javascript - Epic 不在 Redux-Observable 中返回流

Question

我正在测试使用 redux-observable 和一个副项目，我反复遇到这个问题:Uncaught TypeError: combineEpics: one of the provided Epics "handleSearchEpic"不返回流。仔细检查您没有遗漏返回语句!

我已经引用了 redux observable 文档和其他几个在线示例，但我无法确定我可能遗漏了什么。以下是我的行为和相关的史诗。

export const searchContent = query => {
  return {
    type: SEARCH_CONTENT,
    query
  }
}

const returnSearchContent = searchResults => {
  return function(dispatch) {
    dispatch({
      type: RETURN_SEARCH_CONTENT,
      searchResults
    });
  }
}

// Epics
export const handleSearchEpic = action$ => {
  action$.ofType(SEARCH_CONTENT)
    .mergeMap(action => axios.get(`...SOME_API_ENDPOINT`))
    .map(res => returnSearchContent(res))
}

export const rootEpic = combineEpics(
  handleSearchEpic
);

这是应用程序的根目录和商店配置:

const epicMiddleware = createEpicMiddleware(rootEpic);
const store = createStore(Reducer, applyMiddleware(epicMiddleware));

ReactDOM.render(
  <Provider store={store}>
    <App />
  </Provider>,
  document.getElementById('root')
);

Answer 1

您的handleSearchEpic 史诗是一个带有 block 的箭头函数，但实际上并不返回流。

不好

export const handleSearchEpic = action$ => { // <-- start of block
  // vvvvvvv missing return
  action$.ofType(SEARCH_CONTENT)
    .mergeMap(action => axios.get(`...SOME_API_ENDPOINT`))
    .map(res => returnSearchContent(res))
} // <-- end of block

很好

export const handleSearchEpic = action$ => {
  return action$.ofType(SEARCH_CONTENT)
    .mergeMap(action => axios.get(`...SOME_API_ENDPOINT`))
    .map(res => returnSearchContent(res))
}

隐式返回？

或者，您可以删除 block 并使用隐式返回，这可能正是您想要做的？

export const handleSearchEpic = action$ => // <--- no block
  action$.ofType(SEARCH_CONTENT)
    .mergeMap(action => axios.get(`...SOME_API_ENDPOINT`))
    .map(res => returnSearchContent(res));

---

一个非常常见的错误，这就是为什么我添加了您提供的错误消息，但它似乎并没有使解决方案易于理解。有什么建议可以改进错误消息吗？

combineEpics: one of the provided Epics "handleSearchEpic" does not return a stream. Double check you're not missing a return statement!