javascript 承诺 return 待定

Question

我正在创建 javascript 函数以使用 promise 和 async await 获取某些内容。但我得到了 return 的承诺。谁能帮我？这是我的代码

 class NetworkError extends Error {
  constructor(message) {
    super(message);
    this.name = 'NetworkError';
  }
}

const fetchingUserFromInternet = (isOffline) => {
  return new Promise((resolve, reject) => {
    setTimeout(() => {
      if (isOffline) {
        reject(new NetworkError('Gagal mendapatkan data dari internet'), null);
      }
      resolve({ name: 'John', age: 18 });
    }, 500);
  })
  
};


const gettingUserName = async () => {
  const user = await fetchingUserFromInternet(false)
    .then((user) => {
      return user.name
    })
    .catch((rejectedReason) => {
      return rejectedReason
    })
};

module.exports = { fetchingUserFromInternet, gettingUserName, NetworkError };

Answer 1

您的 gettingUserName 方法从不返回任何内容。你应该这样做：

const gettingUserName = async () => {
  try {
      const user = await fetchingUserFromInternet(false)
      return user.name;  
   } catch (e){
      return e;
   }
};

或者这个：

const gettingUserName = () => {
  return fetchingUserFromInternet(false)
    .then((user) => {
      return user.name
    })
    .catch((rejectedReason) => {
      return rejectedReason
    })
};

javascript 承诺 return 待定

javascript promise return pending

javascript

asynchronous

node.js

promise