const fellings = document.getElementById('feelings').value;
console.log(`${baseURL}${ZIP},us&appid=${key}`);
try{
const data = await getWeather(baseURL, ZIP, key,);
const postData= await postData('/addweather', {temperature: data.main.temp ,date:newDate, userResponse: fellings });
} catch(e) {
console.log(e)
} finally {
    updateUI();
}

此外，您不必等待 json 的解析，并且 try catch 应该包含您的请求：

const postData = async ( url = '', data = {}) => {
console.log(`This is what we fetch ${data.temperature}`);
console.log(`This is what we fetch ${data.date}`);
console.log(`This is what we fetch ${data.userResponse}`);
try {
  const response = await fetch(url, {
  method: 'POST', 
  credentials: 'same-origin',
  headers: {
      'Content-Type': 'application/json',
  },
 // Body data type must match "Content-Type" header        
  body: JSON.stringify(data), 
});

    const newData = response.json();
    console.log(`This is the new Data ${newData.temperature}`);
    return newData;
}catch(error){
  console.log("error", error);
}}