异步功能 [英] Async Functions

问题描述

我有一个功能

export const setSearchField = text => dispatch => {
  dispatch({ type: REQUEST_GIFS_PENDING });
  fetch(API_URL + (text || DEFAULT_QUERY) + API_KEY + LIMIT)
    .then(response => response.json())
    .then(items =>
      dispatch({ type: REQUEST_GIFS_SUCCESS, payload: items.data })
    )
    .catch(error => dispatch({ type: REQUEST_GIFS_FAILED, payload: error }));
};

现在，我想将其转换为异步函数，我这样做时无需尝试&捕获，但不起作用

Now i want to convert it to an async function, I did like this without try & catch, but its not working

export const setSearchField = text => dispatch => {
  dispatch({ type: REQUEST_GIFS_PENDING });
  async function fetUsers(){
    const response = await fetch(API_URL + (text || DEFAULT_QUERY) + API_KEY + LIMIT);
    const items = await response.json();
    dispatch({ type: REQUEST_GIFS_SUCCESS, payload: items.data })
  }

我在做什么错了?

推荐答案

代码不起作用的原因是，您永远不会调用已创建的异步函数.但是，没有任何理由要创建内部异步函数；只需使您已经异步的一个即可:

The reason your code doesn't work is that you never call the async function you've created. However, there isn't any reason to make an inner async function; just make the one you already have async:

export const setSearchField = text => async (dispatch) => {
  dispatch({ type: REQUEST_GIFS_PENDING });
  try {
    const response = await fetch(API_URL + (text || DEFAULT_QUERY) + API_KEY + LIMIT);
    const items = await response.json();
    dispatch({ type: REQUEST_GIFS_SUCCESS, payload: items.data })
  } catch (error) {
    dispatch({ type: REQUEST_GIFS_FAILED, payload: error })
  }
}

这篇关于异步功能的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！