1

我正在迭代一个数组并为每个项目进行 REST API 调用，但是我遇到了 js 的异步性质的问题。我正在尝试使用异步/等待，但我认为我没有正确设置它，因为它不会等待响应并返回未定义。

onSearchSuccess = async (response) => {
  const persons = response._embedded.persons_search_collection;
  const personsWithClasses = await persons.reduce(
  (acc, person) => {
    const params = {
      person_id: person.person_id,
      date: '2017-01-05',
      enrollment_status: 3,
      class_status: 2,
    };
    return getClasses( //this function does an GET request and returns the response
      params,
      (classesResponse) => {
        const { classes } = classesResponse._embedded;
        console.log(classes); //logs after the console.log below
        return [...acc, { ...person, classes }];
      },
      () => acc,
    );
  }, []);
console.log(personsWithClasses); //return undefined
}


export const getClasses = (params, success, error) => {
  axios.get(`${uri}/classes`, { params })
  .then(({ data }) => {
    success(data);
  })
  .catch(err => error(err));
};
4

1 回答 1

1

正如我在评论中提到的，reduce如果您调用异步函数，它将无法按您的意愿工作。你可以使用Promise.alland .mapas so（我尽量使用async/await）：

onSearchSuccess = async (response) => {
  const persons = response._embedded.persons_search_collection;
  let personsWithClasses = await Promise.all(persons.map(async (person) => {
    try {
      const classes = await getClasses({
        person_id: person.person_id,
        date: '2017-01-05',
        enrollment_status: 3,
        class_status: 2,
      });

      return {...person, classes};
    } catch(error) {
      // ignore errors if a person wasn't found
      return null;
    }
  }));
  personsWithClasses = personsWithClasses.filter(x => x != null);
  console.log(personsWithClasses);
}


export const getClasses = params => {
  return axios.get(`${uri}/classes`, { params });
};

还要注意我对getClasses. axios.get如果无论如何返回一个承诺，就没有理由让它接受回调。

于 2017-03-01T22:19:20.963 回答